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SSC Placement Paper : SSC Placement Paper General English 16 August 2012

SSC staff selection commission General englsih question and answer

1. To pin one,s faith-

(a) To be sure of some body,s favour -Ans

(b) To be unsure of favour

(c) To bother for friends

(d) To bother for one,s relatives

2. To play fast and loose

(a) To be undependable-Ans

(b) To cheat people .

(c) To hurt some body,s feelings·

(d) To trust others

3 To play on a fiddle-

(a) To play an important role

(b) To play upon a musical instrument

(c) To be busy over trifles-Ans

(d) To be busy over important matter

4. Alma Mater- ,

(a) Mother,s milk

(b) Mother,s concern for the child

(c) The learning that one receives from mother

(cl) Institution where one receives education-Ans

5. To be on the apex

(a) To scale a peak

(b) To be at the highest point-Ans

(c) To punish somebody

(d) To beat somebody-

6. At one,s beck and call

(a) To climb the back

(b) To call from behind

(c) To be always at one,s service or command

(d) Not to care for anybody

7. At one,s wit,s end-

(a Completely confused

(b) To be very witty

(c) To have no sense of humour

(d) To confuse others

8. An axe to grind-

(a) To put an axe in the enemy,s territory

(b) To put an axe in the ground

(c) Not to have any selfish motive

(d) To have a selfish motive

Direction(9-18): Read the following paragraph and answer: Edmunde Burke called the press the Fourth Estate of the realm. I think he did not use this title for the Press thoughtlessly as social ruling group or class. The three Estates or Realms (in England) Lords Spiritual (i.e., the Bishops in the House of Lords), the ,temporal, (i.e. other Lords) and Commons, i. e., the common people). The Press has been rightly called the Fourth Estate as it also. constitutes a ruling group or class like the Lords and Commons. It cannot be denied in a free country that the Press exercises good deal of influence in shaping public opinion and pointing out the weaknesses or defects of society or of Government, and ,in general bringing to light all those good or bad things in society which would have otherwise remained unnoticed. The power is not limited or put under any check. The Press, instead of, being controlled by anyone controls life and thought of a nation: Hence the Press constitutes an Estate by itself.

Obviously. thus power which the Press in an:), country wields depends upon the number of newspaper readers. The opinions . and comments of newspapers can influence. the life of a nation only when they are read, by People. Reading in turn, requires that the general mass of people should be educated. Thus, the spread of education determines the extent of the newspapers. Where readers are few; newspapers must necessarily be few. Their influence, in that case can extend only to a small minority of population. In a country like India, the percentage of literacy is very low and the standard of journalism is n9t very high. So Press has to play the role of a teacher here.

9. Edmunde Burke called the Press

(a) Instrument of Public Opinion

(b) Distributor of news

(c) The Fourth Estate

(d) Lord Temporal

10. The term Fourth Estate stands for

(a) An area of land

(b) Landed Property

(c) Social ruling group or class

(d) Instrument of Power

11. Out of the following the one which is not included in the Three Estates is-

(a) Lords Spiritual

(b) Justices of Peace

(c) Lord Temporal

(d) Commons

12. The Free press docs not perform the function of-

(a) Shaping public opinion

(b) Supporting at all times the official policy

(c) Criticising Government

(d) Exposing social abuses

13. How much power does a Free Pres! possess?

(a) Only that much power which is allowed by the Government of the the country

(b) Unlimited power without any check

(c) Unlimited power subject to the maintenance of la wand order and public morality

(d) No power at all

14. The secret of the Press is-

(a) the money which the newspaper owners can wield

(b The number of newspaper readers

(c) the extent to which it supports official policy

(d) The patronage enjoyed by it of the Government

15. The number of newspaper readers is determined by -

(a) The low price of newspapers

(b) The patronage extended to it by the moneyed people

(c) Education of the general mass of people

(d) The availability of newsprint.

16. The Press exercises power by

(a) Enlisting the support of the people

(b) Keeping watch over the acts of the Government

(c) Controlling life and thought of a nation

(d) Because it is a great moneyedconcern

17. The state of journalism in India

(a) is upto the mark

(b) is rather low

(c) is in its infancy

(d) is not very high

18. The Press has the greatest chances of flourishing in a--

(a) Monarchy

(b) Aristocracy

(c) Democracy

(d) Limited Dictatorship

19-26: Fill in the blanks

19. She has Dot recovered fully---the shock of his failure.

(a) off (b) of (c) from (d) against

20. The master dispensed---the services of his servant.

(a) of (b) with (c) off (d) for

Answers

11. b, 12. b, 13. c, 14. b, 15. c, 16. c, 17. d, 18. c. 19. c, 20. b

21. I look---him as my close friend.

(a) OD (b),up (c) after (d) to

22. My friend is really very good--cricket.

(a) on (b) at (c) in (d) over

23. He has great affection---me.

(a) with (b) on (c) for (d) in

24. He always connives---with his superiors against his colleagues.

(a) on (b) with (c) about (d) at

25. I have been informed that the two brothers have fallen---.

(a) upon (bl through (c) in (d) out

26. Your friend has been convicted---the charge of murder.

(a) upon (b) for (c) on (d) of

Direction: 27-31 : Each word or phrase is followed by four words or phrases. Choose the word or phase which is most nearly the same

27. Pragmatism-

(a) Appearance (b) Obscurantism (c) Practicality (d) Reversion

28. Expeditiously

(a) Rapidly b. easily (c) Vividly d. none of these

29. Precarious

(a) Huge b. uncertain (c) Dangerous d. valuable

30. Vagrant-

(a) Wandering b. Not clear (c) Futile d. None of these

Answers

21. a, 22. b, 23. c, 24. b, 25. d, 26. c, 27. c, 28. a, 29. c, 30. a

31. Valediction

(a) Valid B. Farewell speech (c) Judgement d. None of these

Directions :- Each question is followed by four alternatives. Pick the one which best describe the statement

32. Capable of being approached-

(a) Accessory (b) Easy (c) Accessible (d) Adaptable

33. One who is liked by people-

(a) Samaritan (b) Popular () Philanthropist (d) Misanthropepist

34. No longer in use-

(a) Impracticable (b) Obsolete (c) Absolute (d) Useless

35. A child born after the death of his father-

(a) Posthumous (b) Bastard (c) Kiddy (d) Stepson.

36. One who is present everywhere-

(a) God (b) Omnipotent (c) Omnipresent (d) Visible

37. An office without salary-

(a) Honorary (b) Slavish (c) Sinecure (d) Voluntary

38. A document written by hand-

(a Script (b) Autobiography (c) Manuscript (d) Autography

39. Government by officials-

(a) Oligarchy (b) Bureaucracy (c) Autocracy (d) Democracy

40. A speech made off hand-

(a) Extempore (b) Maiden (c) Lecture (d) Gibberish

1. To pin one,s faith-

(a) To be sure of some body,s favour -Ans

(b) To be unsure of favour

(c) To bother for friends

(d) To bother for one,s relatives

2. To play fast and loose

(a) To be undependable-Ans

(b) To cheat people .

(c) To hurt some body,s feelings·

(d) To trust others

3 To play on a fiddle-

(a) To play an important role

(b) To play upon a musical instrument

(c) To be busy over trifles-Ans

(d) To be busy over important matter

4. Alma Mater- ,

(a) Mother,s milk

(b) Mother,s concern for the child

(c) The learning that one receives from mother

(cl) Institution where one receives education-Ans

5. To be on the apex

(a) To scale a peak

(b) To be at the highest point-Ans

(c) To punish somebody

(d) To beat somebody-

6. At one,s beck and call

(a) To climb the back

(b) To call from behind

(c) To be always at one,s service or command

(d) Not to care for anybody

7. At one,s wit,s end-

(a Completely confused

(b) To be very witty

(c) To have no sense of humour

(d) To confuse others

8. An axe to grind-

(a) To put an axe in the enemy,s territory

(b) To put an axe in the ground

(c) Not to have any selfish motive

(d) To have a selfish motive

Direction(9-18): Read the following paragraph and answer: Edmunde Burke called the press the Fourth Estate of the realm. I think he did not use this title for the Press thoughtlessly as social ruling group or class. The three Estates or Realms (in England) Lords Spiritual (i.e., the Bishops in the House of Lords), the ,temporal, (i.e. other Lords) and Commons, i. e., the common people). The Press has been rightly called the Fourth Estate as it also. constitutes a ruling group or class like the Lords and Commons. It cannot be denied in a free country that the Press exercises good deal of influence in shaping public opinion and pointing out the weaknesses or defects of society or of Government, and ,in general bringing to light all those good or bad things in society which would have otherwise remained unnoticed. The power is not limited or put under any check. The Press, instead of, being controlled by anyone controls life and thought of a nation: Hence the Press constitutes an Estate by itself.

Obviously. thus power which the Press in an:), country wields depends upon the number of newspaper readers. The opinions . and comments of newspapers can influence. the life of a nation only when they are read, by People. Reading in turn, requires that the general mass of people should be educated. Thus, the spread of education determines the extent of the newspapers. Where readers are few; newspapers must necessarily be few. Their influence, in that case can extend only to a small minority of population. In a country like India, the percentage of literacy is very low and the standard of journalism is n9t very high. So Press has to play the role of a teacher here.

9. Edmunde Burke called the Press

(a) Instrument of Public Opinion

(b) Distributor of news

(c) The Fourth Estate

(d) Lord Temporal

10. The term Fourth Estate stands for

(a) An area of land

(b) Landed Property

(c) Social ruling group or class

(d) Instrument of Power

11. Out of the following the one which is not included in the Three Estates is-

(a) Lords Spiritual

(b) Justices of Peace

(c) Lord Temporal

(d) Commons

12. The Free press docs not perform the function of-

(a) Shaping public opinion

(b) Supporting at all times the official policy

(c) Criticising Government

(d) Exposing social abuses

13. How much power does a Free Pres! possess?

(a) Only that much power which is allowed by the Government of the the country

(b) Unlimited power without any check

(c) Unlimited power subject to the maintenance of la wand order and public morality

(d) No power at all

14. The secret of the Press is-

(a) the money which the newspaper owners can wield

(b The number of newspaper readers

(c) the extent to which it supports official policy

(d) The patronage enjoyed by it of the Government

15. The number of newspaper readers is determined by -

(a) The low price of newspapers

(b) The patronage extended to it by the moneyed people

(c) Education of the general mass of people

(d) The availability of newsprint.

16. The Press exercises power by

(a) Enlisting the support of the people

(b) Keeping watch over the acts of the Government

(c) Controlling life and thought of a nation

(d) Because it is a great moneyedconcern

17. The state of journalism in India

(a) is upto the mark

(b) is rather low

(c) is in its infancy

(d) is not very high

18. The Press has the greatest chances of flourishing in a--

(a) Monarchy

(b) Aristocracy

(c) Democracy

(d) Limited Dictatorship

19-26: Fill in the blanks

19. She has Dot recovered fully---the shock of his failure.

(a) off (b) of (c) from (d) against

20. The master dispensed---the services of his servant.

(a) of (b) with (c) off (d) for

Answers

11. b, 12. b, 13. c, 14. b, 15. c, 16. c, 17. d, 18. c. 19. c, 20. b

21. I look---him as my close friend.

(a) OD (b),up (c) after (d) to

22. My friend is really very good--cricket.

(a) on (b) at (c) in (d) over

23. He has great affection---me.

(a) with (b) on (c) for (d) in

24. He always connives---with his superiors against his colleagues.

(a) on (b) with (c) about (d) at

25. I have been informed that the two brothers have fallen---.

(a) upon (bl through (c) in (d) out

26. Your friend has been convicted---the charge of murder.

(a) upon (b) for (c) on (d) of

Direction: 27-31 : Each word or phrase is followed by four words or phrases. Choose the word or phase which is most nearly the same

27. Pragmatism-

(a) Appearance (b) Obscurantism (c) Practicality (d) Reversion

28. Expeditiously

(a) Rapidly b. easily (c) Vividly d. none of these

29. Precarious

(a) Huge b. uncertain (c) Dangerous d. valuable

30. Vagrant-

(a) Wandering b. Not clear (c) Futile d. None of these

Answers

21. a, 22. b, 23. c, 24. b, 25. d, 26. c, 27. c, 28. a, 29. c, 30. a

31. Valediction

(a) Valid B. Farewell speech (c) Judgement d. None of these

Directions :- Each question is followed by four alternatives. Pick the one which best describe the statement

32. Capable of being approached-

(a) Accessory (b) Easy (c) Accessible (d) Adaptable

33. One who is liked by people-

(a) Samaritan (b) Popular () Philanthropist (d) Misanthropepist

34. No longer in use-

(a) Impracticable (b) Obsolete (c) Absolute (d) Useless

35. A child born after the death of his father-

(a) Posthumous (b) Bastard (c) Kiddy (d) Stepson.

36. One who is present everywhere-

(a) God (b) Omnipotent (c) Omnipresent (d) Visible

37. An office without salary-

(a) Honorary (b) Slavish (c) Sinecure (d) Voluntary

38. A document written by hand-

(a Script (b) Autobiography (c) Manuscript (d) Autography

39. Government by officials-

(a) Oligarchy (b) Bureaucracy (c) Autocracy (d) Democracy

40. A speech made off hand-

(a) Extempore (b) Maiden (c) Lecture (d) Gibberish

SSC Placement Paper : SSC Placement Paper General Awareness 16 August 2012

SSC previous year model examination question answer

1The common tree species in Nilgiri Hills is:

(A)Sal

(B)Pine

(C)Eucalyptus

(D)Teak

Answer: Eucalyptus.

2. Which of the following statements on Railway Budget 2011-12 is correct?

(A)There would be a 10% increase in fares for long distance train travel both by AC and NONAC classes

(B)There would be 15% increase in freight rates on all goods other than food grains

(C)There would be 15% increase in passenger fares for all classes for long distance and freights

(D)There would be no increase in fares for both suburban and long distance travel

Answer: D

3. The nuclear reactors which were damaged heavily due to strong Earthquake-cum-Tsunami that hit Japan on March 11, 2011 causing radiation leakage at:

(A)Fukushima

(B)Tokyo

(C)Kyoto

(D)None of them

Answer: Fukushima

4. First Indian Prime Minister to visit Siachen was ?

(A)Rajiv Gandhi

(B)Inder Kumar Gujaral

(C)Mammohan Singh

(D)None of them

Answer: Manmohan Singh

5. Which of the following books has been written by Kishwar Desai?

(A)The Red Devil

(B)Witness the night

(C)Tonight this Savage Rite

(D)Earth and Ashes

Answer: Witness the Night.

6. Which of the following folk / tribal dances is associated with Karnataka?

(A)Yakshagana

(B)Jatra

(C)Veedhi

(D)Jhora

Answer: Yakshagana

7. Who of the following received the Sangeet Natak Academi’s Ustad Bismillah Khan Puraskar for 2009 in theatre?

(A)Omkar Shrikant Dadarkar

(B)Ragini chander sarkar

(C)Abanti Chakraborty and Sukracharjya Rabha

(D)K Nellai Maniknandan

Answer: Abanti Chakraborty and Sukracharjya Rabha

8. Which of the following country did not win any of the FIFA cup in 2002, 2006 and 2010?

(A)Brazil

(B)Argentina

(C)Spain

(D)South Africa

Answer: South Africa

9. Who invented vaccination for small pox?

(A)Sir Frederick Grant Banting

(B)Sir Alexander Fleming

(C)Edward Jenner

(D)Loius Pasteur

Answer: Edward Jenner

10. Who was the first Indian to become the member of British parliament?

(A)Bankim Chandra Chaterjee

(B)W C Banerjee

(C)Dadabhai Naoroji

(D)None of the above

Answer: Dadabhai Naoroji

11. The purchase of shares and bonds of Indian companies by Foreign Institutional Investors is called?

(A)FDI

(B)Portfolio Investment

(C)NRI Investment

(D)Foreign Indirect Investment

Answer: Investment in securities, funds, by FII is Foreign Indirect Investment

12. BT Seed is associated with which among the following?

(A)Rice

(B)Wheat

(C)Cotton

(D)Oil Seeds

Answer: Cotton

13. The headquarters of International Atomic Energy Agency is in ?

(A)Geneva

(B)Paris

(C)Vienna

(D)Washington

Answer: Vienna

14. In the Budget estimates of 2011-12, an allocation of Rs. 400 Crore has been made to bring in second green revolution in East in the rice based cropping system of ____?

(A)Assam and West Bengal

(B)Assam, West Bengal, Odisha, Bihar & Jharkhand

(C)Assam, West Bengal, Odisha, Bihar

(D)Assam, West Bengal, Odisha, Bihar, Jharkhand , Eastern Uttar Pradesh andChhattisgarh

Answer: Assam, West Bengal, Orissa, Bihar, Jharkhand, Eastern Uttar Pradesh and Chattisgarh.

15. In the Budget 2011-12, presented by the Finance Minister on 28.2.2011, the income tax limit for senior citizens has been increased to ?

(A)Rs. 2.50 Lakh

(B)Rs. 2.60 Lakh

(C)Rs. 2.80 Lakh

(D)Rs. 3.00 Lakh

Answer: Rs. 2.50 Lakh. Above 80, its 5 Lakh

16. If the Anglo Indian community does not get adequate representation in the Lok Sabha, two members of the community can be nominated by:

(A)Prime Minister

(B)President

(C)Speaker

(D)President in consultation with Parliament

Answer: President

17. For the election of President of India, a citizen should have completed the age of___?

(A)25 Years

(B)35 Years

(C)30 Years

(D)18 Years

Answer: 35 Years

18. Who said: “Good citizen makes a good state and bad citizen makes a bad state”?

(A)Plato

(B)Aristotle

(C)Rousseau

(D)Laski

Answer: Aristotle

19. Member of parliament will lose his membership if he is continuously absent from sessions for

(A)45 days

(B)60 days

(C)90 days

(D)365 days

Answer: 60 Days

20. In Indian , Residuary Powers are vested in ___?

(A)Union Government

(B)State Government

(C)Both Union and State Government

(D)Local Government

Answer: Union Government

21. Mention the place where Buddha attained enlightment?

(A)Sarnath

(B)Bodhgaya

(C)Kapilvastu

(D)Rajgriha

Answer: Bodhgaya

22. Coronation of Shivaji took place in which year?

(A)1627

(B)1674

(C)1680

(D)1670

Answer: 6 June 1674

23. The system of Dyarchy was introduced in ___?

(A)1909

(B)1919

(C)1935

(D)1945

Answer: 1919, Government of India Act 1919 had introduced the system of Dyarchy to govern the provinces of British India.

24. The editor of Young India and Harijan was ____?

(A)Nehru

(B)Ambedkar

(C)Mahatma Gandhi

(D)Subhash Chandra Bose

Answer: Mahatma Gandhi

25. Who of the following attended all the three round table conferences?

(A)B R Ambedkar

(B)M M Malviya

(C)Vallabh Bhai Patel

(D)Mahatma Gandhi

Answer: B R Ambedkar.

26. Which is the largest living bird on Earth?

(A)Emu

(B)Ostrich

(C)Albatross

(D)Siberian Crane

Answer: Ostrich

27. Rihand Dam project provides irrigation to ____?

(A)Gujarat & Maharastra

(B)Odisha and West Bengal

(C)Uttar Pradesh and Bihar

(D)Kerala and Karnataka

Answer: C

28. The Headquarters of MCF (Master Control Facility) is

(A)Hyderabad

(B)Thumba

(C)Sri Harikota

(D)Hassan

Answer: Hassan , Now Bhopal also.

29. Which is the longest irrigation canal in India?

(A)Sir hind Canal

(B)Yamuna Canal

(C)Indira Gandhi Canal

(D)East Kosi Canal

Answer: Indira Gandhi Canal

30. Which one of the following minerals is found in Monazite Sand?

(A)Potassium

(B)Uranium

(C)Thorium

(D)Sodium

Answer: Thorium

31. Which plant is called ‘Herbal Indian Doctor” ?

(A)Amla

(B)Neem

(C)Tulsi

(D)Mango

Answer: Amla

32. In Coriander, useful parts are?

(A)Roots and leaves

(B)leaves and flowers

(C)leaves and dried fruits

(D)flowers and dried fruits

Answer: leaves and dried fruits

33. The pH of Human Blood is ___?

(A)7.2

(B)7.8

(C)6.6

(D)7.4

Answer: 7.4

34. Which among the following is the largest endocrine gland of country?

(A)Thyroid

(B)Parathyroid

(C)Adrenal

(D)Pituitary

Answer: Thyroid

35. Which amongst the following is the largest mammal?

(A)Elephant

(B)Whale

(C)Dinosaur

(D)Rhinoceros

Answer: Whale (Blue Whale).

36. Which part becomes modified as the tusk of elephant?

(A)Canine

(B)Premolar

(C)Second Incisor

(D)Molar

Answer: second upper incisors

37. Optical fibres are based upon the phenomenon of which of the following?

(A)Interference

(B)Dispersion

(C)Diffraction

(D)Total Internal Reflection

Answer: Total Internal Reflection

38. Now a days, Yellow lamps are frequently used as street lights. Which among the following gases, is used in these lamps?

(A)Sodium

(B)Neon

(C)Hydrogen

(D)Nitrogen

Answer: Sodium

39. Mirage is an example of ____?

(A)Refraction of light

(B)Total Internal Reflection of Light

(C)Refraction and Total Internal Reflection of Light

(D)Dispersion of Light

Answer: Refraction and Total Internal Reflection of Light

40. The phenomenon of light associated with the appearance of blue color of sky is?

(A)Interference

(B)Reflection

(C)Refraction

(D)Scattering

Answer: Scattering

41. In which of the following areas, spreadsheet software is more useful?

(A)Psychology

(B)Publishing

(C)Statistics

(D)Message sending

Answer: Statistics

42. A Groupware is a

(A)Hardware

(B)Software

(C)Network

(D)Firmware

Answer: Groupware is collaborative software . Correct option B

43. Lens is made up of ___?

(A)Pyrex Glass

(B)Flint Glass

(C)Ordinary Glass

(D)Cobalt Glass

Answer: Flint glass

44. The element which is used for vulcanizing rubber is?

(A)Sulfur

(B)Bromine

(C)Silicon

(D)Phosphorus

Answer: Sulfur

45. Which of the following is responsible for extra strength of Pyrex glass?

(A)Potassium carbonate

(B)Lead Oxide

(C)Borax

(D)Ferric Oxide

Answer: Borax

46. The Noble Gas used for the treatment of cancer is ___?

(A)Helium

(B)Argon

(C)Krypton

(D)Radon

Answer: Radon, in radiation therapy,

47. Vasundhara Summit was held in __?

(A)USA

(B)UK

(C)Brazil

(D)Australia

Answer: Brazil, Rio De Janeiro

48. Loktak is a ____?

(A)Valley

(B)Lake

(C)River

(D)Mountain Range

Answer: Lake, in Manipur

49. Which city receives the highest cosmic radiation amongst the following>

(A)Chennai

(B)Mumbai

(C)Kolkata

(D)Delhi

Answer: New Delhi.

50 The first computer made available for commercial use was

A)MANIAC

B)ENIAC

C)UNIVAC

D)EDSAC

Answe-UNIVAC

1The common tree species in Nilgiri Hills is:

(A)Sal

(B)Pine

(C)Eucalyptus

(D)Teak

Answer: Eucalyptus.

2. Which of the following statements on Railway Budget 2011-12 is correct?

(A)There would be a 10% increase in fares for long distance train travel both by AC and NONAC classes

(B)There would be 15% increase in freight rates on all goods other than food grains

(C)There would be 15% increase in passenger fares for all classes for long distance and freights

(D)There would be no increase in fares for both suburban and long distance travel

Answer: D

3. The nuclear reactors which were damaged heavily due to strong Earthquake-cum-Tsunami that hit Japan on March 11, 2011 causing radiation leakage at:

(A)Fukushima

(B)Tokyo

(C)Kyoto

(D)None of them

Answer: Fukushima

4. First Indian Prime Minister to visit Siachen was ?

(A)Rajiv Gandhi

(B)Inder Kumar Gujaral

(C)Mammohan Singh

(D)None of them

Answer: Manmohan Singh

5. Which of the following books has been written by Kishwar Desai?

(A)The Red Devil

(B)Witness the night

(C)Tonight this Savage Rite

(D)Earth and Ashes

Answer: Witness the Night.

6. Which of the following folk / tribal dances is associated with Karnataka?

(A)Yakshagana

(B)Jatra

(C)Veedhi

(D)Jhora

Answer: Yakshagana

7. Who of the following received the Sangeet Natak Academi’s Ustad Bismillah Khan Puraskar for 2009 in theatre?

(A)Omkar Shrikant Dadarkar

(B)Ragini chander sarkar

(C)Abanti Chakraborty and Sukracharjya Rabha

(D)K Nellai Maniknandan

Answer: Abanti Chakraborty and Sukracharjya Rabha

8. Which of the following country did not win any of the FIFA cup in 2002, 2006 and 2010?

(A)Brazil

(B)Argentina

(C)Spain

(D)South Africa

Answer: South Africa

9. Who invented vaccination for small pox?

(A)Sir Frederick Grant Banting

(B)Sir Alexander Fleming

(C)Edward Jenner

(D)Loius Pasteur

Answer: Edward Jenner

10. Who was the first Indian to become the member of British parliament?

(A)Bankim Chandra Chaterjee

(B)W C Banerjee

(C)Dadabhai Naoroji

(D)None of the above

Answer: Dadabhai Naoroji

11. The purchase of shares and bonds of Indian companies by Foreign Institutional Investors is called?

(A)FDI

(B)Portfolio Investment

(C)NRI Investment

(D)Foreign Indirect Investment

Answer: Investment in securities, funds, by FII is Foreign Indirect Investment

12. BT Seed is associated with which among the following?

(A)Rice

(B)Wheat

(C)Cotton

(D)Oil Seeds

Answer: Cotton

13. The headquarters of International Atomic Energy Agency is in ?

(A)Geneva

(B)Paris

(C)Vienna

(D)Washington

Answer: Vienna

14. In the Budget estimates of 2011-12, an allocation of Rs. 400 Crore has been made to bring in second green revolution in East in the rice based cropping system of ____?

(A)Assam and West Bengal

(B)Assam, West Bengal, Odisha, Bihar & Jharkhand

(C)Assam, West Bengal, Odisha, Bihar

(D)Assam, West Bengal, Odisha, Bihar, Jharkhand , Eastern Uttar Pradesh andChhattisgarh

Answer: Assam, West Bengal, Orissa, Bihar, Jharkhand, Eastern Uttar Pradesh and Chattisgarh.

15. In the Budget 2011-12, presented by the Finance Minister on 28.2.2011, the income tax limit for senior citizens has been increased to ?

(A)Rs. 2.50 Lakh

(B)Rs. 2.60 Lakh

(C)Rs. 2.80 Lakh

(D)Rs. 3.00 Lakh

Answer: Rs. 2.50 Lakh. Above 80, its 5 Lakh

16. If the Anglo Indian community does not get adequate representation in the Lok Sabha, two members of the community can be nominated by:

(A)Prime Minister

(B)President

(C)Speaker

(D)President in consultation with Parliament

Answer: President

17. For the election of President of India, a citizen should have completed the age of___?

(A)25 Years

(B)35 Years

(C)30 Years

(D)18 Years

Answer: 35 Years

18. Who said: “Good citizen makes a good state and bad citizen makes a bad state”?

(A)Plato

(B)Aristotle

(C)Rousseau

(D)Laski

Answer: Aristotle

19. Member of parliament will lose his membership if he is continuously absent from sessions for

(A)45 days

(B)60 days

(C)90 days

(D)365 days

Answer: 60 Days

20. In Indian , Residuary Powers are vested in ___?

(A)Union Government

(B)State Government

(C)Both Union and State Government

(D)Local Government

Answer: Union Government

21. Mention the place where Buddha attained enlightment?

(A)Sarnath

(B)Bodhgaya

(C)Kapilvastu

(D)Rajgriha

Answer: Bodhgaya

22. Coronation of Shivaji took place in which year?

(A)1627

(B)1674

(C)1680

(D)1670

Answer: 6 June 1674

23. The system of Dyarchy was introduced in ___?

(A)1909

(B)1919

(C)1935

(D)1945

Answer: 1919, Government of India Act 1919 had introduced the system of Dyarchy to govern the provinces of British India.

24. The editor of Young India and Harijan was ____?

(A)Nehru

(B)Ambedkar

(C)Mahatma Gandhi

(D)Subhash Chandra Bose

Answer: Mahatma Gandhi

25. Who of the following attended all the three round table conferences?

(A)B R Ambedkar

(B)M M Malviya

(C)Vallabh Bhai Patel

(D)Mahatma Gandhi

Answer: B R Ambedkar.

26. Which is the largest living bird on Earth?

(A)Emu

(B)Ostrich

(C)Albatross

(D)Siberian Crane

Answer: Ostrich

27. Rihand Dam project provides irrigation to ____?

(A)Gujarat & Maharastra

(B)Odisha and West Bengal

(C)Uttar Pradesh and Bihar

(D)Kerala and Karnataka

Answer: C

28. The Headquarters of MCF (Master Control Facility) is

(A)Hyderabad

(B)Thumba

(C)Sri Harikota

(D)Hassan

Answer: Hassan , Now Bhopal also.

29. Which is the longest irrigation canal in India?

(A)Sir hind Canal

(B)Yamuna Canal

(C)Indira Gandhi Canal

(D)East Kosi Canal

Answer: Indira Gandhi Canal

30. Which one of the following minerals is found in Monazite Sand?

(A)Potassium

(B)Uranium

(C)Thorium

(D)Sodium

Answer: Thorium

31. Which plant is called ‘Herbal Indian Doctor” ?

(A)Amla

(B)Neem

(C)Tulsi

(D)Mango

Answer: Amla

32. In Coriander, useful parts are?

(A)Roots and leaves

(B)leaves and flowers

(C)leaves and dried fruits

(D)flowers and dried fruits

Answer: leaves and dried fruits

33. The pH of Human Blood is ___?

(A)7.2

(B)7.8

(C)6.6

(D)7.4

Answer: 7.4

34. Which among the following is the largest endocrine gland of country?

(A)Thyroid

(B)Parathyroid

(C)Adrenal

(D)Pituitary

Answer: Thyroid

35. Which amongst the following is the largest mammal?

(A)Elephant

(B)Whale

(C)Dinosaur

(D)Rhinoceros

Answer: Whale (Blue Whale).

36. Which part becomes modified as the tusk of elephant?

(A)Canine

(B)Premolar

(C)Second Incisor

(D)Molar

Answer: second upper incisors

37. Optical fibres are based upon the phenomenon of which of the following?

(A)Interference

(B)Dispersion

(C)Diffraction

(D)Total Internal Reflection

Answer: Total Internal Reflection

38. Now a days, Yellow lamps are frequently used as street lights. Which among the following gases, is used in these lamps?

(A)Sodium

(B)Neon

(C)Hydrogen

(D)Nitrogen

Answer: Sodium

39. Mirage is an example of ____?

(A)Refraction of light

(B)Total Internal Reflection of Light

(C)Refraction and Total Internal Reflection of Light

(D)Dispersion of Light

Answer: Refraction and Total Internal Reflection of Light

40. The phenomenon of light associated with the appearance of blue color of sky is?

(A)Interference

(B)Reflection

(C)Refraction

(D)Scattering

Answer: Scattering

41. In which of the following areas, spreadsheet software is more useful?

(A)Psychology

(B)Publishing

(C)Statistics

(D)Message sending

Answer: Statistics

42. A Groupware is a

(A)Hardware

(B)Software

(C)Network

(D)Firmware

Answer: Groupware is collaborative software . Correct option B

43. Lens is made up of ___?

(A)Pyrex Glass

(B)Flint Glass

(C)Ordinary Glass

(D)Cobalt Glass

Answer: Flint glass

44. The element which is used for vulcanizing rubber is?

(A)Sulfur

(B)Bromine

(C)Silicon

(D)Phosphorus

Answer: Sulfur

45. Which of the following is responsible for extra strength of Pyrex glass?

(A)Potassium carbonate

(B)Lead Oxide

(C)Borax

(D)Ferric Oxide

Answer: Borax

46. The Noble Gas used for the treatment of cancer is ___?

(A)Helium

(B)Argon

(C)Krypton

(D)Radon

Answer: Radon, in radiation therapy,

47. Vasundhara Summit was held in __?

(A)USA

(B)UK

(C)Brazil

(D)Australia

Answer: Brazil, Rio De Janeiro

48. Loktak is a ____?

(A)Valley

(B)Lake

(C)River

(D)Mountain Range

Answer: Lake, in Manipur

49. Which city receives the highest cosmic radiation amongst the following>

(A)Chennai

(B)Mumbai

(C)Kolkata

(D)Delhi

Answer: New Delhi.

50 The first computer made available for commercial use was

A)MANIAC

B)ENIAC

C)UNIVAC

D)EDSAC

Answe-UNIVAC

SSC Placement Paper : SSC Placement paper 16 August 2012

Arithmetical Ability

1. The product of 2 numbers is 1575 and their quotient is 9/7. Then the sum of the numbers is –

a. 74

b. 78

c. 80

d. 90

Ans : Let the numbers be x and y .

∴ xy = 1575

And x/y = 9/7

∴ xy/x/y = 1575/9/7

∴ y2 = 1225

∴ y = 35 and x = 45

∴ The sum of the numbers = 45+35

= 80

2. The value of (81)3.6 * (9)2.7/ (81)4.2 * (3) is __

a. 3

b. 6

c. 9

d. 8.2

Ans : (81)3.6 * (9)2.7/(81)4.2 * (3) = (3)14.4 * (3)5.4/(3)16.8 * (3)

= 314.4+5.4-16.8-1

= 32

= 9

3. √6+√6+√6+… is equal to –

a. 2

b. 5

c. 4

d. 3

Ans : Let √6+√6+√6+…. be x.

∴ x =√6+x

∴ x2 = 6+x

∴ x2 + x – 6 = 0

∴ (x-3) (x+2) = 0

∴ x = 3

4. The sum of the squares of two natural consecutive odd numbers is 394. The sum of the numbers is –

a. 24

b. 32

c. 40

d. 28

Ans : Let the consecutive odd numbers be x and (x+2)

∵ x2 + (x + 2)2 = 394

∴ x2 + x2 + 4x + 4 = 394

∴ 2 x2 +4x – 390 = 0

∴ x2 + 2x – 195 = 0

∴ (x +15) (x-13) = 0

∴ x = 13

∴ Required sum = 13 +15 =28

5. When (6767 +67) is divided by 68, the remainder is-

a. 1

b. 63

c. 66

d. 67

Ans: (6767+ 67) = 67(6766 + 166)

As 66 is an even number

∴ 6766 is an even number

∴ (6766 + 1) is perfectly divisible by (67 + 1)

i.e. 68

∴ The remainder = 67

6. In a division sum, the divisor is 4 times the quotient and twice the remainder. If a and b are respectively the divisor and the dividend then-

a. 4b-a2 /a =3

b. 4b-2a /a2 =2

c. (a+1)2 = 4b

d. A(a+2)/b = 4

Ans : As divisor is a, and dividend is b.

∴ Quotient = a/4

And Remainder = a/2

∴ b = a * a/4 + a/2

∴ 4b = a2 + 2a

∴ a(a+2)/b = 4

7. If 738 A6A is divisible by 11, then the value of A is-

a. 6

b. 3

c. 9

d. 1

Ans : As 738A6 A is divisible by 11.

∴ A + A + 3 = 6 + 8 + 7

∴ A = 9

8. The east positive integer that should be subtracted from 3011 * 3012 so that the difference is perfect square is-

a. 3009

b. 3010

c. 3011

d. 3012

Ans : ∵ 3011 * 3012 = 3011 (3011 + 1)

= (3011)2 + 3011

∴ Required least number = 3011

9. P, Q, R are employed to do a work for Rs. 5750. P and Q together finished 19/23 of work and Q and R together finished 8/23 of work. Wage of Q, in rupees, is-

a. 2850

b. 3750

c. 2750

d. 1000

Ans : Work done by Q = 19/23 + 8/23 – 1

4/23

∴ Wage of Q = 4/23 * 5750

= Rs. 1000

10. A can do a piece of work in 24 day, B in 32 days and C in 64 days. All begin to do it together, but A leaves after 6 days and B leaves 6 days before the completion of the work. How many days did the work last?

a. 15

b. 20

c. 18

d. 30

Ans : Work done by A = 6/24 = ¼

Work done by B = (x-6)/32

(where x is no. of days in which work is competed)

∵ ¼ + x – 6/32 + x/64 = 1

∴ 16 + 24 – 12 + x /64 = 1

∴ 3x + 4 = 64

∴ x = 60/3 = 20 days

11. The square root of (0.75)3 /1-0.75 + [0.75 + 90.75)2 +1] is-

a. 1

b. 2

c. 3

d. 4

Ans : The square root of

[(0.75)3/1 – 0.75 + {0.75 + (0.75)2 + 1}]

= √1.6875 + 2.3125

=√4

= 2

12. Given that √4096 = 64, the value of √4096 + √40.96 +√0.004096 is-

a. 70.4

b. 70.464

c. 71.104

d. 71.4

Ans : Given Exp. = √4096 + √40.96 +√0.004096

= 64 + 6.4 + 0.064

= 70.464

13. By selling an article at 3/4th of the marked price, there is a gain of 25%. The ratio of the marked price and the cost price is-

a. 5 : 3

b. 3 : 5

c. 3 : 4

d. 4 : 3

Ans : Let of M.P. be Rs. x.

∴ S.P. = Rs.3x/4

and C.P. = 3x/4 * 100/125

= Rs.3x/5

∴ required ratio = x: 3x/5

= 5:3

14. A and B earn in the ratio 2:1. They spend in the ratio 5:3 and save in the ratio 4:1. If the total monthly savings of both A and B are Rs.5,000, the monthly income of B is-

a. Rs.7,000

b. Rs.14,000

c. Rs.5,000

d. Rs.10,000

Ans : Let the monthly income of B be Rs. x.

∴ Monthly income of A = Rs. 2x and

Saving of A =5000 * 4/(4 + 1)

= Rs. 4000

Saving of B = Rs.1000

∵ 2x – 4000/x-1000 = 5/3

⇒ 6x – 12000 = 5x – 5000

∴ x = Rs.7000

15. The ratio of the sum of two numbers and their difference is 5:1. The ratio of the greater number to the smaller number is-

a. 2 : 3

b. 3 : 2

c. 5 : 1

d. 1 : 5

Ans : Let the numbers be x and y.

∵ x + y/ x – y = 5/1

⇒ 5x – 5y = x + y

⇒ 4x = 6y

⇒ x/y = 6/4

x : y = 3 : 2

16. A cistern has 3 pipes A, B and C. A and B can fill it in 3 and 4 hours respectively, and C can empty it in 1 hour. If the pipes are opened at 3 p.m., 4 p.m. and 5 p.m. respectively on the same day, the cistern will be empty at-

a. 7.12 p.m.

b. 7.15 p.m.

c. 7.10 p.m.

d. 7.18 p.m.

Ans : Let the cistern be emptied at x p.m.

∵ x -3/3 + x – 4/4 = x - 5/1

⇒ 4x – 12 + 3x – 12/12 = x – 5/1

⇒ 7x – 24 = 12x – 60

⇒ 5x = 36

∴ x = 36/5

= 7 hr. + 12 min.

= 7.12 p.m.

17. If A works alone, he would take 4 days more to complete the job than if both A and B worked together. If B worked alone, he would take 16 days more to complete the job than if A and B work together. How many days would they take to complete the work if both of them worked together?

a. 10 days

b. 12 days

c. 6 days

d. 8 days

Let x days are taken when they work together.

∴ Time taken by A to complete the work = (x + 4) days

And Time taken by B to complete the work = (x + 16) days

∵ 1/x = 1/x + 4 + 1/x + 16

∴ 1/x = x + 16 + x + 4/(x + 4) (x +16)

∴ 2x2 + 20x = x2 + 20x +64

∴ x2 = 64 (8)2

∴ x = 8 days

18. 250 men can finish a work in 20 days working 5 hours a day. To finish the work within 10 days working 8 hours a day, the minimum number of men required is-

a. 310

b. 300

c. 313

d. 312

Ans : Required no. of men = 250 * 5 * 20/8 * 10

=312.5

313

19. 2 men and 5 women can do a work in 12 days. 5 men 2 women can do that work in 9 days. Only 3 women can finish the same work in-

a. 36 days

b. 21 days

c. 30 days

d. 42 days

∵ (2m + 5w) * 12 = (5m + 2w) * 9

⇒ 24m + 60w = 45m + 18w

⇒ 21m = 42w

⇒ 1m = 2w

∴ 2m + 5w = 9w

∴ required no. of days = 9 * 12/3

= 36

20. While selling, a businessman allows 40% discount on the marked price and there is a loss of 30%. If it is sold at the marked price, profit per cent will be –

a. 10%

b. 20%

c. 16 2/3%

d. 16 1/3%

Ans : Let the M.P. be Rs.100.

∴ S.P = (100-40) = Rs.60

and C. P. = 60 * 100/(100-30) = Rs.600/7

∴ Reqd. % profit = 100-600/7 /600/7 * 100%

= (700 – 600)/7 * 600 * 100 * 7%

= 16 2/3%

21. Successive discount of 10% , 20% and 50% will be equivalent to a single discount of-

a. 36%

b. 64%

c. 80%

d. 56%

Ans : Equivalent to a single discount

= [100 - (100 – 10) (100 – 20) (100 – 50)/100*100]%

= [100 - 90*80*50/10000]%

= [100-36]%

= 64%

22. A retailer offers the following discount schemes for buyers on an article-

I. Two successive discounts of 10%

II. A discount of 12% followed by a discount of 8%.

III. Successive discounts of 15% and 5%

IV. A discount of 20%

The selling price will be minimum under the scheme-

a. I

b. II

c. III

d. IV

Ans : From (i) single discount = [10 + 10 – 10*10/100] % = 19%

From (ii) single discount = [12 + 8 -12*8/100] % = 19.04%

From (iii) single discount = [15 + 5 – 15*5/100] % = 19.25%

From (iv) single discount = 20%

∴ The S.P. will be minimum under the scheme IV.

23. Of three numbers, the second is thrice the first and the third number is three-fourth of the first. If the average of the three numbers is 114, the largest number is –

a. 72

b. 216

c. 354

d. 726

Ans : Let the first number be x.

∴ Second number = 3x and Third number = 3x/4

∵ x + 3x + 3x/4 = 3*114

⇒ 19x/4 = 342

∴ x = 342*4/19 = 72

∴ The largest number = 3*72

= 216

24. A car covers 1/5 of the distance from A to B at the speed of 8 km/hour, 1/10 of the distance at 25 km per hour and the remaining at the speed of 20 km per hour. Find the average speed of the whole journey-

a. 12.625 km/hr

b. 13.625 km/hr

c. 14.625 km/hr

d. 15.625 km/hr

Ans : If the whole journey be x km. The total time taken

= (x/5/8 + x/10/25 + 7x/10/20) hrs

= (x/40 + x/250 + 7x/200) hrs

= 25x + 4x + 35x/1000

= 64x/1000 hrs

∴ Average speed = x/64x/1000

= 15.625 km/hr

25. The average of 3 numbers is 154. The first number is twice the second and the second number is twice the third. The first number is-

a. 264

b. 132

c. 88

d. 66

Ans : Let the first number be x.

∴ Second number = x/2 and Third number = x/4

∵ x + x/2 + x/4 = 3 x 154

⇒ 7x/4 = 462

∴ x = 462*4/7

=264

26. The average salary of all the staff in an office of a corporate house is Rs. 5,000. The average salary of the officers is Rs. 14,000 and that of the rest is Rs. 4,000. If the total number of staff is 500, the number of officers is –

a. 10

b. 15

c. 25

d. 50

Ans : Let the number of officers be x.

∵ 5000*500 = 14000x + 4000(500-x)

∴ 2500000 =14000x + 2000000 – 4000x

∴ x = 500000/10000

= 50

27. The average marks of 40 students in an English exam are 72. Later it is found that three marks 64, 62 and 84 were wrongly entered as 60, 65 and 73. The average after mistakes were rectified is-

a. 70

b. 72

c. 71.9

d. 72.1

Ans : Correct average

= 40*72 + (64 + 62 +84) – 68 – 65 – 73/40

= 2880 + 210 – 206/40 = 2884/40

= 72.1

28. A and B are two alloys of gold and copper prepared by mixing metals in the ratio 7:2 and 7:11 respectively. If equal quantities of the alloys are melted to form a third alloy C, the ratio of gold and copper in C will be-

a. 5 : 7

b. 5 : 9

c. 7 : 5

d. 9 : 5

Ans : Quantity of gold in A = 7/9* wt. of A

Quantity of gold in B = 7/18* wt. of B

If 1 kg of each A and B are mixed to form third alloys C.

Then quantity of gold in 2 kg C = 7/9 + 7/18

= 7/6 kg

And Quantity of copper in 2 kg C = 2 – 7/6

= 5/6 kg

∴ Required ratio = 7/6 : 5/6 = 7 : 5

29. In a laboratory, two bottles contain mixture of acid and water in the ratio 2 : 5 in the first bottle and 7 : 3 in the second. The ratio in which the contents of these two bottles be mixed such that the new mixture has acid and water in the ratio 2 : 3 is-

a. 4 : 15

b. 9 : 8

c. 21 : 8

d. 1 : 2

Ans : Quantity of acid in first bottle = 2/7 x mix.

and Quantity of acid in second bottle = 7/10 x mix.

If x and 1 volumes are taken from I and II bottle respectively to form new mixture.

Then, (2/7 x + 7/10 * 1)/(5x/7 +3/10 * 1) = 2/3

⇒ 6x/7 21/10 = 10x/7 +6/10

⇒ 4x/7 = 15/10

∴ x = 15/10*7/4 = 21/8

∴ Required ratio = x : 1

= 21 : 8

30. A mixture contains 80% acid and rest water. Part of the mixture that should be removed and replaced by same amount of water to make the ratio of acid and water 4 : 3 is-

a. 1/3 rd

b. 3/7 th

c. 2/3 rd

d. 2/7 th

Ans : Let the initial wt. of mixture be 1 kg and x kg of mixture is taken out and replaced by same amount of water.

∵ Amt. of acid/Amt. of water = 0.8 – 0.8x/ (0.2 – 0.2x + x) = 4/3

⇒ 2.4 – 2.4x = 0.8 + 3.2x

⇒ 5.6x = 1.6

∴ x = 1.6/5.6 = 2/7th part

31. An employer reduces the number of his employees in the ratio 9 : 8 and increases their wages in the ratio 14 : 15. If the original wage bill was Rs. 189,900, find the ratio in which the wage bill is decreased-

a. 20 : 21

b. 21 : 20

c. 20 : 19

d. 19 : 21

Ans : Let the initial number of employees be 9x and the employer gives Rs. 14y as wage to each.

∵ 9x * 14y =18900

∴ xy = 150 and The later bill = 8x*15y = 120xy

= 120*150 = 18000

∴ Required ratio = 18000 : 18900

= 20 : 21

32. The batting average for 40 innings of a cricketer is 50 runs. His highest score exceeds his lowest score by 172 runs. If these two innings are excluded, the average of the remaining 38 innings is 48 runs. The highest score of the player is-

a. 165

b. 170

c. 172

d. 174

Ans : Let the max. number of runs be x.

∴ The lowest score = (x-172)

∵ 40*50 = 38*48 + x + (x-172)

⇒ 2000 = 1824 + 2x – 172

∴ x= 174

33. Walking at 3 km per hour, Pintu reaches his school 5 minutes late. If he walks at 4 km per hour he will be 5 minutes early. The distance of Pintu’s school from his house is-

a. 1 ½ Km

b. 2 Km

c. 2 ½ Km

d. 5 Km

Ans : Distance of the school from the house

= 4*3/(4 – 3) * 5 +5/60 km

= 12*1/6

=2 km

34. Nitin bought some oranges at Rs. 40 a dozen and an equal number at Rs.30 a dozen. He sold them at Rs. 45 a dozen and made a profit of Rs.480. The number of oranges, he bought, was-

a. 48 dozen

b. 60 dozen

c. 72 dozen

d. 84 dozen

Ans : Let the number of oranges bought be x.

∴ 45x/12 – 70x/24 = 480

⇒ (45-35)/12 = 480

⇒ x = 480*12/10 = 576

= 48*12 = 48 dozen

35. A man buys two chairs for a total cost of Rs.900. By selling one for 4/5 of its cost and the other for 5/4 of its cost, he makes a profit of Rs.90 on the whole transaction. The cost of the lower priced chair is-

a. Rs.360

b. Rs.400

c. Rs.420

d. Rs.300

Ans : Let the cost price of 1 chair be Rs. x.

∴ C.P. of other chair = Rs. (900-x)

∵ 4/5 x + 5/4 (900-x) = 900 + 90

⇒ 4/5 x +1125 -5x/4 = 990

∴ 9x/20 = 135

∴ x = 135*20/9

= Rs. 300

∴ C.P. of the lower priced chair is Rs. 300.

36. By selling 100 oranges, a vendor gains the selling price of 20 oranges. His gain per cent is-

a. 20

b. 25

c. 30

d. 32

Ans : Let the S.P. of 100 Oranges be Rs. x.

∴ S.P. of 20 oranges = x/100 *20 = Rs. x/5

∴C.P. of 100 oranges = x - x/5

= Rs. 4x/5

∴ Reqd. Profit % = x/5 * 100*5/4x %

= 25%

37. 60% of the cost price of an article is equal to 50% of its selling price. Then the percentage of profit or loss on the cost price is-

a. 20% loss

b. 16 2/3% profit

c. 20% profit

d. 10% loss

Ans : Let the cost price be Rs. 100.

∵ S.P * 50/100 = 100*16/100

∴ S.P = 60*100/50

= Rs. 120

∴ Reqd. % profit = (120 – 100)% = 20%

38. Maninder bought two horses at Rs.40,000 each. He sold one horse at 15% gain, but had to sell the second horse at a loss. If he hand suffered a loss of Rs.3,600 on the whole transaction, then the selling price of the second horse is-

a. Rs.30,000

b. Rs.30,200

c. Rs.30,300

d. Rs.30,400

Ans: C.P. of two horses = 2*40000 = Rs. 80000

and S.P. of two horse = 80000 – 3600 = Rs. 76400

∴ S.P. of the other horse = 76400 – 46000 = 30400

39. A fruit-seller buys x guavas for Rs.y and sells y guavas for Rs. x. If x>y, then he made-

a. x2 – y2 / xy % loss

b. x2 – y2 / xy % gain

c. x2 – y2 / y2 % loss

d. x2 – y2 / y2 * 100% gain

Ans : C.P. of 1 guava = Rs. y/x [x>y]

and S.P. of 1 guava = Rs. x/y

∴ Reqd. Gain% = x/y – y/x/ y/x * 100%

= x2 - y2 /y2 *100%

40. A jar contain 10 red marbles and 30 green ones. How many red marbles must be added to the jar so that 60% of the marbles will be red?

a. 25

b. 30

c. 35

d. 40

Ans : Let after adding x red marbles, the red marbles with be 60% of the total.

∵ (10+x)/ (10 + x) + 30 = 60/100

⇒ 10 + x/40 + x = 3/5

⇒ 50 + 5x = 120 + 3x

x = 70/2 = 35

41. If a number multiplied by 25% of itself gives a number which is 200% more than the number, then the number is –

a. 12

b. 16

c. 35

d. 24

Ans : Let the number be x.

∵ x*25x/100 = x + 200x/100

⇒ x2/4 = 3x

⇒ x2 - 12x = 0

⇒ x – 12 = 0

∴ x = 12

42. The value of an article depreciates every year at the rate of 10% of its value. If the present value of the article is Rs.729, then its worth 3 years ago was-

a. Rs.1250

b. Rs.1000

c. Rs.1125

d. Rs.1200

Ans : Let the worth 3 years ago be Rs. x.

∵ 729 = x (1 – 10/100)3

⇒ 729 = x*9*9*9/10*10*10

∴ x = Rs. 1000

43. The price of onions has been increased by 50%. In order to keep the expenditure on onions the same the percentage of reduction in consumption has to be-

a. 50%

b. 33 1/3%

c. 33%

d. 30%

Ans : Reqd. Percentage of reduction

= 50*100/(100 + 50) %

= 5000/150 %

= 33 1/3%

44. A took two loans altogether of Rs.1200 from B and C. B claimed 14% simple interest per annum, while C claimed 15% per annum. The total interest paid by A in one year was Rs.172. Then, A borrowed-

a. Rs.800 from C

b. Rs.625 from C

c. Rs.400 from B

d. Rs.800 from B

Ans : If A borrowed Rs. x from B. and A borrowed Rs. Rs. (1200 – x) from C.

∵ (1200 – x)*15*1/100 + x*14*1/100

⇒ 18000 – 15x + 14x = 172*100

x = Rs. 800

45. If a regular polygon has each of its angles equal to 3/5 times of two right angles, then the number of side is-

a. 3

b. 5

c. 6

d. 8

Ans : If the number of sides a regular polygon be n.

Then (2n-4)/n = 2*3/5

⇒ (2n – 4)*5 = 6n

∴ n = 5

46. A square is of area 200 sq. m. A new square is formed in such a way that the length of its diagonal is √2 times of the diagonal of the given square. The the area of the new square formed is-

a. 200√2 sq.m

b. 400√2 sq.m

c. 400 sq.m

d. 800 sq.m

Ans : Length of the diagonal of Ist square

= √2*200

= 20 m

∴ Length of the diagonal of new square = 20√2m

∴ Area of the new square = ½*(20.√2)2 = 400 sq. m

47. The heights of a cone, cylinder and hemisphere are equal. If their radii are in the ratio 2 : 3 : 1, then the ratio of their volumes is-

a. 2 : 9 : 2

b. 4 : 9 : 1

c. 4 : 27 : 2

d. 2 : 3 : 1

Ans : Ratio of their volumes [cone : cylinder : hemisphere]

= 1/3(2)2h : (3)2*h : 2/3(1)2.h

= 4/3 : 9 : 2/3 = 4 : 27 : 2

48. A motor-boat can travel at 10 km/hr in still water. It travelled 91 km downstream in a river and then returned to the same place, taking altogether 2 hours. Find the rate of flow of river-

a. 3 km/hr

b. 4 km/hr

c. 2 km/hr

d. 5 km/hr

Ans : Let the rate of flow of river be x km/hr.

∵ 91/(10 + x) + 91/(10 – x) = 20

⇒ 91(10 – x + 10 + x)(10 + x) (10 – x) = 20

⇒ 91*20 = 20(100 – x2)

⇒ x2 = 9 = (3)2

∴ x = 3 km/hr

49. A man driving at 3/4th of his original speed reaches his destination 20 minutes later than the usual time. Then the usual time is-

a. 45 minutes

b. 60 minutes

c. 75 minutes

d. 120 minutes

Ans : Let the original speed be x km/hr and the usual time be y hours.

∵ x * y = ¾ x(y+1/3)

∴ 4y = 3y + 1

∴ y = 1 hr = 60 minutes

50. A motor-boat, travelling at the same speed, can cover 25 km upstream and 39 km downstream in 8 hours. At the same speed, it can travel 35 km upstream and 52 km downstream in 11 hours. The speed of the stream is –

a. 2 km/hr

b. 3 km/hr

c. 4 km/hr

d. 5 km/hr

Ans : Let the speeds of motor boat and the stream be x and y km/hr respectively.

∵ 39/x + y + 25/x – y = 8 …(1)

and 52/x + y + 35/x – y = 11 …(2)

Solving equations (1) and (2), we get-

∴ 100 - 105/x – y = 32 – 33

∴ x – y = 5

and x + y = 13

∴ y = 4 km/hr

51. If a sum of money placed at compound interest, compounded annually, doubles itself in 5 years, then the same amount of money will be 8 times of itself in-

a. 25 years

b. 20 years

c. 15 years

d. 10 years

Ans : Required time = 5log 8/log 2

= 5*3log 2/log 2

= 15 years

52. A person has left an amount of Rs.1,20,000 to be divided between his 2 son aged 14 years and 12 years such that they get equal amounts when each attains 18 years of age. If the amount gets a simple interest of 5% per annum, the younger son’s share at present is-

a. Rs.48,800

b. Rs.57,600

c. Rs.62,400

d. Rs.84,400

Ans : Let the present share of the younger son be Rs. x.

∴ The share of the elder son = Rs. (120000 – x)

∵ x + x*6*5/100

= (120000 – x) + (120000 – x)*4*5/100

⇒ 130x/100 = (120000 –x)*120/100

⇒ 13x = 1440000 – 12x

⇒ 25x = 1440000

∴ x = Rs. 57600

53. If the simple interest on Rs.x at a rate of a% for m years is same as that on Rs. y at a rate of a2% for m2 years, then x : y is equal to-

a. m : a

b. am : 1

c. 1/m : 1/a

d. 1/am : 1

∵ x*a*m/100 = y*a2*m2/100

⇒ x/y = am/1

∴ x : y = am : 1

54. Base of a right prism is an equilateral triangle of side 6 cm. If the volume of the prism is 108√3 cc, its height is-

a. 9 cm

b. 10 cm

c. 11 cm

d. 12 cm

Ans : Height of the prism = 108√3*4/√3*(6)2 = 12cm

55. If a + 1/a + 2 = 0, then the value of (a37 – 1/a100) is-

a. 0

b. -2

c. 1

d. 2

Ans : ∵ a + 1/a + 2 = 0

⇒ a2 + 1 +2a = 0

⇒ (a + 1)2 = 0

⇒ a +1 = 0

∴ a = -1

∴ a37 -1/a100 = (-1) – (1) = -2

56. The value of k for which the graphs of (k-1) x+y-2 = 0 and (2-k) x -3y + 1 = 0 are parallel is-

a. ½

b. -1/2

c. 2

d. -2

Ans : (k – 1) x + y – 2 = 0

∴ y = (1 – k) x + 2 ….(1)

and (2 –k) x – 3y – 1 = 0

3y = (2 – k) x +1

Y = 2 – k/3 x + 1/3 ….(2)

∵ m1 = m2

⇒ 1 – k = 2 – k/3

⇒ 3 – 3k = 2 - k

∴ k = 1/2

57. If a2 + b2 + c2 = 2 (a-b-c) – 3, then the value of (a – b + c) is-

a. -1

b. 3

c. 1

d. -2

Ans : a2 + b2 + c2 = 2(a – b – c) – 3

⇒ a2 - 2a + 1 + b2 + 2b + 1 + c2 + 2c + 1 = 0

⇒ (a – 1)2 + (b + 1)2 + (c + 1)2 = 0

⇒ a – 1 = 0, b + 1 = 0, c + 1 = 0

⇒ a = 1, b = -1, c = -1

∴ a + b – c = 1 – 1 + 1 =1

58. If x2 + 3x + 1 = 0, then the value of x3 + 1/x3 is-

a. -18

b. 18

c. 36

d. -36

Ans : ∵ x2 + 3x + 1 = 0

⇒ x + 3 + 1/x = 0

⇒ x + 1/x = -3

⇒ (x + 1/x)3 = (-3)3

⇒ x3 + 1/x3 + 3 (-3) = -27

∴ x3 + 1/ x3 = -18

59. If xa, xb, xc = 1, then the value of a3 + b3 + c3 is –

a. 9

b. abc

c. a + b + c

d. 3abc

Ans : ∵ xa. xb. xc = 1

⇒ xa + b + c = x0

⇒ a +b + c =0

∴ a3 + b3 + c3 = 3abc

60. Base of a right pyramid is a square, length of diagonal of the base is 24√2 m. If the volume of the pyramid is 1728 cu.m, its height is-

a. 7 m

b. 8 m

c. 9 m

d. 10 m

Ans : Area of the base of the pyramid

= ½ (24√2)2 = 576m2

∴ Height of pyramid + 1728*3/576 = 9m

61. The height of a right circular cone and the radius of its circular base are respectively 9 cm and 3 cm. The cone is cut by a plane parallel to its base so as to divide it into two parts. The volume of the frustum (i.e., the lower part) of the cone is 44 cubic cm. The radius of the upper circular surface of the frustum (taking = 22-7) is-

a. 3√12 cm

b. 3√13 cm

c. 3√6 cm

d. 3√20 cm

Ans : Let the radius of the upper circular part of the frustum be r cm.

(Picture)

Then r/3 = x/9 = AC/AD

∴ x = 3r (where AC = x cm)

⇒ /3h(r12 + r1r2+r22 = 44

⇒ /3(9 – x)*(9 + 3r + r2) = 44

⇒ (9 – x)*(9 + 3r + r2) = 44*3*7/22 = 42

⇒ 81 + 27r + 9r2 – 9x – 3rx - r2x = 42 On putting x = 3r,

⇒ 81 + 27r + 9r2 - 9r2 - 3r3 – 27r = 42

⇒ 3r3 = 39

∴ r = 3√13 cm

62. The ratio of radii of two right circular cylinder is 2 : 3 and their heights are in the ratio 5 : 4. The ratio of their curved surface area is-

a. 5 : 6

b. 3 : 4

c. 4 : 5

d. 2 : 3

Ans : Required ratio = 2*2r*5h/2*3r*4h = 5 : 6

63. A solid cylinder has total surface area of 462 sq.cm. Curved surface area is 1/3rd of its total surface area. The volume of the cylinder is-

a. 530 cm3

b. 536 cm3

c. 539 cm3

d. 545 cm3

Ans : ∵ 2r(r + h) = 462

and 2rh = 1/3*462 = 154

∴ r + h/h = 462/154 =3

∴ r + h = 3h

∴ r = 2h

∴ 2*2h2 = 154

∴ h2 = 154*7/22*4 = 49/4

= (7/2)2

∴ h = 7/2 cm

and r = 7 cm

∴ Volume of the cylinder

= 22/7*49*7/2

= 539 cm3

64. A cylinder and a cone have equal radii of their bases and equal heights. If their curved surface areas are in the ratio 8 : 5, the ratio of their radius and height is-

a. 1 : 2

b. 1 : 3

c. 2 : 3

d. 3 : 4

Ans : Let the radius and height of each are r and h respectively.

∵ 2rh/r√h2 + r2 = 8/5

∴ 10h = 8√r2 + h2

⇒ 100h2 =64r2 + 64h2

∴ h2 = 64 r2/36 = (4/3 r)2

⇒ h = 4/3r

∴ r : h = 3 : 4

65. A solid is hemispherical at the bottom and conical above. If the surface areas of the two parts are equal, then the ratio of radius and height of its conical part is-

a. 1 : 3

b. 1 :1

c. √3 : 1

d. 1 : √3

Ans :

∵ rl = 2r2

⇒ l = 2r

⇒ √h2 + r2 = 2r

⇒ h2 = 3r2

∴ r : h = r/h

= 1 : √3

66. If O is the circumcentre of ∆ ABC and ∠OBC = 350, then the ∠BAC is equal to-

a. 550

b. 1100

c. 700

d. 350

Ans : ∴ ∠BOC = 1800 – (350 + 350) = 1100

∴ ∠BAC = 1/2 * 1100 = 550

67. If I is the incentre of ∆ ABC and ∠ BIC = 1350, then ∆ ABC is-

a. Acute angled

b. Equilateral

c. Right angled

d. Obtuse angled

Ans :

∠BIC = 1350

⇒ B/2 + C/2 = 1800 – 1350 = 450

⇒ ∠B + ∠C = 900

∴ ∠A = 1800 – (∠B + ∠C) = 900

i.e., ∆ ABC is a right angled.

68. If sin2 (∝ + β/2) is-

a. 1

b. -1

c. 0

d. 0.5

∵ sin2 ∝ + sin2 β = 2

⇒ 1 – cos2 ∝ 1 - cos2 β = 2

⇒ cos2 ∝ cos2 β = 0

⇒ cos ∝ = 0

and cos β = 0

⇒ ∝ = /2 and β = /2

∴ cos (∝ + β/2) = cos [/2 + /2 / 2]

= cos /2 = 0

69. The length of a shadow of a vertical tower is 1/√3 times its height. The angle of elevation of the Sun is-

a. 300

b. 450

c. 600

d. 900

Ans :

∵ tan θ = h/1√3 h = √3 = tan 600

70. The graphs of x +2y =3 and 3x-2y = 1 meet the Y-axis at two points having distance-

a. 8/3 units

b. 4/3 units

c. 1 unit

d. 2 units

Ans : When the graphs meet the Y-axis at two points.

Then, [x + 2y = 3] at x = 0 ⇒ [0, y1 =3/2]

[3x - 2y = 1] at x = 0

and i.e., [0, y2 = -1/2]

Required distance = (y1 – y2)

= 3/2 – (- ½) = 2 units

71. If x+1/16x = 1, then the value of 64x3 + 1/64x3 is-

a. 4

b. 52

c. 64

d. 76

∵ x + 1/16x = 1

⇒ 16x2 – 16x + 1 =0

⇒ 16x2 – 16x + 4 = 3

⇒ (4x – 2)2 = 3

⇒ 4x = 2 + √3

⇒ 64x3 = (2±√3)3

=8 + 3√3 + 6√3 (2 + √3)

= 26 + 15√3

∴ 64x3 + 1/64x3 = (26 + 15√3) + 1/ (26 + 15√3)

= (26 + 15√3) + 26 - 15√3/676 -675

=52

72. If a, b, c, are three non-zero real numbers such that a + b + c = 0, and b2 ≠ ca, then the value of a2 + b2 + c2/ b2 –ca is-

a. 3

b. 2

c. 0

d. 1

∵ a + b + c = 0

⇒ a + c = -b

⇒ a2 + c2 = b2 -2ac

⇒ a2 + b2 + c2 = 2b2 – 2ac

∴ a2 + b2 + c2/ b2 ac = 2

73. If a4 + a2 b2 + b4 = 8 and a2 + ab + b2 = 4, then the value of ab is-

a. -1

b. 0

c. 2

d. 1

∵ a4 + a2 b2 + b4/ a2 + ab + b2 = 8/4

⇒ (a2 + b2)2 – (ab) 2/ (a2 + b2 + ab) = 2

⇒ a2 – ab + b2 = 2 ….(1)

and a2 + ab + b2 = 4 …..(2)

⇒ 2ab = 2

⇒ ab = 1

74. If a = 25, b = 15, c = -10, then the value of a3 + b3 + c3 – 3abc/ (a-b)2 + (b-c)2 + (c-a)2 is-

a. 30

b. -15

c. -30

d. 15

∵ a3 + b3 + c3 – 3abc

= (25)3 + (15)3 + (-10)3 – 3*25*15*(-10)

=15625 + 3375 – 1000 + 11250 = 29250

and (a - b)2 + (b – c)2 + (c – a)2

= (10)2 + (25)2 + (-35)2

= (10)2 + 625 + 1225

= 1950

∴ Required value = 29250/1950 =15

75. A, B, C are three points on a circle. The tangent at A meets BC produced at T, ∠BTA = 400, ∠CAT = 440. The angle subtended by BC at the centre of the circle is-

a. 840

b. 920

c. 960

d. 1040

Ans : (Picture)

∠ACB = 400 + 440 = 840

∴ ∠ACO = 900 - 440 = 460 = ∠OAC

⇒ ∠OCB = ∠ACB - ∠ACO

= 840 - 460 = 380 = ∠OBC

∴ ∠BOC = 1800 – (∠OCB + ∠OBC)

= 1800 – (380 + 380) = 1040

76. If the length of a chord of a circle at a distance of 12 cm from the Centre is 10 cm, then the diameter of the circle is-

a. 13 cm

b. 15 cm

c. 26 cm

d. 30 cm

Ans : (Picture)

∵ OA = √OM2 + AM2

= √122 + 52 = 13

∴ Diameter of the circle = 2*OA

= 2*13 = 26cm

77. In ∆ ABC, P and Q are the middle points of the sides AB and AC respectively. R is a point on the segment PQ such that PR : RQ = 1 : 2. If PR = 2 cm, then BC =

a. 4 cm

b. 2 cm

c. 12 cm

d. 6 cm

∵ PR/RQ = ½

But, PR = 2cm

RQ = 2*PR

= 4cm

(Picture)

∴ PQ = PR + RQ

= 2 + 4 = 6 cm

∴ BC = 2*PQ = 12CM

78. If tan θ tan 2θ = 1, then the value of sin2 2θ + tan2 is equal to –

a. ¾

b. 10/3

c. 3 ¾

d. 3

∵ tan θ * tan 2θ = 1

⇒ tan θ * 2 tan θ/1 – tan2 θ = 1

⇒ 2 tan2 θ = 1 - tan2 θ

⇒ 3 tan2 θ = 1

⇒ tan θ = 1/√3 = tan 300

∴ θ = 300

∴ sin2 2θ + tan2 2θ = sin2 600 + tan2 600

= ¾ + 3

=3 ¾

79. The value of cot /20 cot 3/20 cot 5/20 cot 7/20 cot 9/20 is-

a. -1

b. ½

c. 0

d. 1

Ans : Given Exp.

= cot /20. cot 3/20. cot 5/20. cot 7/20. cot 9/20

= cot 90. cot 270. cot 450. cot 630. cot 810

= cot 90. cot 270*1*tan 270. tan 90.

= cot 90. cot 270*1*1/ cot 270*1/ cot 90

=1

80. If sin θ+ cos θ = 17/13, 0<θ<900, then the value of sin θ- cos θ is-

a. 5/17

b. 3/19

c. 7/10

d. 7/13

∵ sin θ + cos θ = 17/13

⇒ sin2 θ + cos2 θ + 2sin θ.cos θ = 289/169

∴ 2sin θ.cos θ = 289 /169 -1

= 289 -169/169 = 120/169

∴ sin2 θ + cos2 θ – 2 sin2 θ.cos θ = 1 – 120/169

= 49/169

⇒ (sin θ – cos θ)2 = (7/13)2

∴ sin θ – cos θ = 7/13

S.S.C. Combined Graduate Level Exam

(Tier-II) 2011

Solved Paper – I

Arithmetical Ability

1. The product of 2 numbers is 1575 and their quotient is 9/7. Then the sum of the numbers is –

a. 74

b. 78

c. 80

d. 90

Ans : Let the numbers be x and y .

∴ xy = 1575

And x/y = 9/7

∴ xy/x/y = 1575/9/7

∴ y2 = 1225

∴ y = 35 and x = 45

∴ The sum of the numbers = 45+35

= 80

2. The value of (81)3.6 * (9)2.7/ (81)4.2 * (3) is __

a. 3

b. 6

c. 9

d. 8.2

Ans : (81)3.6 * (9)2.7/(81)4.2 * (3) = (3)14.4 * (3)5.4/(3)16.8 * (3)

= 314.4+5.4-16.8-1

= 32

= 9

3. √6+√6+√6+… is equal to –

a. 2

b. 5

c. 4

d. 3

Ans : Let √6+√6+√6+…. be x.

∴ x =√6+x

∴ x2 = 6+x

∴ x2 + x – 6 = 0

∴ (x-3) (x+2) = 0

∴ x = 3

4. The sum of the squares of two natural consecutive odd numbers is 394. The sum of the numbers is –

a. 24

b. 32

c. 40

d. 28

Ans : Let the consecutive odd numbers be x and (x+2)

∵ x2 + (x + 2)2 = 394

∴ x2 + x2 + 4x + 4 = 394

∴ 2 x2 +4x – 390 = 0

∴ x2 + 2x – 195 = 0

∴ (x +15) (x-13) = 0

∴ x = 13

∴ Required sum = 13 +15 =28

5. When (6767 +67) is divided by 68, the remainder is-

a. 1

b. 63

c. 66

d. 67

Ans: (6767+ 67) = 67(6766 + 166)

As 66 is an even number

∴ 6766 is an even number

∴ (6766 + 1) is perfectly divisible by (67 + 1)

i.e. 68

∴ The remainder = 67

6. In a division sum, the divisor is 4 times the quotient and twice the remainder. If a and b are respectively the divisor and the dividend then-

a. 4b-a2 /a =3

b. 4b-2a /a2 =2

c. (a+1)2 = 4b

d. A(a+2)/b = 4

Ans : As divisor is a, and dividend is b.

∴ Quotient = a/4

And Remainder = a/2

∴ b = a * a/4 + a/2

∴ 4b = a2 + 2a

∴ a(a+2)/b = 4

7. If 738 A6A is divisible by 11, then the value of A is-

a. 6

b. 3

c. 9

d. 1

Ans : As 738A6 A is divisible by 11.

∴ A + A + 3 = 6 + 8 + 7

∴ A = 9

8. The east positive integer that should be subtracted from 3011 * 3012 so that the difference is perfect square is-

a. 3009

b. 3010

c. 3011

d. 3012

Ans : ∵ 3011 * 3012 = 3011 (3011 + 1)

= (3011)2 + 3011

∴ Required least number = 3011

9. P, Q, R are employed to do a work for Rs. 5750. P and Q together finished 19/23 of work and Q and R together finished 8/23 of work. Wage of Q, in rupees, is-

a. 2850

b. 3750

c. 2750

d. 1000

Ans : Work done by Q = 19/23 + 8/23 – 1

4/23

∴ Wage of Q = 4/23 * 5750

= Rs. 1000

10. A can do a piece of work in 24 day, B in 32 days and C in 64 days. All begin to do it together, but A leaves after 6 days and B leaves 6 days before the completion of the work. How many days did the work last?

a. 15

b. 20

c. 18

d. 30

Ans : Work done by A = 6/24 = ¼

Work done by B = (x-6)/32

(where x is no. of days in which work is competed)

∵ ¼ + x – 6/32 + x/64 = 1

∴ 16 + 24 – 12 + x /64 = 1

∴ 3x + 4 = 64

∴ x = 60/3 = 20 days

11. The square root of (0.75)3 /1-0.75 + [0.75 + 90.75)2 +1] is-

a. 1

b. 2

c. 3

d. 4

Ans : The square root of

[(0.75)3/1 – 0.75 + {0.75 + (0.75)2 + 1}]

= √1.6875 + 2.3125

=√4

= 2

12. Given that √4096 = 64, the value of √4096 + √40.96 +√0.004096 is-

a. 70.4

b. 70.464

c. 71.104

d. 71.4

Ans : Given Exp. = √4096 + √40.96 +√0.004096

= 64 + 6.4 + 0.064

= 70.464

13. By selling an article at 3/4th of the marked price, there is a gain of 25%. The ratio of the marked price and the cost price is-

a. 5 : 3

b. 3 : 5

c. 3 : 4

d. 4 : 3

Ans : Let of M.P. be Rs. x.

∴ S.P. = Rs.3x/4

and C.P. = 3x/4 * 100/125

= Rs.3x/5

∴ required ratio = x: 3x/5

= 5:3

14. A and B earn in the ratio 2:1. They spend in the ratio 5:3 and save in the ratio 4:1. If the total monthly savings of both A and B are Rs.5,000, the monthly income of B is-

a. Rs.7,000

b. Rs.14,000

c. Rs.5,000

d. Rs.10,000

Ans : Let the monthly income of B be Rs. x.

∴ Monthly income of A = Rs. 2x and

Saving of A =5000 * 4/(4 + 1)

= Rs. 4000

Saving of B = Rs.1000

∵ 2x – 4000/x-1000 = 5/3

⇒ 6x – 12000 = 5x – 5000

∴ x = Rs.7000

15. The ratio of the sum of two numbers and their difference is 5:1. The ratio of the greater number to the smaller number is-

a. 2 : 3

b. 3 : 2

c. 5 : 1

d. 1 : 5

Ans : Let the numbers be x and y.

∵ x + y/ x – y = 5/1

⇒ 5x – 5y = x + y

⇒ 4x = 6y

⇒ x/y = 6/4

x : y = 3 : 2

16. A cistern has 3 pipes A, B and C. A and B can fill it in 3 and 4 hours respectively, and C can empty it in 1 hour. If the pipes are opened at 3 p.m., 4 p.m. and 5 p.m. respectively on the same day, the cistern will be empty at-

a. 7.12 p.m.

b. 7.15 p.m.

c. 7.10 p.m.

d. 7.18 p.m.

Ans : Let the cistern be emptied at x p.m.

∵ x -3/3 + x – 4/4 = x - 5/1

⇒ 4x – 12 + 3x – 12/12 = x – 5/1

⇒ 7x – 24 = 12x – 60

⇒ 5x = 36

∴ x = 36/5

= 7 hr. + 12 min.

= 7.12 p.m.

17. If A works alone, he would take 4 days more to complete the job than if both A and B worked together. If B worked alone, he would take 16 days more to complete the job than if A and B work together. How many days would they take to complete the work if both of them worked together?

a. 10 days

b. 12 days

c. 6 days

d. 8 days

Let x days are taken when they work together.

∴ Time taken by A to complete the work = (x + 4) days

And Time taken by B to complete the work = (x + 16) days

∵ 1/x = 1/x + 4 + 1/x + 16

∴ 1/x = x + 16 + x + 4/(x + 4) (x +16)

∴ 2x2 + 20x = x2 + 20x +64

∴ x2 = 64 (8)2

∴ x = 8 days

18. 250 men can finish a work in 20 days working 5 hours a day. To finish the work within 10 days working 8 hours a day, the minimum number of men required is-

a. 310

b. 300

c. 313

d. 312

Ans : Required no. of men = 250 * 5 * 20/8 * 10

=312.5

313

19. 2 men and 5 women can do a work in 12 days. 5 men 2 women can do that work in 9 days. Only 3 women can finish the same work in-

a. 36 days

b. 21 days

c. 30 days

d. 42 days

∵ (2m + 5w) * 12 = (5m + 2w) * 9

⇒ 24m + 60w = 45m + 18w

⇒ 21m = 42w

⇒ 1m = 2w

∴ 2m + 5w = 9w

∴ required no. of days = 9 * 12/3

= 36

20. While selling, a businessman allows 40% discount on the marked price and there is a loss of 30%. If it is sold at the marked price, profit per cent will be –

a. 10%

b. 20%

c. 16 2/3%

d. 16 1/3%

Ans : Let the M.P. be Rs.100.

∴ S.P = (100-40) = Rs.60

and C. P. = 60 * 100/(100-30) = Rs.600/7

∴ Reqd. % profit = 100-600/7 /600/7 * 100%

= (700 – 600)/7 * 600 * 100 * 7%

= 16 2/3%

21. Successive discount of 10% , 20% and 50% will be equivalent to a single discount of-

a. 36%

b. 64%

c. 80%

d. 56%

Ans : Equivalent to a single discount

= [100 - (100 – 10) (100 – 20) (100 – 50)/100*100]%

= [100 - 90*80*50/10000]%

= [100-36]%

= 64%

22. A retailer offers the following discount schemes for buyers on an article-

I. Two successive discounts of 10%

II. A discount of 12% followed by a discount of 8%.

III. Successive discounts of 15% and 5%

IV. A discount of 20%

The selling price will be minimum under the scheme-

a. I

b. II

c. III

d. IV

Ans : From (i) single discount = [10 + 10 – 10*10/100] % = 19%

From (ii) single discount = [12 + 8 -12*8/100] % = 19.04%

From (iii) single discount = [15 + 5 – 15*5/100] % = 19.25%

From (iv) single discount = 20%

∴ The S.P. will be minimum under the scheme IV.

23. Of three numbers, the second is thrice the first and the third number is three-fourth of the first. If the average of the three numbers is 114, the largest number is –

a. 72

b. 216

c. 354

d. 726

Ans : Let the first number be x.

∴ Second number = 3x and Third number = 3x/4

∵ x + 3x + 3x/4 = 3*114

⇒ 19x/4 = 342

∴ x = 342*4/19 = 72

∴ The largest number = 3*72

= 216

24. A car covers 1/5 of the distance from A to B at the speed of 8 km/hour, 1/10 of the distance at 25 km per hour and the remaining at the speed of 20 km per hour. Find the average speed of the whole journey-

a. 12.625 km/hr

b. 13.625 km/hr

c. 14.625 km/hr

d. 15.625 km/hr

Ans : If the whole journey be x km. The total time taken

= (x/5/8 + x/10/25 + 7x/10/20) hrs

= (x/40 + x/250 + 7x/200) hrs

= 25x + 4x + 35x/1000

= 64x/1000 hrs

∴ Average speed = x/64x/1000

= 15.625 km/hr

25. The average of 3 numbers is 154. The first number is twice the second and the second number is twice the third. The first number is-

a. 264

b. 132

c. 88

d. 66

Ans : Let the first number be x.

∴ Second number = x/2 and Third number = x/4

∵ x + x/2 + x/4 = 3 x 154

⇒ 7x/4 = 462

∴ x = 462*4/7

=264

26. The average salary of all the staff in an office of a corporate house is Rs. 5,000. The average salary of the officers is Rs. 14,000 and that of the rest is Rs. 4,000. If the total number of staff is 500, the number of officers is –

a. 10

b. 15

c. 25

d. 50

Ans : Let the number of officers be x.

∵ 5000*500 = 14000x + 4000(500-x)

∴ 2500000 =14000x + 2000000 – 4000x

∴ x = 500000/10000

= 50

27. The average marks of 40 students in an English exam are 72. Later it is found that three marks 64, 62 and 84 were wrongly entered as 60, 65 and 73. The average after mistakes were rectified is-

a. 70

b. 72

c. 71.9

d. 72.1

Ans : Correct average

= 40*72 + (64 + 62 +84) – 68 – 65 – 73/40

= 2880 + 210 – 206/40 = 2884/40

= 72.1

28. A and B are two alloys of gold and copper prepared by mixing metals in the ratio 7:2 and 7:11 respectively. If equal quantities of the alloys are melted to form a third alloy C, the ratio of gold and copper in C will be-

a. 5 : 7

b. 5 : 9

c. 7 : 5

d. 9 : 5

Ans : Quantity of gold in A = 7/9* wt. of A

Quantity of gold in B = 7/18* wt. of B

If 1 kg of each A and B are mixed to form third alloys C.

Then quantity of gold in 2 kg C = 7/9 + 7/18

= 7/6 kg

And Quantity of copper in 2 kg C = 2 – 7/6

= 5/6 kg

∴ Required ratio = 7/6 : 5/6 = 7 : 5

29. In a laboratory, two bottles contain mixture of acid and water in the ratio 2 : 5 in the first bottle and 7 : 3 in the second. The ratio in which the contents of these two bottles be mixed such that the new mixture has acid and water in the ratio 2 : 3 is-

a. 4 : 15

b. 9 : 8

c. 21 : 8

d. 1 : 2

Ans : Quantity of acid in first bottle = 2/7 x mix.

and Quantity of acid in second bottle = 7/10 x mix.

If x and 1 volumes are taken from I and II bottle respectively to form new mixture.

Then, (2/7 x + 7/10 * 1)/(5x/7 +3/10 * 1) = 2/3

⇒ 6x/7 21/10 = 10x/7 +6/10

⇒ 4x/7 = 15/10

∴ x = 15/10*7/4 = 21/8

∴ Required ratio = x : 1

= 21 : 8

30. A mixture contains 80% acid and rest water. Part of the mixture that should be removed and replaced by same amount of water to make the ratio of acid and water 4 : 3 is-

a. 1/3 rd

b. 3/7 th

c. 2/3 rd

d. 2/7 th

Ans : Let the initial wt. of mixture be 1 kg and x kg of mixture is taken out and replaced by same amount of water.

∵ Amt. of acid/Amt. of water = 0.8 – 0.8x/ (0.2 – 0.2x + x) = 4/3

⇒ 2.4 – 2.4x = 0.8 + 3.2x

⇒ 5.6x = 1.6

∴ x = 1.6/5.6 = 2/7th part

31. An employer reduces the number of his employees in the ratio 9 : 8 and increases their wages in the ratio 14 : 15. If the original wage bill was Rs. 189,900, find the ratio in which the wage bill is decreased-

a. 20 : 21

b. 21 : 20

c. 20 : 19

d. 19 : 21

Ans : Let the initial number of employees be 9x and the employer gives Rs. 14y as wage to each.

∵ 9x * 14y =18900

∴ xy = 150 and The later bill = 8x*15y = 120xy

= 120*150 = 18000

∴ Required ratio = 18000 : 18900

= 20 : 21

32. The batting average for 40 innings of a cricketer is 50 runs. His highest score exceeds his lowest score by 172 runs. If these two innings are excluded, the average of the remaining 38 innings is 48 runs. The highest score of the player is-

a. 165

b. 170

c. 172

d. 174

Ans : Let the max. number of runs be x.

∴ The lowest score = (x-172)

∵ 40*50 = 38*48 + x + (x-172)

⇒ 2000 = 1824 + 2x – 172

∴ x= 174

33. Walking at 3 km per hour, Pintu reaches his school 5 minutes late. If he walks at 4 km per hour he will be 5 minutes early. The distance of Pintu’s school from his house is-

a. 1 ½ Km

b. 2 Km

c. 2 ½ Km

d. 5 Km

Ans : Distance of the school from the house

= 4*3/(4 – 3) * 5 +5/60 km

= 12*1/6

=2 km

34. Nitin bought some oranges at Rs. 40 a dozen and an equal number at Rs.30 a dozen. He sold them at Rs. 45 a dozen and made a profit of Rs.480. The number of oranges, he bought, was-

a. 48 dozen

b. 60 dozen

c. 72 dozen

d. 84 dozen

Ans : Let the number of oranges bought be x.

∴ 45x/12 – 70x/24 = 480

⇒ (45-35)/12 = 480

⇒ x = 480*12/10 = 576

= 48*12 = 48 dozen

35. A man buys two chairs for a total cost of Rs.900. By selling one for 4/5 of its cost and the other for 5/4 of its cost, he makes a profit of Rs.90 on the whole transaction. The cost of the lower priced chair is-

a. Rs.360

b. Rs.400

c. Rs.420

d. Rs.300

Ans : Let the cost price of 1 chair be Rs. x.

∴ C.P. of other chair = Rs. (900-x)

∵ 4/5 x + 5/4 (900-x) = 900 + 90

⇒ 4/5 x +1125 -5x/4 = 990

∴ 9x/20 = 135

∴ x = 135*20/9

= Rs. 300

∴ C.P. of the lower priced chair is Rs. 300.

36. By selling 100 oranges, a vendor gains the selling price of 20 oranges. His gain per cent is-

a. 20

b. 25

c. 30

d. 32

Ans : Let the S.P. of 100 Oranges be Rs. x.

∴ S.P. of 20 oranges = x/100 *20 = Rs. x/5

∴C.P. of 100 oranges = x - x/5

= Rs. 4x/5

∴ Reqd. Profit % = x/5 * 100*5/4x %

= 25%

37. 60% of the cost price of an article is equal to 50% of its selling price. Then the percentage of profit or loss on the cost price is-

a. 20% loss

b. 16 2/3% profit

c. 20% profit

d. 10% loss

Ans : Let the cost price be Rs. 100.

∵ S.P * 50/100 = 100*16/100

∴ S.P = 60*100/50

= Rs. 120

∴ Reqd. % profit = (120 – 100)% = 20%

38. Maninder bought two horses at Rs.40,000 each. He sold one horse at 15% gain, but had to sell the second horse at a loss. If he hand suffered a loss of Rs.3,600 on the whole transaction, then the selling price of the second horse is-

a. Rs.30,000

b. Rs.30,200

c. Rs.30,300

d. Rs.30,400

Ans: C.P. of two horses = 2*40000 = Rs. 80000

and S.P. of two horse = 80000 – 3600 = Rs. 76400

∴ S.P. of the other horse = 76400 – 46000 = 30400

39. A fruit-seller buys x guavas for Rs.y and sells y guavas for Rs. x. If x>y, then he made-

a. x2 – y2 / xy % loss

b. x2 – y2 / xy % gain

c. x2 – y2 / y2 % loss

d. x2 – y2 / y2 * 100% gain

Ans : C.P. of 1 guava = Rs. y/x [x>y]

and S.P. of 1 guava = Rs. x/y

∴ Reqd. Gain% = x/y – y/x/ y/x * 100%

= x2 - y2 /y2 *100%

40. A jar contain 10 red marbles and 30 green ones. How many red marbles must be added to the jar so that 60% of the marbles will be red?

a. 25

b. 30

c. 35

d. 40

Ans : Let after adding x red marbles, the red marbles with be 60% of the total.

∵ (10+x)/ (10 + x) + 30 = 60/100

⇒ 10 + x/40 + x = 3/5

⇒ 50 + 5x = 120 + 3x

x = 70/2 = 35

41. If a number multiplied by 25% of itself gives a number which is 200% more than the number, then the number is –

a. 12

b. 16

c. 35

d. 24

Ans : Let the number be x.

∵ x*25x/100 = x + 200x/100

⇒ x2/4 = 3x

⇒ x2 - 12x = 0

⇒ x – 12 = 0

∴ x = 12

42. The value of an article depreciates every year at the rate of 10% of its value. If the present value of the article is Rs.729, then its worth 3 years ago was-

a. Rs.1250

b. Rs.1000

c. Rs.1125

d. Rs.1200

Ans : Let the worth 3 years ago be Rs. x.

∵ 729 = x (1 – 10/100)3

⇒ 729 = x*9*9*9/10*10*10

∴ x = Rs. 1000

43. The price of onions has been increased by 50%. In order to keep the expenditure on onions the same the percentage of reduction in consumption has to be-

a. 50%

b. 33 1/3%

c. 33%

d. 30%

Ans : Reqd. Percentage of reduction

= 50*100/(100 + 50) %

= 5000/150 %

= 33 1/3%

44. A took two loans altogether of Rs.1200 from B and C. B claimed 14% simple interest per annum, while C claimed 15% per annum. The total interest paid by A in one year was Rs.172. Then, A borrowed-

a. Rs.800 from C

b. Rs.625 from C

c. Rs.400 from B

d. Rs.800 from B

Ans : If A borrowed Rs. x from B. and A borrowed Rs. Rs. (1200 – x) from C.

∵ (1200 – x)*15*1/100 + x*14*1/100

⇒ 18000 – 15x + 14x = 172*100

x = Rs. 800

45. If a regular polygon has each of its angles equal to 3/5 times of two right angles, then the number of side is-

a. 3

b. 5

c. 6

d. 8

Ans : If the number of sides a regular polygon be n.

Then (2n-4)/n = 2*3/5

⇒ (2n – 4)*5 = 6n

∴ n = 5

46. A square is of area 200 sq. m. A new square is formed in such a way that the length of its diagonal is √2 times of the diagonal of the given square. The the area of the new square formed is-

a. 200√2 sq.m

b. 400√2 sq.m

c. 400 sq.m

d. 800 sq.m

Ans : Length of the diagonal of Ist square

= √2*200

= 20 m

∴ Length of the diagonal of new square = 20√2m

∴ Area of the new square = ½*(20.√2)2 = 400 sq. m

47. The heights of a cone, cylinder and hemisphere are equal. If their radii are in the ratio 2 : 3 : 1, then the ratio of their volumes is-

a. 2 : 9 : 2

b. 4 : 9 : 1

c. 4 : 27 : 2

d. 2 : 3 : 1

Ans : Ratio of their volumes [cone : cylinder : hemisphere]

= 1/3(2)2h : (3)2*h : 2/3(1)2.h

= 4/3 : 9 : 2/3 = 4 : 27 : 2

48. A motor-boat can travel at 10 km/hr in still water. It travelled 91 km downstream in a river and then returned to the same place, taking altogether 2 hours. Find the rate of flow of river-

a. 3 km/hr

b. 4 km/hr

c. 2 km/hr

d. 5 km/hr

Ans : Let the rate of flow of river be x km/hr.

∵ 91/(10 + x) + 91/(10 – x) = 20

⇒ 91(10 – x + 10 + x)(10 + x) (10 – x) = 20

⇒ 91*20 = 20(100 – x2)

⇒ x2 = 9 = (3)2

∴ x = 3 km/hr

49. A man driving at 3/4th of his original speed reaches his destination 20 minutes later than the usual time. Then the usual time is-

a. 45 minutes

b. 60 minutes

c. 75 minutes

d. 120 minutes

Ans : Let the original speed be x km/hr and the usual time be y hours.

∵ x * y = ¾ x(y+1/3)

∴ 4y = 3y + 1

∴ y = 1 hr = 60 minutes

50. A motor-boat, travelling at the same speed, can cover 25 km upstream and 39 km downstream in 8 hours. At the same speed, it can travel 35 km upstream and 52 km downstream in 11 hours. The speed of the stream is –

a. 2 km/hr

b. 3 km/hr

c. 4 km/hr

d. 5 km/hr

Ans : Let the speeds of motor boat and the stream be x and y km/hr respectively.

∵ 39/x + y + 25/x – y = 8 …(1)

and 52/x + y + 35/x – y = 11 …(2)

Solving equations (1) and (2), we get-

∴ 100 - 105/x – y = 32 – 33

∴ x – y = 5

and x + y = 13

∴ y = 4 km/hr

51. If a sum of money placed at compound interest, compounded annually, doubles itself in 5 years, then the same amount of money will be 8 times of itself in-

a. 25 years

b. 20 years

c. 15 years

d. 10 years

Ans : Required time = 5log 8/log 2

= 5*3log 2/log 2

= 15 years

52. A person has left an amount of Rs.1,20,000 to be divided between his 2 son aged 14 years and 12 years such that they get equal amounts when each attains 18 years of age. If the amount gets a simple interest of 5% per annum, the younger son’s share at present is-

a. Rs.48,800

b. Rs.57,600

c. Rs.62,400

d. Rs.84,400

Ans : Let the present share of the younger son be Rs. x.

∴ The share of the elder son = Rs. (120000 – x)

∵ x + x*6*5/100

= (120000 – x) + (120000 – x)*4*5/100

⇒ 130x/100 = (120000 –x)*120/100

⇒ 13x = 1440000 – 12x

⇒ 25x = 1440000

∴ x = Rs. 57600

53. If the simple interest on Rs.x at a rate of a% for m years is same as that on Rs. y at a rate of a2% for m2 years, then x : y is equal to-

a. m : a

b. am : 1

c. 1/m : 1/a

d. 1/am : 1

∵ x*a*m/100 = y*a2*m2/100

⇒ x/y = am/1

∴ x : y = am : 1

54. Base of a right prism is an equilateral triangle of side 6 cm. If the volume of the prism is 108√3 cc, its height is-

a. 9 cm

b. 10 cm

c. 11 cm

d. 12 cm

Ans : Height of the prism = 108√3*4/√3*(6)2 = 12cm

55. If a + 1/a + 2 = 0, then the value of (a37 – 1/a100) is-

a. 0

b. -2

c. 1

d. 2

Ans : ∵ a + 1/a + 2 = 0

⇒ a2 + 1 +2a = 0

⇒ (a + 1)2 = 0

⇒ a +1 = 0

∴ a = -1

∴ a37 -1/a100 = (-1) – (1) = -2

56. The value of k for which the graphs of (k-1) x+y-2 = 0 and (2-k) x -3y + 1 = 0 are parallel is-

a. ½

b. -1/2

c. 2

d. -2

Ans : (k – 1) x + y – 2 = 0

∴ y = (1 – k) x + 2 ….(1)

and (2 –k) x – 3y – 1 = 0

3y = (2 – k) x +1

Y = 2 – k/3 x + 1/3 ….(2)

∵ m1 = m2

⇒ 1 – k = 2 – k/3

⇒ 3 – 3k = 2 - k

∴ k = 1/2

57. If a2 + b2 + c2 = 2 (a-b-c) – 3, then the value of (a – b + c) is-

a. -1

b. 3

c. 1

d. -2

Ans : a2 + b2 + c2 = 2(a – b – c) – 3

⇒ a2 - 2a + 1 + b2 + 2b + 1 + c2 + 2c + 1 = 0

⇒ (a – 1)2 + (b + 1)2 + (c + 1)2 = 0

⇒ a – 1 = 0, b + 1 = 0, c + 1 = 0

⇒ a = 1, b = -1, c = -1

∴ a + b – c = 1 – 1 + 1 =1

58. If x2 + 3x + 1 = 0, then the value of x3 + 1/x3 is-

a. -18

b. 18

c. 36

d. -36

Ans : ∵ x2 + 3x + 1 = 0

⇒ x + 3 + 1/x = 0

⇒ x + 1/x = -3

⇒ (x + 1/x)3 = (-3)3

⇒ x3 + 1/x3 + 3 (-3) = -27

∴ x3 + 1/ x3 = -18

59. If xa, xb, xc = 1, then the value of a3 + b3 + c3 is –

a. 9

b. abc

c. a + b + c

d. 3abc

Ans : ∵ xa. xb. xc = 1

⇒ xa + b + c = x0

⇒ a +b + c =0

∴ a3 + b3 + c3 = 3abc

60. Base of a right pyramid is a square, length of diagonal of the base is 24√2 m. If the volume of the pyramid is 1728 cu.m, its height is-

a. 7 m

b. 8 m

c. 9 m

d. 10 m

Ans : Area of the base of the pyramid

= ½ (24√2)2 = 576m2

∴ Height of pyramid + 1728*3/576 = 9m

61. The height of a right circular cone and the radius of its circular base are respectively 9 cm and 3 cm. The cone is cut by a plane parallel to its base so as to divide it into two parts. The volume of the frustum (i.e., the lower part) of the cone is 44 cubic cm. The radius of the upper circular surface of the frustum (taking = 22-7) is-

a. 3√12 cm

b. 3√13 cm

c. 3√6 cm

d. 3√20 cm

Ans : Let the radius of the upper circular part of the frustum be r cm.

(Picture)

Then r/3 = x/9 = AC/AD

∴ x = 3r (where AC = x cm)

⇒ /3h(r12 + r1r2+r22 = 44

⇒ /3(9 – x)*(9 + 3r + r2) = 44

⇒ (9 – x)*(9 + 3r + r2) = 44*3*7/22 = 42

⇒ 81 + 27r + 9r2 – 9x – 3rx - r2x = 42 On putting x = 3r,

⇒ 81 + 27r + 9r2 - 9r2 - 3r3 – 27r = 42

⇒ 3r3 = 39

∴ r = 3√13 cm

62. The ratio of radii of two right circular cylinder is 2 : 3 and their heights are in the ratio 5 : 4. The ratio of their curved surface area is-

a. 5 : 6

b. 3 : 4

c. 4 : 5

d. 2 : 3

Ans : Required ratio = 2*2r*5h/2*3r*4h = 5 : 6

63. A solid cylinder has total surface area of 462 sq.cm. Curved surface area is 1/3rd of its total surface area. The volume of the cylinder is-

a. 530 cm3

b. 536 cm3

c. 539 cm3

d. 545 cm3

Ans : ∵ 2r(r + h) = 462

and 2rh = 1/3*462 = 154

∴ r + h/h = 462/154 =3

∴ r + h = 3h

∴ r = 2h

∴ 2*2h2 = 154

∴ h2 = 154*7/22*4 = 49/4

= (7/2)2

∴ h = 7/2 cm

and r = 7 cm

∴ Volume of the cylinder

= 22/7*49*7/2

= 539 cm3

64. A cylinder and a cone have equal radii of their bases and equal heights. If their curved surface areas are in the ratio 8 : 5, the ratio of their radius and height is-

a. 1 : 2

b. 1 : 3

c. 2 : 3

d. 3 : 4

Ans : Let the radius and height of each are r and h respectively.

∵ 2rh/r√h2 + r2 = 8/5

∴ 10h = 8√r2 + h2

⇒ 100h2 =64r2 + 64h2

∴ h2 = 64 r2/36 = (4/3 r)2

⇒ h = 4/3r

∴ r : h = 3 : 4

65. A solid is hemispherical at the bottom and conical above. If the surface areas of the two parts are equal, then the ratio of radius and height of its conical part is-

a. 1 : 3

b. 1 :1

c. √3 : 1

d. 1 : √3

Ans : (Picture)

∵ rl = 2r2

⇒ l = 2r

⇒ √h2 + r2 = 2r

⇒ h2 = 3r2

∴ r : h = r/h

= 1 : √3

66. If O is the circumcentre of ∆ ABC and ∠OBC = 350, then the ∠BAC is equal to-

a. 550

b. 1100

c. 700

d. 350

Ans : ∴ ∠BOC = 1800 – (350 + 350) = 1100

(Picture)

∴ ∠BAC = 1/2 * 1100 = 550

67. If I is the incentre of ∆ ABC and ∠ BIC = 1350, then ∆ ABC is-

a. Acute angled

b. Equilateral

c. Right angled

d. Obtuse angled

Ans : (Picture)

∠BIC = 1350

⇒ B/2 + C/2 = 1800 – 1350 = 450

⇒ ∠B + ∠C = 900

∴ ∠A = 1800 – (∠B + ∠C) = 900

i.e., ∆ ABC is a right angled.

68. If sin2 (∝ + β/2) is-

a. 1

b. -1

c. 0

d. 0.5

∵ sin2 ∝ + sin2 β = 2

⇒ 1 – cos2 ∝ 1 - cos2 β = 2

⇒ cos2 ∝ cos2 β = 0

⇒ cos ∝ = 0

and cos β = 0

⇒ ∝ = /2 and β = /2

∴ cos (∝ + β/2) = cos [/2 + /2 / 2]

= cos /2 = 0

69. The length of a shadow of a vertical tower is 1/√3 times its height. The angle of elevation of the Sun is-

a. 300

b. 450

c. 600

d. 900

Ans : (Picture)

∵ tan θ = h/1√3 h = √3 = tan 600

70. The graphs of x +2y =3 and 3x-2y = 1 meet the Y-axis at two points having distance-

a. 8/3 units

b. 4/3 units

c. 1 unit

d. 2 units

Ans : When the graphs meet the Y-axis at two points.

Then, [x + 2y = 3] at x = 0 ⇒ [0, y1 =3/2]

[3x - 2y = 1] at x = 0

and i.e., [0, y2 = -1/2]

Required distance = (y1 – y2)

= 3/2 – (- ½) = 2 units

71. If x+1/16x = 1, then the value of 64x3 + 1/64x3 is-

a. 4

b. 52

c. 64

d. 76

∵ x + 1/16x = 1

⇒ 16x2 – 16x + 1 =0

⇒ 16x2 – 16x + 4 = 3

⇒ (4x – 2)2 = 3

⇒ 4x = 2 + √3

⇒ 64x3 = (2±√3)3

=8 + 3√3 + 6√3 (2 + √3)

= 26 + 15√3

∴ 64x3 + 1/64x3 = (26 + 15√3) + 1/ (26 + 15√3)

= (26 + 15√3) + 26 - 15√3/676 -675

=52

72. If a, b, c, are three non-zero real numbers such that a + b + c = 0, and b2 ≠ ca, then the value of a2 + b2 + c2/ b2 –ca is-

a. 3

b. 2

c. 0

d. 1

∵ a + b + c = 0

⇒ a + c = -b

⇒ a2 + c2 = b2 -2ac

⇒ a2 + b2 + c2 = 2b2 – 2ac

∴ a2 + b2 + c2/ b2 ac = 2

73. If a4 + a2 b2 + b4 = 8 and a2 + ab + b2 = 4, then the value of ab is-

a. -1

b. 0

c. 2

d. 1

∵ a4 + a2 b2 + b4/ a2 + ab + b2 = 8/4

⇒ (a2 + b2)2 – (ab) 2/ (a2 + b2 + ab) = 2

⇒ a2 – ab + b2 = 2 ….(1)

and a2 + ab + b2 = 4 …..(2)

⇒ 2ab = 2

⇒ ab = 1

74. If a = 25, b = 15, c = -10, then the value of a3 + b3 + c3 – 3abc/ (a-b)2 + (b-c)2 + (c-a)2 is-

a. 30

b. -15

c. -30

d. 15

∵ a3 + b3 + c3 – 3abc

= (25)3 + (15)3 + (-10)3 – 3*25*15*(-10)

=15625 + 3375 – 1000 + 11250 = 29250

and (a - b)2 + (b – c)2 + (c – a)2

= (10)2 + (25)2 + (-35)2

= (10)2 + 625 + 1225

= 1950

∴ Required value = 29250/1950 =15

75. A, B, C are three points on a circle. The tangent at A meets BC produced at T, ∠BTA = 400, ∠CAT = 440. The angle subtended by BC at the centre of the circle is-

a. 840

b. 920

c. 960

d. 1040

Ans : (Picture)

∠ACB = 400 + 440 = 840

∴ ∠ACO = 900 - 440 = 460 = ∠OAC

⇒ ∠OCB = ∠ACB - ∠ACO

= 840 - 460 = 380 = ∠OBC

∴ ∠BOC = 1800 – (∠OCB + ∠OBC)

= 1800 – (380 + 380) = 1040

76. If the length of a chord of a circle at a distance of 12 cm from the Centre is 10 cm, then the diameter of the circle is-

a. 13 cm

b. 15 cm

c. 26 cm

d. 30 cm

Ans : (Picture)

∵ OA = √OM2 + AM2

= √122 + 52 = 13

∴ Diameter of the circle = 2*OA

= 2*13 = 26cm

77. In ∆ ABC, P and Q are the middle points of the sides AB and AC respectively. R is a point on the segment PQ such that PR : RQ = 1 : 2. If PR = 2 cm, then BC =

a. 4 cm

b. 2 cm

c. 12 cm

d. 6 cm

∵ PR/RQ = ½

But, PR = 2cm

RQ = 2*PR

= 4cm

(Picture)

∴ PQ = PR + RQ

= 2 + 4 = 6 cm

∴ BC = 2*PQ = 12CM

78. If tan θ tan 2θ = 1, then the value of sin2 2θ + tan2 is equal to –

a. ¾

b. 10/3

c. 3 ¾

d. 3

∵ tan θ * tan 2θ = 1

⇒ tan θ * 2 tan θ/1 – tan2 θ = 1

⇒ 2 tan2 θ = 1 - tan2 θ

⇒ 3 tan2 θ = 1

⇒ tan θ = 1/√3 = tan 300

∴ θ = 300

∴ sin2 2θ + tan2 2θ = sin2 600 + tan2 600

= ¾ + 3

=3 ¾

79. The value of cot /20 cot 3/20 cot 5/20 cot 7/20 cot 9/20 is-

a. -1

b. ½

c. 0

d. 1

Ans : Given Exp.

= cot /20. cot 3/20. cot 5/20. cot 7/20. cot 9/20

= cot 90. cot 270. cot 450. cot 630. cot 810

= cot 90. cot 270*1*tan 270. tan 90.

= cot 90. cot 270*1*1/ cot 270*1/ cot 90

=1

80. If sin θ+ cos θ = 17/13, 0<θ<900, then the value of sin θ- cos θ is-

a. 5/17

b. 3/19

c. 7/10

d. 7/13

∵ sin θ + cos θ = 17/13

⇒ sin2 θ + cos2 θ + 2sin θ.cos θ = 289/169

∴ 2sin θ.cos θ = 289 /169 -1

= 289 -169/169 = 120/169

∴ sin2 θ + cos2 θ – 2 sin2 θ.cos θ = 1 – 120/169

= 49/169

⇒ (sin θ – cos θ)2 = (7/13)2

∴ sin θ – cos θ = 7/13

S.S.C. Combined Graduate Level Exam

(Tier-II) 2011

Solved Paper – I

Arithmetical Ability

1. The product of 2 numbers is 1575 and their quotient is 9/7. Then the sum of the numbers is –

a. 74

b. 78

c. 80

d. 90

Ans : Let the numbers be x and y .

∴ xy = 1575

And x/y = 9/7

∴ xy/x/y = 1575/9/7

∴ y2 = 1225

∴ y = 35 and x = 45

∴ The sum of the numbers = 45+35

= 80

2. The value of (81)3.6 * (9)2.7/ (81)4.2 * (3) is __

a. 3

b. 6

c. 9

d. 8.2

Ans : (81)3.6 * (9)2.7/(81)4.2 * (3) = (3)14.4 * (3)5.4/(3)16.8 * (3)

= 314.4+5.4-16.8-1

= 32

= 9

3. √6+√6+√6+… is equal to –

a. 2

b. 5

c. 4

d. 3

Ans : Let √6+√6+√6+…. be x.

∴ x =√6+x

∴ x2 = 6+x

∴ x2 + x – 6 = 0

∴ (x-3) (x+2) = 0

∴ x = 3

4. The sum of the squares of two natural consecutive odd numbers is 394. The sum of the numbers is –

a. 24

b. 32

c. 40

d. 28

Ans : Let the consecutive odd numbers be x and (x+2)

∵ x2 + (x + 2)2 = 394

∴ x2 + x2 + 4x + 4 = 394

∴ 2 x2 +4x – 390 = 0

∴ x2 + 2x – 195 = 0

∴ (x +15) (x-13) = 0

∴ x = 13

∴ Required sum = 13 +15 =28

5. When (6767 +67) is divided by 68, the remainder is-

a. 1

b. 63

c. 66

d. 67

Ans: (6767+ 67) = 67(6766 + 166)

As 66 is an even number

∴ 6766 is an even number

∴ (6766 + 1) is perfectly divisible by (67 + 1)

i.e. 68

∴ The remainder = 67

6. In a division sum, the divisor is 4 times the quotient and twice the remainder. If a and b are respectively the divisor and the dividend then-

a. 4b-a2 /a =3

b. 4b-2a /a2 =2

c. (a+1)2 = 4b

d. A(a+2)/b = 4

Ans : As divisor is a, and dividend is b.

∴ Quotient = a/4

And Remainder = a/2

∴ b = a * a/4 + a/2

∴ 4b = a2 + 2a

∴ a(a+2)/b = 4

7. If 738 A6A is divisible by 11, then the value of A is-

a. 6

b. 3

c. 9

d. 1

Ans : As 738A6 A is divisible by 11.

∴ A + A + 3 = 6 + 8 + 7

∴ A = 9

8. The east positive integer that should be subtracted from 3011 * 3012 so that the difference is perfect square is-

a. 3009

b. 3010

c. 3011

d. 3012

Ans : ∵ 3011 * 3012 = 3011 (3011 + 1)

= (3011)2 + 3011

∴ Required least number = 3011

9. P, Q, R are employed to do a work for Rs. 5750. P and Q together finished 19/23 of work and Q and R together finished 8/23 of work. Wage of Q, in rupees, is-

a. 2850

b. 3750

c. 2750

d. 1000

Ans : Work done by Q = 19/23 + 8/23 – 1

4/23

∴ Wage of Q = 4/23 * 5750

= Rs. 1000

10. A can do a piece of work in 24 day, B in 32 days and C in 64 days. All begin to do it together, but A leaves after 6 days and B leaves 6 days before the completion of the work. How many days did the work last?

a. 15

b. 20

c. 18

d. 30

Ans : Work done by A = 6/24 = ¼

Work done by B = (x-6)/32

(where x is no. of days in which work is competed)

∵ ¼ + x – 6/32 + x/64 = 1

∴ 16 + 24 – 12 + x /64 = 1

∴ 3x + 4 = 64

∴ x = 60/3 = 20 days

11. The square root of (0.75)3 /1-0.75 + [0.75 + 90.75)2 +1] is-

a. 1

b. 2

c. 3

d. 4

Ans : The square root of

[(0.75)3/1 – 0.75 + {0.75 + (0.75)2 + 1}]

= √1.6875 + 2.3125

=√4

= 2

12. Given that √4096 = 64, the value of √4096 + √40.96 +√0.004096 is-

a. 70.4

b. 70.464

c. 71.104

d. 71.4

Ans : Given Exp. = √4096 + √40.96 +√0.004096

= 64 + 6.4 + 0.064

= 70.464

13. By selling an article at 3/4th of the marked price, there is a gain of 25%. The ratio of the marked price and the cost price is-

a. 5 : 3

b. 3 : 5

c. 3 : 4

d. 4 : 3

Ans : Let of M.P. be Rs. x.

∴ S.P. = Rs.3x/4

and C.P. = 3x/4 * 100/125

= Rs.3x/5

∴ required ratio = x: 3x/5

= 5:3

14. A and B earn in the ratio 2:1. They spend in the ratio 5:3 and save in the ratio 4:1. If the total monthly savings of both A and B are Rs.5,000, the monthly income of B is-

a. Rs.7,000

b. Rs.14,000

c. Rs.5,000

d. Rs.10,000

Ans : Let the monthly income of B be Rs. x.

∴ Monthly income of A = Rs. 2x and

Saving of A =5000 * 4/(4 + 1)

= Rs. 4000

Saving of B = Rs.1000

∵ 2x – 4000/x-1000 = 5/3

⇒ 6x – 12000 = 5x – 5000

∴ x = Rs.7000

15. The ratio of the sum of two numbers and their difference is 5:1. The ratio of the greater number to the smaller number is-

a. 2 : 3

b. 3 : 2

c. 5 : 1

d. 1 : 5

Ans : Let the numbers be x and y.

∵ x + y/ x – y = 5/1

⇒ 5x – 5y = x + y

⇒ 4x = 6y

⇒ x/y = 6/4

x : y = 3 : 2

16. A cistern has 3 pipes A, B and C. A and B can fill it in 3 and 4 hours respectively, and C can empty it in 1 hour. If the pipes are opened at 3 p.m., 4 p.m. and 5 p.m. respectively on the same day, the cistern will be empty at-

a. 7.12 p.m.

b. 7.15 p.m.

c. 7.10 p.m.

d. 7.18 p.m.

Ans : Let the cistern be emptied at x p.m.

∵ x -3/3 + x – 4/4 = x - 5/1

⇒ 4x – 12 + 3x – 12/12 = x – 5/1

⇒ 7x – 24 = 12x – 60

⇒ 5x = 36

∴ x = 36/5

= 7 hr. + 12 min.

= 7.12 p.m.

17. If A works alone, he would take 4 days more to complete the job than if both A and B worked together. If B worked alone, he would take 16 days more to complete the job than if A and B work together. How many days would they take to complete the work if both of them worked together?

a. 10 days

b. 12 days

c. 6 days

d. 8 days

Let x days are taken when they work together.

∴ Time taken by A to complete the work = (x + 4) days

And Time taken by B to complete the work = (x + 16) days

∵ 1/x = 1/x + 4 + 1/x + 16

∴ 1/x = x + 16 + x + 4/(x + 4) (x +16)

∴ 2x2 + 20x = x2 + 20x +64

∴ x2 = 64 (8)2

∴ x = 8 days

18. 250 men can finish a work in 20 days working 5 hours a day. To finish the work within 10 days working 8 hours a day, the minimum number of men required is-

a. 310

b. 300

c. 313

d. 312

Ans : Required no. of men = 250 * 5 * 20/8 * 10

=312.5

313

19. 2 men and 5 women can do a work in 12 days. 5 men 2 women can do that work in 9 days. Only 3 women can finish the same work in-

a. 36 days

b. 21 days

c. 30 days

d. 42 days

∵ (2m + 5w) * 12 = (5m + 2w) * 9

⇒ 24m + 60w = 45m + 18w

⇒ 21m = 42w

⇒ 1m = 2w

∴ 2m + 5w = 9w

∴ required no. of days = 9 * 12/3

= 36

20. While selling, a businessman allows 40% discount on the marked price and there is a loss of 30%. If it is sold at the marked price, profit per cent will be –

a. 10%

b. 20%

c. 16 2/3%

d. 16 1/3%

Ans : Let the M.P. be Rs.100.

∴ S.P = (100-40) = Rs.60

and C. P. = 60 * 100/(100-30) = Rs.600/7

∴ Reqd. % profit = 100-600/7 /600/7 * 100%

= (700 – 600)/7 * 600 * 100 * 7%

= 16 2/3%

21. Successive discount of 10% , 20% and 50% will be equivalent to a single discount of-

a. 36%

b. 64%

c. 80%

d. 56%

Ans : Equivalent to a single discount

= [100 - (100 – 10) (100 – 20) (100 – 50)/100*100]%

= [100 - 90*80*50/10000]%

= [100-36]%

= 64%

22. A retailer offers the following discount schemes for buyers on an article-

I. Two successive discounts of 10%

II. A discount of 12% followed by a discount of 8%.

III. Successive discounts of 15% and 5%

IV. A discount of 20%

The selling price will be minimum under the scheme-

a. I

b. II

c. III

d. IV

Ans : From (i) single discount = [10 + 10 – 10*10/100] % = 19%

From (ii) single discount = [12 + 8 -12*8/100] % = 19.04%

From (iii) single discount = [15 + 5 – 15*5/100] % = 19.25%

From (iv) single discount = 20%

∴ The S.P. will be minimum under the scheme IV.

23. Of three numbers, the second is thrice the first and the third number is three-fourth of the first. If the average of the three numbers is 114, the largest number is –

a. 72

b. 216

c. 354

d. 726

Ans : Let the first number be x.

∴ Second number = 3x and Third number = 3x/4

∵ x + 3x + 3x/4 = 3*114

⇒ 19x/4 = 342

∴ x = 342*4/19 = 72

∴ The largest number = 3*72

= 216

24. A car covers 1/5 of the distance from A to B at the speed of 8 km/hour, 1/10 of the distance at 25 km per hour and the remaining at the speed of 20 km per hour. Find the average speed of the whole journey-

a. 12.625 km/hr

b. 13.625 km/hr

c. 14.625 km/hr

d. 15.625 km/hr

Ans : If the whole journey be x km. The total time taken

= (x/5/8 + x/10/25 + 7x/10/20) hrs

= (x/40 + x/250 + 7x/200) hrs

= 25x + 4x + 35x/1000

= 64x/1000 hrs

∴ Average speed = x/64x/1000

= 15.625 km/hr

25. The average of 3 numbers is 154. The first number is twice the second and the second number is twice the third. The first number is-

a. 264

b. 132

c. 88

d. 66

Ans : Let the first number be x.

∴ Second number = x/2 and Third number = x/4

∵ x + x/2 + x/4 = 3 x 154

⇒ 7x/4 = 462

∴ x = 462*4/7

=264

26. The average salary of all the staff in an office of a corporate house is Rs. 5,000. The average salary of the officers is Rs. 14,000 and that of the rest is Rs. 4,000. If the total number of staff is 500, the number of officers is –

a. 10

b. 15

c. 25

d. 50

Ans : Let the number of officers be x.

∵ 5000*500 = 14000x + 4000(500-x)

∴ 2500000 =14000x + 2000000 – 4000x

∴ x = 500000/10000

= 50

27. The average marks of 40 students in an English exam are 72. Later it is found that three marks 64, 62 and 84 were wrongly entered as 60, 65 and 73. The average after mistakes were rectified is-

a. 70

b. 72

c. 71.9

d. 72.1

Ans : Correct average

= 40*72 + (64 + 62 +84) – 68 – 65 – 73/40

= 2880 + 210 – 206/40 = 2884/40

= 72.1

28. A and B are two alloys of gold and copper prepared by mixing metals in the ratio 7:2 and 7:11 respectively. If equal quantities of the alloys are melted to form a third alloy C, the ratio of gold and copper in C will be-

a. 5 : 7

b. 5 : 9

c. 7 : 5

d. 9 : 5

Ans : Quantity of gold in A = 7/9* wt. of A

Quantity of gold in B = 7/18* wt. of B

If 1 kg of each A and B are mixed to form third alloys C.

Then quantity of gold in 2 kg C = 7/9 + 7/18

= 7/6 kg

And Quantity of copper in 2 kg C = 2 – 7/6

= 5/6 kg

∴ Required ratio = 7/6 : 5/6 = 7 : 5

29. In a laboratory, two bottles contain mixture of acid and water in the ratio 2 : 5 in the first bottle and 7 : 3 in the second. The ratio in which the contents of these two bottles be mixed such that the new mixture has acid and water in the ratio 2 : 3 is-

a. 4 : 15

b. 9 : 8

c. 21 : 8

d. 1 : 2

Ans : Quantity of acid in first bottle = 2/7 x mix.

and Quantity of acid in second bottle = 7/10 x mix.

If x and 1 volumes are taken from I and II bottle respectively to form new mixture.

Then, (2/7 x + 7/10 * 1)/(5x/7 +3/10 * 1) = 2/3

⇒ 6x/7 21/10 = 10x/7 +6/10

⇒ 4x/7 = 15/10

∴ x = 15/10*7/4 = 21/8

∴ Required ratio = x : 1

= 21 : 8

30. A mixture contains 80% acid and rest water. Part of the mixture that should be removed and replaced by same amount of water to make the ratio of acid and water 4 : 3 is-

a. 1/3 rd

b. 3/7 th

c. 2/3 rd

d. 2/7 th

Ans : Let the initial wt. of mixture be 1 kg and x kg of mixture is taken out and replaced by same amount of water.

∵ Amt. of acid/Amt. of water = 0.8 – 0.8x/ (0.2 – 0.2x + x) = 4/3

⇒ 2.4 – 2.4x = 0.8 + 3.2x

⇒ 5.6x = 1.6

∴ x = 1.6/5.6 = 2/7th part

31. An employer reduces the number of his employees in the ratio 9 : 8 and increases their wages in the ratio 14 : 15. If the original wage bill was Rs. 189,900, find the ratio in which the wage bill is decreased-

a. 20 : 21

b. 21 : 20

c. 20 : 19

d. 19 : 21

Ans : Let the initial number of employees be 9x and the employer gives Rs. 14y as wage to each.

∵ 9x * 14y =18900

∴ xy = 150 and The later bill = 8x*15y = 120xy

= 120*150 = 18000

∴ Required ratio = 18000 : 18900

= 20 : 21

32. The batting average for 40 innings of a cricketer is 50 runs. His highest score exceeds his lowest score by 172 runs. If these two innings are excluded, the average of the remaining 38 innings is 48 runs. The highest score of the player is-

a. 165

b. 170

c. 172

d. 174

Ans : Let the max. number of runs be x.

∴ The lowest score = (x-172)

∵ 40*50 = 38*48 + x + (x-172)

⇒ 2000 = 1824 + 2x – 172

∴ x= 174

33. Walking at 3 km per hour, Pintu reaches his school 5 minutes late. If he walks at 4 km per hour he will be 5 minutes early. The distance of Pintu’s school from his house is-

a. 1 ½ Km

b. 2 Km

c. 2 ½ Km

d. 5 Km

Ans : Distance of the school from the house

= 4*3/(4 – 3) * 5 +5/60 km

= 12*1/6

=2 km

34. Nitin bought some oranges at Rs. 40 a dozen and an equal number at Rs.30 a dozen. He sold them at Rs. 45 a dozen and made a profit of Rs.480. The number of oranges, he bought, was-

a. 48 dozen

b. 60 dozen

c. 72 dozen

d. 84 dozen

Ans : Let the number of oranges bought be x.

∴ 45x/12 – 70x/24 = 480

⇒ (45-35)/12 = 480

⇒ x = 480*12/10 = 576

= 48*12 = 48 dozen

35. A man buys two chairs for a total cost of Rs.900. By selling one for 4/5 of its cost and the other for 5/4 of its cost, he makes a profit of Rs.90 on the whole transaction. The cost of the lower priced chair is-

a. Rs.360

b. Rs.400

c. Rs.420

d. Rs.300

Ans : Let the cost price of 1 chair be Rs. x.

∴ C.P. of other chair = Rs. (900-x)

∵ 4/5 x + 5/4 (900-x) = 900 + 90

⇒ 4/5 x +1125 -5x/4 = 990

∴ 9x/20 = 135

∴ x = 135*20/9

= Rs. 300

∴ C.P. of the lower priced chair is Rs. 300.

36. By selling 100 oranges, a vendor gains the selling price of 20 oranges. His gain per cent is-

a. 20

b. 25

c. 30

d. 32

Ans : Let the S.P. of 100 Oranges be Rs. x.

∴ S.P. of 20 oranges = x/100 *20 = Rs. x/5

∴C.P. of 100 oranges = x - x/5

= Rs. 4x/5

∴ Reqd. Profit % = x/5 * 100*5/4x %

= 25%

37. 60% of the cost price of an article is equal to 50% of its selling price. Then the percentage of profit or loss on the cost price is-

a. 20% loss

b. 16 2/3% profit

c. 20% profit

d. 10% loss

Ans : Let the cost price be Rs. 100.

∵ S.P * 50/100 = 100*16/100

∴ S.P = 60*100/50

= Rs. 120

∴ Reqd. % profit = (120 – 100)% = 20%

38. Maninder bought two horses at Rs.40,000 each. He sold one horse at 15% gain, but had to sell the second horse at a loss. If he hand suffered a loss of Rs.3,600 on the whole transaction, then the selling price of the second horse is-

a. Rs.30,000

b. Rs.30,200

c. Rs.30,300

d. Rs.30,400

Ans: C.P. of two horses = 2*40000 = Rs. 80000

and S.P. of two horse = 80000 – 3600 = Rs. 76400

∴ S.P. of the other horse = 76400 – 46000 = 30400

39. A fruit-seller buys x guavas for Rs.y and sells y guavas for Rs. x. If x>y, then he made-

a. x2 – y2 / xy % loss

b. x2 – y2 / xy % gain

c. x2 – y2 / y2 % loss

d. x2 – y2 / y2 * 100% gain

Ans : C.P. of 1 guava = Rs. y/x [x>y]

and S.P. of 1 guava = Rs. x/y

∴ Reqd. Gain% = x/y – y/x/ y/x * 100%

= x2 - y2 /y2 *100%

40. A jar contain 10 red marbles and 30 green ones. How many red marbles must be added to the jar so that 60% of the marbles will be red?

a. 25

b. 30

c. 35

d. 40

Ans : Let after adding x red marbles, the red marbles with be 60% of the total.

∵ (10+x)/ (10 + x) + 30 = 60/100

⇒ 10 + x/40 + x = 3/5

⇒ 50 + 5x = 120 + 3x

x = 70/2 = 35

41. If a number multiplied by 25% of itself gives a number which is 200% more than the number, then the number is –

a. 12

b. 16

c. 35

d. 24

Ans : Let the number be x.

∵ x*25x/100 = x + 200x/100

⇒ x2/4 = 3x

⇒ x2 - 12x = 0

⇒ x – 12 = 0

∴ x = 12

42. The value of an article depreciates every year at the rate of 10% of its value. If the present value of the article is Rs.729, then its worth 3 years ago was-

a. Rs.1250

b. Rs.1000

c. Rs.1125

d. Rs.1200

Ans : Let the worth 3 years ago be Rs. x.

∵ 729 = x (1 – 10/100)3

⇒ 729 = x*9*9*9/10*10*10

∴ x = Rs. 1000

43. The price of onions has been increased by 50%. In order to keep the expenditure on onions the same the percentage of reduction in consumption has to be-

a. 50%

b. 33 1/3%

c. 33%

d. 30%

Ans : Reqd. Percentage of reduction

= 50*100/(100 + 50) %

= 5000/150 %

= 33 1/3%

44. A took two loans altogether of Rs.1200 from B and C. B claimed 14% simple interest per annum, while C claimed 15% per annum. The total interest paid by A in one year was Rs.172. Then, A borrowed-

a. Rs.800 from C

b. Rs.625 from C

c. Rs.400 from B

d. Rs.800 from B

Ans : If A borrowed Rs. x from B. and A borrowed Rs. Rs. (1200 – x) from C.

∵ (1200 – x)*15*1/100 + x*14*1/100

⇒ 18000 – 15x + 14x = 172*100

x = Rs. 800

45. If a regular polygon has each of its angles equal to 3/5 times of two right angles, then the number of side is-

a. 3

b. 5

c. 6

d. 8

Ans : If the number of sides a regular polygon be n.

Then (2n-4)/n = 2*3/5

⇒ (2n – 4)*5 = 6n

∴ n = 5

46. A square is of area 200 sq. m. A new square is formed in such a way that the length of its diagonal is √2 times of the diagonal of the given square. The the area of the new square formed is-

a. 200√2 sq.m

b. 400√2 sq.m

c. 400 sq.m

d. 800 sq.m

Ans : Length of the diagonal of Ist square

= √2*200

= 20 m

∴ Length of the diagonal of new square = 20√2m

∴ Area of the new square = ½*(20.√2)2 = 400 sq. m

47. The heights of a cone, cylinder and hemisphere are equal. If their radii are in the ratio 2 : 3 : 1, then the ratio of their volumes is-

a. 2 : 9 : 2

b. 4 : 9 : 1

c. 4 : 27 : 2

d. 2 : 3 : 1

Ans : Ratio of their volumes [cone : cylinder : hemisphere]

= 1/3(2)2h : (3)2*h : 2/3(1)2.h

= 4/3 : 9 : 2/3 = 4 : 27 : 2

48. A motor-boat can travel at 10 km/hr in still water. It travelled 91 km downstream in a river and then returned to the same place, taking altogether 2 hours. Find the rate of flow of river-

a. 3 km/hr

b. 4 km/hr

c. 2 km/hr

d. 5 km/hr

Ans : Let the rate of flow of river be x km/hr.

∵ 91/(10 + x) + 91/(10 – x) = 20

⇒ 91(10 – x + 10 + x)(10 + x) (10 – x) = 20

⇒ 91*20 = 20(100 – x2)

⇒ x2 = 9 = (3)2

∴ x = 3 km/hr

49. A man driving at 3/4th of his original speed reaches his destination 20 minutes later than the usual time. Then the usual time is-

a. 45 minutes

b. 60 minutes

c. 75 minutes

d. 120 minutes

Ans : Let the original speed be x km/hr and the usual time be y hours.

∵ x * y = ¾ x(y+1/3)

∴ 4y = 3y + 1

∴ y = 1 hr = 60 minutes

50. A motor-boat, travelling at the same speed, can cover 25 km upstream and 39 km downstream in 8 hours. At the same speed, it can travel 35 km upstream and 52 km downstream in 11 hours. The speed of the stream is –

a. 2 km/hr

b. 3 km/hr

c. 4 km/hr

d. 5 km/hr

Ans : Let the speeds of motor boat and the stream be x and y km/hr respectively.

∵ 39/x + y + 25/x – y = 8 …(1)

and 52/x + y + 35/x – y = 11 …(2)

Solving equations (1) and (2), we get-

∴ 100 - 105/x – y = 32 – 33

∴ x – y = 5

and x + y = 13

∴ y = 4 km/hr

51. If a sum of money placed at compound interest, compounded annually, doubles itself in 5 years, then the same amount of money will be 8 times of itself in-

a. 25 years

b. 20 years

c. 15 years

d. 10 years

Ans : Required time = 5log 8/log 2

= 5*3log 2/log 2

= 15 years

52. A person has left an amount of Rs.1,20,000 to be divided between his 2 son aged 14 years and 12 years such that they get equal amounts when each attains 18 years of age. If the amount gets a simple interest of 5% per annum, the younger son’s share at present is-

a. Rs.48,800

b. Rs.57,600

c. Rs.62,400

d. Rs.84,400

Ans : Let the present share of the younger son be Rs. x.

∴ The share of the elder son = Rs. (120000 – x)

∵ x + x*6*5/100

= (120000 – x) + (120000 – x)*4*5/100

⇒ 130x/100 = (120000 –x)*120/100

⇒ 13x = 1440000 – 12x

⇒ 25x = 1440000

∴ x = Rs. 57600

53. If the simple interest on Rs.x at a rate of a% for m years is same as that on Rs. y at a rate of a2% for m2 years, then x : y is equal to-

a. m : a

b. am : 1

c. 1/m : 1/a

d. 1/am : 1

∵ x*a*m/100 = y*a2*m2/100

⇒ x/y = am/1

∴ x : y = am : 1

54. Base of a right prism is an equilateral triangle of side 6 cm. If the volume of the prism is 108√3 cc, its height is-

a. 9 cm

b. 10 cm

c. 11 cm

d. 12 cm

Ans : Height of the prism = 108√3*4/√3*(6)2 = 12cm

55. If a + 1/a + 2 = 0, then the value of (a37 – 1/a100) is-

a. 0

b. -2

c. 1

d. 2

Ans : ∵ a + 1/a + 2 = 0

⇒ a2 + 1 +2a = 0

⇒ (a + 1)2 = 0

⇒ a +1 = 0

∴ a = -1

∴ a37 -1/a100 = (-1) – (1) = -2

56. The value of k for which the graphs of (k-1) x+y-2 = 0 and (2-k) x -3y + 1 = 0 are parallel is-

a. ½

b. -1/2

c. 2

d. -2

Ans : (k – 1) x + y – 2 = 0

∴ y = (1 – k) x + 2 ….(1)

and (2 –k) x – 3y – 1 = 0

3y = (2 – k) x +1

Y = 2 – k/3 x + 1/3 ….(2)

∵ m1 = m2

⇒ 1 – k = 2 – k/3

⇒ 3 – 3k = 2 - k

∴ k = 1/2

57. If a2 + b2 + c2 = 2 (a-b-c) – 3, then the value of (a – b + c) is-

a. -1

b. 3

c. 1

d. -2

Ans : a2 + b2 + c2 = 2(a – b – c) – 3

⇒ a2 - 2a + 1 + b2 + 2b + 1 + c2 + 2c + 1 = 0

⇒ (a – 1)2 + (b + 1)2 + (c + 1)2 = 0

⇒ a – 1 = 0, b + 1 = 0, c + 1 = 0

⇒ a = 1, b = -1, c = -1

∴ a + b – c = 1 – 1 + 1 =1

58. If x2 + 3x + 1 = 0, then the value of x3 + 1/x3 is-

a. -18

b. 18

c. 36

d. -36

Ans : ∵ x2 + 3x + 1 = 0

⇒ x + 3 + 1/x = 0

⇒ x + 1/x = -3

⇒ (x + 1/x)3 = (-3)3

⇒ x3 + 1/x3 + 3 (-3) = -27

∴ x3 + 1/ x3 = -18

59. If xa, xb, xc = 1, then the value of a3 + b3 + c3 is –

a. 9

b. abc

c. a + b + c

d. 3abc

Ans : ∵ xa. xb. xc = 1

⇒ xa + b + c = x0

⇒ a +b + c =0

∴ a3 + b3 + c3 = 3abc

60. Base of a right pyramid is a square, length of diagonal of the base is 24√2 m. If the volume of the pyramid is 1728 cu.m, its height is-

a. 7 m

b. 8 m

c. 9 m

d. 10 m

Ans : Area of the base of the pyramid

= ½ (24√2)2 = 576m2

∴ Height of pyramid + 1728*3/576 = 9m

61. The height of a right circular cone and the radius of its circular base are respectively 9 cm and 3 cm. The cone is cut by a plane parallel to its base so as to divide it into two parts. The volume of the frustum (i.e., the lower part) of the cone is 44 cubic cm. The radius of the upper circular surface of the frustum (taking = 22-7) is-

a. 3√12 cm

b. 3√13 cm

c. 3√6 cm

d. 3√20 cm

Ans : Let the radius of the upper circular part of the frustum be r cm.

(Picture)

Then r/3 = x/9 = AC/AD

∴ x = 3r (where AC = x cm)

⇒ /3h(r12 + r1r2+r22 = 44

⇒ /3(9 – x)*(9 + 3r + r2) = 44

⇒ (9 – x)*(9 + 3r + r2) = 44*3*7/22 = 42

⇒ 81 + 27r + 9r2 – 9x – 3rx - r2x = 42 On putting x = 3r,

⇒ 81 + 27r + 9r2 - 9r2 - 3r3 – 27r = 42

⇒ 3r3 = 39

∴ r = 3√13 cm

62. The ratio of radii of two right circular cylinder is 2 : 3 and their heights are in the ratio 5 : 4. The ratio of their curved surface area is-

a. 5 : 6

b. 3 : 4

c. 4 : 5

d. 2 : 3

Ans : Required ratio = 2*2r*5h/2*3r*4h = 5 : 6

63. A solid cylinder has total surface area of 462 sq.cm. Curved surface area is 1/3rd of its total surface area. The volume of the cylinder is-

a. 530 cm3

b. 536 cm3

c. 539 cm3

d. 545 cm3

Ans : ∵ 2r(r + h) = 462

and 2rh = 1/3*462 = 154

∴ r + h/h = 462/154 =3

∴ r + h = 3h

∴ r = 2h

∴ 2*2h2 = 154

∴ h2 = 154*7/22*4 = 49/4

= (7/2)2

∴ h = 7/2 cm

and r = 7 cm

∴ Volume of the cylinder

= 22/7*49*7/2

= 539 cm3

64. A cylinder and a cone have equal radii of their bases and equal heights. If their curved surface areas are in the ratio 8 : 5, the ratio of their radius and height is-

a. 1 : 2

b. 1 : 3

c. 2 : 3

d. 3 : 4

Ans : Let the radius and height of each are r and h respectively.

∵ 2rh/r√h2 + r2 = 8/5

∴ 10h = 8√r2 + h2

⇒ 100h2 =64r2 + 64h2

∴ h2 = 64 r2/36 = (4/3 r)2

⇒ h = 4/3r

∴ r : h = 3 : 4

65. A solid is hemispherical at the bottom and conical above. If the surface areas of the two parts are equal, then the ratio of radius and height of its conical part is-

a. 1 : 3

b. 1 :1

c. √3 : 1

d. 1 : √3

Ans : (Picture)

∵ rl = 2r2

⇒ l = 2r

⇒ √h2 + r2 = 2r

⇒ h2 = 3r2

∴ r : h = r/h

= 1 : √3

66. If O is the circumcentre of ∆ ABC and ∠OBC = 350, then the ∠BAC is equal to-

a. 550

b. 1100

c. 700

d. 350

Ans : ∴ ∠BOC = 1800 – (350 + 350) = 1100

(Picture)

∴ ∠BAC = 1/2 * 1100 = 550

67. If I is the incentre of ∆ ABC and ∠ BIC = 1350, then ∆ ABC is-

a. Acute angled

b. Equilateral

c. Right angled

d. Obtuse angled

Ans : (Picture)

∠BIC = 1350

⇒ B/2 + C/2 = 1800 – 1350 = 450

⇒ ∠B + ∠C = 900

∴ ∠A = 1800 – (∠B + ∠C) = 900

i.e., ∆ ABC is a right angled.

68. If sin2 (∝ + β/2) is-

a. 1

b. -1

c. 0

d. 0.5

∵ sin2 ∝ + sin2 β = 2

⇒ 1 – cos2 ∝ 1 - cos2 β = 2

⇒ cos2 ∝ cos2 β = 0

⇒ cos ∝ = 0

and cos β = 0

⇒ ∝ = /2 and β = /2

∴ cos (∝ + β/2) = cos [/2 + /2 / 2]

= cos /2 = 0

69. The length of a shadow of a vertical tower is 1/√3 times its height. The angle of elevation of the Sun is-

a. 300

b. 450

c. 600

d. 900

Ans : (Picture)

∵ tan θ = h/1√3 h = √3 = tan 600

70. The graphs of x +2y =3 and 3x-2y = 1 meet the Y-axis at two points having distance-

a. 8/3 units

b. 4/3 units

c. 1 unit

d. 2 units

Ans : When the graphs meet the Y-axis at two points.

Then, [x + 2y = 3] at x = 0 ⇒ [0, y1 =3/2]

[3x - 2y = 1] at x = 0

and i.e., [0, y2 = -1/2]

Required distance = (y1 – y2)

= 3/2 – (- ½) = 2 units

71. If x+1/16x = 1, then the value of 64x3 + 1/64x3 is-

a. 4

b. 52

c. 64

d. 76

∵ x + 1/16x = 1

⇒ 16x2 – 16x + 1 =0

⇒ 16x2 – 16x + 4 = 3

⇒ (4x – 2)2 = 3

⇒ 4x = 2 + √3

⇒ 64x3 = (2±√3)3

=8 + 3√3 + 6√3 (2 + √3)

= 26 + 15√3

∴ 64x3 + 1/64x3 = (26 + 15√3) + 1/ (26 + 15√3)

= (26 + 15√3) + 26 - 15√3/676 -675

=52

72. If a, b, c, are three non-zero real numbers such that a + b + c = 0, and b2 ≠ ca, then the value of a2 + b2 + c2/ b2 –ca is-

a. 3

b. 2

c. 0

d. 1

∵ a + b + c = 0

⇒ a + c = -b

⇒ a2 + c2 = b2 -2ac

⇒ a2 + b2 + c2 = 2b2 – 2ac

∴ a2 + b2 + c2/ b2 ac = 2

73. If a4 + a2 b2 + b4 = 8 and a2 + ab + b2 = 4, then the value of ab is-

a. -1

b. 0

c. 2

d. 1

∵ a4 + a2 b2 + b4/ a2 + ab + b2 = 8/4

⇒ (a2 + b2)2 – (ab) 2/ (a2 + b2 + ab) = 2

⇒ a2 – ab + b2 = 2 ….(1)

and a2 + ab + b2 = 4 …..(2)

⇒ 2ab = 2

⇒ ab = 1

74. If a = 25, b = 15, c = -10, then the value of a3 + b3 + c3 – 3abc/ (a-b)2 + (b-c)2 + (c-a)2 is-

a. 30

b. -15

c. -30

d. 15

∵ a3 + b3 + c3 – 3abc

= (25)3 + (15)3 + (-10)3 – 3*25*15*(-10)

=15625 + 3375 – 1000 + 11250 = 29250

and (a - b)2 + (b – c)2 + (c – a)2

= (10)2 + (25)2 + (-35)2

= (10)2 + 625 + 1225

= 1950

∴ Required value = 29250/1950 =15

75. A, B, C are three points on a circle. The tangent at A meets BC produced at T, ∠BTA = 400, ∠CAT = 440. The angle subtended by BC at the centre of the circle is-

a. 840

b. 920

c. 960

d. 1040

Ans : (Picture)

∠ACB = 400 + 440 = 840

∴ ∠ACO = 900 - 440 = 460 = ∠OAC

⇒ ∠OCB = ∠ACB - ∠ACO

= 840 - 460 = 380 = ∠OBC

∴ ∠BOC = 1800 – (∠OCB + ∠OBC)

= 1800 – (380 + 380) = 1040

76. If the length of a chord of a circle at a distance of 12 cm from the Centre is 10 cm, then the diameter of the circle is-

a. 13 cm

b. 15 cm

c. 26 cm

d. 30 cm

Ans : (Picture)

∵ OA = √OM2 + AM2

= √122 + 52 = 13

∴ Diameter of the circle = 2*OA

= 2*13 = 26cm

77. In ∆ ABC, P and Q are the middle points of the sides AB and AC respectively. R is a point on the segment PQ such that PR : RQ = 1 : 2. If PR = 2 cm, then BC =

a. 4 cm

b. 2 cm

c. 12 cm

d. 6 cm

∵ PR/RQ = ½

But, PR = 2cm

RQ = 2*PR

= 4cm

(Picture)

∴ PQ = PR + RQ

= 2 + 4 = 6 cm

∴ BC = 2*PQ = 12CM

78. If tan θ tan 2θ = 1, then the value of sin2 2θ + tan2 is equal to –

a. ¾

b. 10/3

c. 3 ¾

d. 3

∵ tan θ * tan 2θ = 1

⇒ tan θ * 2 tan θ/1 – tan2 θ = 1

⇒ 2 tan2 θ = 1 - tan2 θ

⇒ 3 tan2 θ = 1

⇒ tan θ = 1/√3 = tan 300

∴ θ = 300

∴ sin2 2θ + tan2 2θ = sin2 600 + tan2 600

= ¾ + 3

=3 ¾

79. The value of cot /20 cot 3/20 cot 5/20 cot 7/20 cot 9/20 is-

a. -1

b. ½

c. 0

d. 1

Ans : Given Exp.

= cot /20. cot 3/20. cot 5/20. cot 7/20. cot 9/20

= cot 90. cot 270. cot 450. cot 630. cot 810

= cot 90. cot 270*1*tan 270. tan 90.

= cot 90. cot 270*1*1/ cot 270*1/ cot 90

=1

80. If sin θ+ cos θ = 17/13, 0<θ<900, then the value of sin θ- cos θ is-

a. 5/17

b. 3/19

c. 7/10

d. 7/13

∵ sin θ + cos θ = 17/13

⇒ sin2 θ + cos2 θ + 2sin θ.cos θ = 289/169

∴ 2sin θ.cos θ = 289 /169 -1

= 289 -169/169 = 120/169

∴ sin2 θ + cos2 θ – 2 sin2 θ.cos θ = 1 – 120/169

= 49/169

⇒ (sin θ – cos θ)2 = (7/13)2

∴ sin θ – cos θ = 7/13

S.S.C. Combined Graduate Level Exam

(Tier-II) 2011

Solved Paper – I

Arithmetical Ability

1. The product of 2 numbers is 1575 and their quotient is 9/7. Then the sum of the numbers is –

a. 74

b. 78

c. 80

d. 90

Ans : Let the numbers be x and y .

∴ xy = 1575

And x/y = 9/7

∴ xy/x/y = 1575/9/7

∴ y2 = 1225

∴ y = 35 and x = 45

∴ The sum of the numbers = 45+35

= 80

2. The value of (81)3.6 * (9)2.7/ (81)4.2 * (3) is __

a. 3

b. 6

c. 9

d. 8.2

Ans : (81)3.6 * (9)2.7/(81)4.2 * (3) = (3)14.4 * (3)5.4/(3)16.8 * (3)

= 314.4+5.4-16.8-1

= 32

= 9

3. √6+√6+√6+… is equal to –

a. 2

b. 5

c. 4

d. 3

Ans : Let √6+√6+√6+…. be x.

∴ x =√6+x

∴ x2 = 6+x

∴ x2 + x – 6 = 0

∴ (x-3) (x+2) = 0

∴ x = 3

4. The sum of the squares of two natural consecutive odd numbers is 394. The sum of the numbers is –

a. 24

b. 32

c. 40

d. 28

Ans : Let the consecutive odd numbers be x and (x+2)

∵ x2 + (x + 2)2 = 394

∴ x2 + x2 + 4x + 4 = 394

∴ 2 x2 +4x – 390 = 0

∴ x2 + 2x – 195 = 0

∴ (x +15) (x-13) = 0

∴ x = 13

∴ Required sum = 13 +15 =28

5. When (6767 +67) is divided by 68, the remainder is-

a. 1

b. 63

c. 66

d. 67

Ans: (6767+ 67) = 67(6766 + 166)

As 66 is an even number

∴ 6766 is an even number

∴ (6766 + 1) is perfectly divisible by (67 + 1)

i.e. 68

∴ The remainder = 67

6. In a division sum, the divisor is 4 times the quotient and twice the remainder. If a and b are respectively the divisor and the dividend then-

a. 4b-a2 /a =3

b. 4b-2a /a2 =2

c. (a+1)2 = 4b

d. A(a+2)/b = 4

Ans : As divisor is a, and dividend is b.

∴ Quotient = a/4

And Remainder = a/2

∴ b = a * a/4 + a/2

∴ 4b = a2 + 2a

∴ a(a+2)/b = 4

7. If 738 A6A is divisible by 11, then the value of A is-

a. 6

b. 3

c. 9

d. 1

Ans : As 738A6 A is divisible by 11.

∴ A + A + 3 = 6 + 8 + 7

∴ A = 9

8. The east positive integer that should be subtracted from 3011 * 3012 so that the difference is perfect square is-

a. 3009

b. 3010

c. 3011

d. 3012

Ans : ∵ 3011 * 3012 = 3011 (3011 + 1)

= (3011)2 + 3011

∴ Required least number = 3011

9. P, Q, R are employed to do a work for Rs. 5750. P and Q together finished 19/23 of work and Q and R together finished 8/23 of work. Wage of Q, in rupees, is-

a. 2850

b. 3750

c. 2750

d. 1000

Ans : Work done by Q = 19/23 + 8/23 – 1

4/23

∴ Wage of Q = 4/23 * 5750

= Rs. 1000

10. A can do a piece of work in 24 day, B in 32 days and C in 64 days. All begin to do it together, but A leaves after 6 days and B leaves 6 days before the completion of the work. How many days did the work last?

a. 15

b. 20

c. 18

d. 30

Ans : Work done by A = 6/24 = ¼

Work done by B = (x-6)/32

(where x is no. of days in which work is competed)

∵ ¼ + x – 6/32 + x/64 = 1

∴ 16 + 24 – 12 + x /64 = 1

∴ 3x + 4 = 64

∴ x = 60/3 = 20 days

11. The square root of (0.75)3 /1-0.75 + [0.75 + 90.75)2 +1] is-

a. 1

b. 2

c. 3

d. 4

Ans : The square root of

[(0.75)3/1 – 0.75 + {0.75 + (0.75)2 + 1}]

= √1.6875 + 2.3125

=√4

= 2

12. Given that √4096 = 64, the value of √4096 + √40.96 +√0.004096 is-

a. 70.4

b. 70.464

c. 71.104

d. 71.4

Ans : Given Exp. = √4096 + √40.96 +√0.004096

= 64 + 6.4 + 0.064

= 70.464

13. By selling an article at 3/4th of the marked price, there is a gain of 25%. The ratio of the marked price and the cost price is-

a. 5 : 3

b. 3 : 5

c. 3 : 4

d. 4 : 3

Ans : Let of M.P. be Rs. x.

∴ S.P. = Rs.3x/4

and C.P. = 3x/4 * 100/125

= Rs.3x/5

∴ required ratio = x: 3x/5

= 5:3

14. A and B earn in the ratio 2:1. They spend in the ratio 5:3 and save in the ratio 4:1. If the total monthly savings of both A and B are Rs.5,000, the monthly income of B is-

a. Rs.7,000

b. Rs.14,000

c. Rs.5,000

d. Rs.10,000

Ans : Let the monthly income of B be Rs. x.

∴ Monthly income of A = Rs. 2x and

Saving of A =5000 * 4/(4 + 1)

= Rs. 4000

Saving of B = Rs.1000

∵ 2x – 4000/x-1000 = 5/3

⇒ 6x – 12000 = 5x – 5000

∴ x = Rs.7000

15. The ratio of the sum of two numbers and their difference is 5:1. The ratio of the greater number to the smaller number is-

a. 2 : 3

b. 3 : 2

c. 5 : 1

d. 1 : 5

Ans : Let the numbers be x and y.

∵ x + y/ x – y = 5/1

⇒ 5x – 5y = x + y

⇒ 4x = 6y

⇒ x/y = 6/4

x : y = 3 : 2

16. A cistern has 3 pipes A, B and C. A and B can fill it in 3 and 4 hours respectively, and C can empty it in 1 hour. If the pipes are opened at 3 p.m., 4 p.m. and 5 p.m. respectively on the same day, the cistern will be empty at-

a. 7.12 p.m.

b. 7.15 p.m.

c. 7.10 p.m.

d. 7.18 p.m.

Ans : Let the cistern be emptied at x p.m.

∵ x -3/3 + x – 4/4 = x - 5/1

⇒ 4x – 12 + 3x – 12/12 = x – 5/1

⇒ 7x – 24 = 12x – 60

⇒ 5x = 36

∴ x = 36/5

= 7 hr. + 12 min.

= 7.12 p.m.

17. If A works alone, he would take 4 days more to complete the job than if both A and B worked together. If B worked alone, he would take 16 days more to complete the job than if A and B work together. How many days would they take to complete the work if both of them worked together?

a. 10 days

b. 12 days

c. 6 days

d. 8 days

Let x days are taken when they work together.

∴ Time taken by A to complete the work = (x + 4) days

And Time taken by B to complete the work = (x + 16) days

∵ 1/x = 1/x + 4 + 1/x + 16

∴ 1/x = x + 16 + x + 4/(x + 4) (x +16)

∴ 2x2 + 20x = x2 + 20x +64

∴ x2 = 64 (8)2

∴ x = 8 days

18. 250 men can finish a work in 20 days working 5 hours a day. To finish the work within 10 days working 8 hours a day, the minimum number of men required is-

a. 310

b. 300

c. 313

d. 312

Ans : Required no. of men = 250 * 5 * 20/8 * 10

=312.5

313

19. 2 men and 5 women can do a work in 12 days. 5 men 2 women can do that work in 9 days. Only 3 women can finish the same work in-

a. 36 days

b. 21 days

c. 30 days

d. 42 days

∵ (2m + 5w) * 12 = (5m + 2w) * 9

⇒ 24m + 60w = 45m + 18w

⇒ 21m = 42w

⇒ 1m = 2w

∴ 2m + 5w = 9w

∴ required no. of days = 9 * 12/3

= 36

20. While selling, a businessman allows 40% discount on the marked price and there is a loss of 30%. If it is sold at the marked price, profit per cent will be –

a. 10%

b. 20%

c. 16 2/3%

d. 16 1/3%

Ans : Let the M.P. be Rs.100.

∴ S.P = (100-40) = Rs.60

and C. P. = 60 * 100/(100-30) = Rs.600/7

∴ Reqd. % profit = 100-600/7 /600/7 * 100%

= (700 – 600)/7 * 600 * 100 * 7%

= 16 2/3%

21. Successive discount of 10% , 20% and 50% will be equivalent to a single discount of-

a. 36%

b. 64%

c. 80%

d. 56%

Ans : Equivalent to a single discount

= [100 - (100 – 10) (100 – 20) (100 – 50)/100*100]%

= [100 - 90*80*50/10000]%

= [100-36]%

= 64%

22. A retailer offers the following discount schemes for buyers on an article-

I. Two successive discounts of 10%

II. A discount of 12% followed by a discount of 8%.

III. Successive discounts of 15% and 5%

IV. A discount of 20%

The selling price will be minimum under the scheme-

a. I

b. II

c. III

d. IV

Ans : From (i) single discount = [10 + 10 – 10*10/100] % = 19%

From (ii) single discount = [12 + 8 -12*8/100] % = 19.04%

From (iii) single discount = [15 + 5 – 15*5/100] % = 19.25%

From (iv) single discount = 20%

∴ The S.P. will be minimum under the scheme IV.

23. Of three numbers, the second is thrice the first and the third number is three-fourth of the first. If the average of the three numbers is 114, the largest number is –

a. 72

b. 216

c. 354

d. 726

Ans : Let the first number be x.

∴ Second number = 3x and Third number = 3x/4

∵ x + 3x + 3x/4 = 3*114

⇒ 19x/4 = 342

∴ x = 342*4/19 = 72

∴ The largest number = 3*72

= 216

24. A car covers 1/5 of the distance from A to B at the speed of 8 km/hour, 1/10 of the distance at 25 km per hour and the remaining at the speed of 20 km per hour. Find the average speed of the whole journey-

a. 12.625 km/hr

b. 13.625 km/hr

c. 14.625 km/hr

d. 15.625 km/hr

Ans : If the whole journey be x km. The total time taken

= (x/5/8 + x/10/25 + 7x/10/20) hrs

= (x/40 + x/250 + 7x/200) hrs

= 25x + 4x + 35x/1000

= 64x/1000 hrs

∴ Average speed = x/64x/1000

= 15.625 km/hr

25. The average of 3 numbers is 154. The first number is twice the second and the second number is twice the third. The first number is-

a. 264

b. 132

c. 88

d. 66

Ans : Let the first number be x.

∴ Second number = x/2 and Third number = x/4

∵ x + x/2 + x/4 = 3 x 154

⇒ 7x/4 = 462

∴ x = 462*4/7

=264

26. The average salary of all the staff in an office of a corporate house is Rs. 5,000. The average salary of the officers is Rs. 14,000 and that of the rest is Rs. 4,000. If the total number of staff is 500, the number of officers is –

a. 10

b. 15

c. 25

d. 50

Ans : Let the number of officers be x.

∵ 5000*500 = 14000x + 4000(500-x)

∴ 2500000 =14000x + 2000000 – 4000x

∴ x = 500000/10000

= 50

27. The average marks of 40 students in an English exam are 72. Later it is found that three marks 64, 62 and 84 were wrongly entered as 60, 65 and 73. The average after mistakes were rectified is-

a. 70

b. 72

c. 71.9

d. 72.1

Ans : Correct average

= 40*72 + (64 + 62 +84) – 68 – 65 – 73/40

= 2880 + 210 – 206/40 = 2884/40

= 72.1

28. A and B are two alloys of gold and copper prepared by mixing metals in the ratio 7:2 and 7:11 respectively. If equal quantities of the alloys are melted to form a third alloy C, the ratio of gold and copper in C will be-

a. 5 : 7

b. 5 : 9

c. 7 : 5

d. 9 : 5

Ans : Quantity of gold in A = 7/9* wt. of A

Quantity of gold in B = 7/18* wt. of B

If 1 kg of each A and B are mixed to form third alloys C.

Then quantity of gold in 2 kg C = 7/9 + 7/18

= 7/6 kg

And Quantity of copper in 2 kg C = 2 – 7/6

= 5/6 kg

∴ Required ratio = 7/6 : 5/6 = 7 : 5

29. In a laboratory, two bottles contain mixture of acid and water in the ratio 2 : 5 in the first bottle and 7 : 3 in the second. The ratio in which the contents of these two bottles be mixed such that the new mixture has acid and water in the ratio 2 : 3 is-

a. 4 : 15

b. 9 : 8

c. 21 : 8

d. 1 : 2

Ans : Quantity of acid in first bottle = 2/7 x mix.

and Quantity of acid in second bottle = 7/10 x mix.

If x and 1 volumes are taken from I and II bottle respectively to form new mixture.

Then, (2/7 x + 7/10 * 1)/(5x/7 +3/10 * 1) = 2/3

⇒ 6x/7 21/10 = 10x/7 +6/10

⇒ 4x/7 = 15/10

∴ x = 15/10*7/4 = 21/8

∴ Required ratio = x : 1

= 21 : 8

30. A mixture contains 80% acid and rest water. Part of the mixture that should be removed and replaced by same amount of water to make the ratio of acid and water 4 : 3 is-

a. 1/3 rd

b. 3/7 th

c. 2/3 rd

d. 2/7 th

Ans : Let the initial wt. of mixture be 1 kg and x kg of mixture is taken out and replaced by same amount of water.

∵ Amt. of acid/Amt. of water = 0.8 – 0.8x/ (0.2 – 0.2x + x) = 4/3

⇒ 2.4 – 2.4x = 0.8 + 3.2x

⇒ 5.6x = 1.6

∴ x = 1.6/5.6 = 2/7th part

31. An employer reduces the number of his employees in the ratio 9 : 8 and increases their wages in the ratio 14 : 15. If the original wage bill was Rs. 189,900, find the ratio in which the wage bill is decreased-

a. 20 : 21

b. 21 : 20

c. 20 : 19

d. 19 : 21

Ans : Let the initial number of employees be 9x and the employer gives Rs. 14y as wage to each.

∵ 9x * 14y =18900

∴ xy = 150 and The later bill = 8x*15y = 120xy

= 120*150 = 18000

∴ Required ratio = 18000 : 18900

= 20 : 21

32. The batting average for 40 innings of a cricketer is 50 runs. His highest score exceeds his lowest score by 172 runs. If these two innings are excluded, the average of the remaining 38 innings is 48 runs. The highest score of the player is-

a. 165

b. 170

c. 172

d. 174

Ans : Let the max. number of runs be x.

∴ The lowest score = (x-172)

∵ 40*50 = 38*48 + x + (x-172)

⇒ 2000 = 1824 + 2x – 172

∴ x= 174

33. Walking at 3 km per hour, Pintu reaches his school 5 minutes late. If he walks at 4 km per hour he will be 5 minutes early. The distance of Pintu’s school from his house is-

a. 1 ½ Km

b. 2 Km

c. 2 ½ Km

d. 5 Km

Ans : Distance of the school from the house

= 4*3/(4 – 3) * 5 +5/60 km

= 12*1/6

=2 km

34. Nitin bought some oranges at Rs. 40 a dozen and an equal number at Rs.30 a dozen. He sold them at Rs. 45 a dozen and made a profit of Rs.480. The number of oranges, he bought, was-

a. 48 dozen

b. 60 dozen

c. 72 dozen

d. 84 dozen

Ans : Let the number of oranges bought be x.

∴ 45x/12 – 70x/24 = 480

⇒ (45-35)/12 = 480

⇒ x = 480*12/10 = 576

= 48*12 = 48 dozen

35. A man buys two chairs for a total cost of Rs.900. By selling one for 4/5 of its cost and the other for 5/4 of its cost, he makes a profit of Rs.90 on the whole transaction. The cost of the lower priced chair is-

a. Rs.360

b. Rs.400

c. Rs.420

d. Rs.300

Ans : Let the cost price of 1 chair be Rs. x.

∴ C.P. of other chair = Rs. (900-x)

∵ 4/5 x + 5/4 (900-x) = 900 + 90

⇒ 4/5 x +1125 -5x/4 = 990

∴ 9x/20 = 135

∴ x = 135*20/9

= Rs. 300

∴ C.P. of the lower priced chair is Rs. 300.

36. By selling 100 oranges, a vendor gains the selling price of 20 oranges. His gain per cent is-

a. 20

b. 25

c. 30

d. 32

Ans : Let the S.P. of 100 Oranges be Rs. x.

∴ S.P. of 20 oranges = x/100 *20 = Rs. x/5

∴C.P. of 100 oranges = x - x/5

= Rs. 4x/5

∴ Reqd. Profit % = x/5 * 100*5/4x %

= 25%

37. 60% of the cost price of an article is equal to 50% of its selling price. Then the percentage of profit or loss on the cost price is-

a. 20% loss

b. 16 2/3% profit

c. 20% profit

d. 10% loss

Ans : Let the cost price be Rs. 100.

∵ S.P * 50/100 = 100*16/100

∴ S.P = 60*100/50

= Rs. 120

∴ Reqd. % profit = (120 – 100)% = 20%

38. Maninder bought two horses at Rs.40,000 each. He sold one horse at 15% gain, but had to sell the second horse at a loss. If he hand suffered a loss of Rs.3,600 on the whole transaction, then the selling price of the second horse is-

a. Rs.30,000

b. Rs.30,200

c. Rs.30,300

d. Rs.30,400

Ans: C.P. of two horses = 2*40000 = Rs. 80000

and S.P. of two horse = 80000 – 3600 = Rs. 76400

∴ S.P. of the other horse = 76400 – 46000 = 30400

39. A fruit-seller buys x guavas for Rs.y and sells y guavas for Rs. x. If x>y, then he made-

a. x2 – y2 / xy % loss

b. x2 – y2 / xy % gain

c. x2 – y2 / y2 % loss

d. x2 – y2 / y2 * 100% gain

Ans : C.P. of 1 guava = Rs. y/x [x>y]

and S.P. of 1 guava = Rs. x/y

∴ Reqd. Gain% = x/y – y/x/ y/x * 100%

= x2 - y2 /y2 *100%

40. A jar contain 10 red marbles and 30 green ones. How many red marbles must be added to the jar so that 60% of the marbles will be red?

a. 25

b. 30

c. 35

d. 40

Ans : Let after adding x red marbles, the red marbles with be 60% of the total.

∵ (10+x)/ (10 + x) + 30 = 60/100

⇒ 10 + x/40 + x = 3/5

⇒ 50 + 5x = 120 + 3x

x = 70/2 = 35

41. If a number multiplied by 25% of itself gives a number which is 200% more than the number, then the number is –

a. 12

b. 16

c. 35

d. 24

Ans : Let the number be x.

∵ x*25x/100 = x + 200x/100

⇒ x2/4 = 3x

⇒ x2 - 12x = 0

⇒ x – 12 = 0

∴ x = 12

42. The value of an article depreciates every year at the rate of 10% of its value. If the present value of the article is Rs.729, then its worth 3 years ago was-

a. Rs.1250

b. Rs.1000

c. Rs.1125

d. Rs.1200

Ans : Let the worth 3 years ago be Rs. x.

∵ 729 = x (1 – 10/100)3

⇒ 729 = x*9*9*9/10*10*10

∴ x = Rs. 1000

43. The price of onions has been increased by 50%. In order to keep the expenditure on onions the same the percentage of reduction in consumption has to be-

a. 50%

b. 33 1/3%

c. 33%

d. 30%

Ans : Reqd. Percentage of reduction

= 50*100/(100 + 50) %

= 5000/150 %

= 33 1/3%

44. A took two loans altogether of Rs.1200 from B and C. B claimed 14% simple interest per annum, while C claimed 15% per annum. The total interest paid by A in one year was Rs.172. Then, A borrowed-

a. Rs.800 from C

b. Rs.625 from C

c. Rs.400 from B

d. Rs.800 from B

Ans : If A borrowed Rs. x from B. and A borrowed Rs. Rs. (1200 – x) from C.

∵ (1200 – x)*15*1/100 + x*14*1/100

⇒ 18000 – 15x + 14x = 172*100

x = Rs. 800

45. If a regular polygon has each of its angles equal to 3/5 times of two right angles, then the number of side is-

a. 3

b. 5

c. 6

d. 8

Ans : If the number of sides a regular polygon be n.

Then (2n-4)/n = 2*3/5

⇒ (2n – 4)*5 = 6n

∴ n = 5

46. A square is of area 200 sq. m. A new square is formed in such a way that the length of its diagonal is √2 times of the diagonal of the given square. The the area of the new square formed is-

a. 200√2 sq.m

b. 400√2 sq.m

c. 400 sq.m

d. 800 sq.m

Ans : Length of the diagonal of Ist square

= √2*200

= 20 m

∴ Length of the diagonal of new square = 20√2m

∴ Area of the new square = ½*(20.√2)2 = 400 sq. m

47. The heights of a cone, cylinder and hemisphere are equal. If their radii are in the ratio 2 : 3 : 1, then the ratio of their volumes is-

a. 2 : 9 : 2

b. 4 : 9 : 1

c. 4 : 27 : 2

d. 2 : 3 : 1

Ans : Ratio of their volumes [cone : cylinder : hemisphere]

= 1/3(2)2h : (3)2*h : 2/3(1)2.h

= 4/3 : 9 : 2/3 = 4 : 27 : 2

48. A motor-boat can travel at 10 km/hr in still water. It travelled 91 km downstream in a river and then returned to the same place, taking altogether 2 hours. Find the rate of flow of river-

a. 3 km/hr

b. 4 km/hr

c. 2 km/hr

d. 5 km/hr

Ans : Let the rate of flow of river be x km/hr.

∵ 91/(10 + x) + 91/(10 – x) = 20

⇒ 91(10 – x + 10 + x)(10 + x) (10 – x) = 20

⇒ 91*20 = 20(100 – x2)

⇒ x2 = 9 = (3)2

∴ x = 3 km/hr

49. A man driving at 3/4th of his original speed reaches his destination 20 minutes later than the usual time. Then the usual time is-

a. 45 minutes

b. 60 minutes

c. 75 minutes

d. 120 minutes

Ans : Let the original speed be x km/hr and the usual time be y hours.

∵ x * y = ¾ x(y+1/3)

∴ 4y = 3y + 1

∴ y = 1 hr = 60 minutes

50. A motor-boat, travelling at the same speed, can cover 25 km upstream and 39 km downstream in 8 hours. At the same speed, it can travel 35 km upstream and 52 km downstream in 11 hours. The speed of the stream is –

a. 2 km/hr

b. 3 km/hr

c. 4 km/hr

d. 5 km/hr

Ans : Let the speeds of motor boat and the stream be x and y km/hr respectively.

∵ 39/x + y + 25/x – y = 8 …(1)

and 52/x + y + 35/x – y = 11 …(2)

Solving equations (1) and (2), we get-

∴ 100 - 105/x – y = 32 – 33

∴ x – y = 5

and x + y = 13

∴ y = 4 km/hr

51. If a sum of money placed at compound interest, compounded annually, doubles itself in 5 years, then the same amount of money will be 8 times of itself in-

a. 25 years

b. 20 years

c. 15 years

d. 10 years

Ans : Required time = 5log 8/log 2

= 5*3log 2/log 2

= 15 years

52. A person has left an amount of Rs.1,20,000 to be divided between his 2 son aged 14 years and 12 years such that they get equal amounts when each attains 18 years of age. If the amount gets a simple interest of 5% per annum, the younger son’s share at present is-

a. Rs.48,800

b. Rs.57,600

c. Rs.62,400

d. Rs.84,400

Ans : Let the present share of the younger son be Rs. x.

∴ The share of the elder son = Rs. (120000 – x)

∵ x + x*6*5/100

= (120000 – x) + (120000 – x)*4*5/100

⇒ 130x/100 = (120000 –x)*120/100

⇒ 13x = 1440000 – 12x

⇒ 25x = 1440000

∴ x = Rs. 57600

53. If the simple interest on Rs.x at a rate of a% for m years is same as that on Rs. y at a rate of a2% for m2 years, then x : y is equal to-

a. m : a

b. am : 1

c. 1/m : 1/a

d. 1/am : 1

∵ x*a*m/100 = y*a2*m2/100

⇒ x/y = am/1

∴ x : y = am : 1

54. Base of a right prism is an equilateral triangle of side 6 cm. If the volume of the prism is 108√3 cc, its height is-

a. 9 cm

b. 10 cm

c. 11 cm

d. 12 cm

Ans : Height of the prism = 108√3*4/√3*(6)2 = 12cm

55. If a + 1/a + 2 = 0, then the value of (a37 – 1/a100) is-

a. 0

b. -2

c. 1

d. 2

Ans : ∵ a + 1/a + 2 = 0

⇒ a2 + 1 +2a = 0

⇒ (a + 1)2 = 0

⇒ a +1 = 0

∴ a = -1

∴ a37 -1/a100 = (-1) – (1) = -2

56. The value of k for which the graphs of (k-1) x+y-2 = 0 and (2-k) x -3y + 1 = 0 are parallel is-

a. ½

b. -1/2

c. 2

d. -2

Ans : (k – 1) x + y – 2 = 0

∴ y = (1 – k) x + 2 ….(1)

and (2 –k) x – 3y – 1 = 0

3y = (2 – k) x +1

Y = 2 – k/3 x + 1/3 ….(2)

∵ m1 = m2

⇒ 1 – k = 2 – k/3

⇒ 3 – 3k = 2 - k

∴ k = 1/2

57. If a2 + b2 + c2 = 2 (a-b-c) – 3, then the value of (a – b + c) is-

a. -1

b. 3

c. 1

d. -2

Ans : a2 + b2 + c2 = 2(a – b – c) – 3

⇒ a2 - 2a + 1 + b2 + 2b + 1 + c2 + 2c + 1 = 0

⇒ (a – 1)2 + (b + 1)2 + (c + 1)2 = 0

⇒ a – 1 = 0, b + 1 = 0, c + 1 = 0

⇒ a = 1, b = -1, c = -1

∴ a + b – c = 1 – 1 + 1 =1

58. If x2 + 3x + 1 = 0, then the value of x3 + 1/x3 is-

a. -18

b. 18

c. 36

d. -36

Ans : ∵ x2 + 3x + 1 = 0

⇒ x + 3 + 1/x = 0

⇒ x + 1/x = -3

⇒ (x + 1/x)3 = (-3)3

⇒ x3 + 1/x3 + 3 (-3) = -27

∴ x3 + 1/ x3 = -18

59. If xa, xb, xc = 1, then the value of a3 + b3 + c3 is –

a. 9

b. abc

c. a + b + c

d. 3abc

Ans : ∵ xa. xb. xc = 1

⇒ xa + b + c = x0

⇒ a +b + c =0

∴ a3 + b3 + c3 = 3abc

60. Base of a right pyramid is a square, length of diagonal of the base is 24√2 m. If the volume of the pyramid is 1728 cu.m, its height is-

a. 7 m

b. 8 m

c. 9 m

d. 10 m

Ans : Area of the base of the pyramid

= ½ (24√2)2 = 576m2

∴ Height of pyramid + 1728*3/576 = 9m

61. The height of a right circular cone and the radius of its circular base are respectively 9 cm and 3 cm. The cone is cut by a plane parallel to its base so as to divide it into two parts. The volume of the frustum (i.e., the lower part) of the cone is 44 cubic cm. The radius of the upper circular surface of the frustum (taking = 22-7) is-

a. 3√12 cm

b. 3√13 cm

c. 3√6 cm

d. 3√20 cm

Ans : Let the radius of the upper circular part of the frustum be r cm.

(Picture)

Then r/3 = x/9 = AC/AD

∴ x = 3r (where AC = x cm)

⇒ /3h(r12 + r1r2+r22 = 44

⇒ /3(9 – x)*(9 + 3r + r2) = 44

⇒ (9 – x)*(9 + 3r + r2) = 44*3*7/22 = 42

⇒ 81 + 27r + 9r2 – 9x – 3rx - r2x = 42 On putting x = 3r,

⇒ 81 + 27r + 9r2 - 9r2 - 3r3 – 27r = 42

⇒ 3r3 = 39

∴ r = 3√13 cm

62. The ratio of radii of two right circular cylinder is 2 : 3 and their heights are in the ratio 5 : 4. The ratio of their curved surface area is-

a. 5 : 6

b. 3 : 4

c. 4 : 5

d. 2 : 3

Ans : Required ratio = 2*2r*5h/2*3r*4h = 5 : 6

63. A solid cylinder has total surface area of 462 sq.cm. Curved surface area is 1/3rd of its total surface area. The volume of the cylinder is-

a. 530 cm3

b. 536 cm3

c. 539 cm3

d. 545 cm3

Ans : ∵ 2r(r + h) = 462

and 2rh = 1/3*462 = 154

∴ r + h/h = 462/154 =3

∴ r + h = 3h

∴ r = 2h

∴ 2*2h2 = 154

∴ h2 = 154*7/22*4 = 49/4

= (7/2)2

∴ h = 7/2 cm

and r = 7 cm

∴ Volume of the cylinder

= 22/7*49*7/2

= 539 cm3

64. A cylinder and a cone have equal radii of their bases and equal heights. If their curved surface areas are in the ratio 8 : 5, the ratio of their radius and height is-

a. 1 : 2

b. 1 : 3

c. 2 : 3

d. 3 : 4

Ans : Let the radius and height of each are r and h respectively.

∵ 2rh/r√h2 + r2 = 8/5

∴ 10h = 8√r2 + h2

⇒ 100h2 =64r2 + 64h2

∴ h2 = 64 r2/36 = (4/3 r)2

⇒ h = 4/3r

∴ r : h = 3 : 4

65. A solid is hemispherical at the bottom and conical above. If the surface areas of the two parts are equal, then the ratio of radius and height of its conical part is-

a. 1 : 3

b. 1 :1

c. √3 : 1

d. 1 : √3

Ans : (Picture)

∵ rl = 2r2

⇒ l = 2r

⇒ √h2 + r2 = 2r

⇒ h2 = 3r2

∴ r : h = r/h

= 1 : √3

66. If O is the circumcentre of ∆ ABC and ∠OBC = 350, then the ∠BAC is equal to-

a. 550

b. 1100

c. 700

d. 350

Ans : ∴ ∠BOC = 1800 – (350 + 350) = 1100

(Picture)

∴ ∠BAC = 1/2 * 1100 = 550

67. If I is the incentre of ∆ ABC and ∠ BIC = 1350, then ∆ ABC is-

a. Acute angled

b. Equilateral

c. Right angled

d. Obtuse angled

Ans : (Picture)

∠BIC = 1350

⇒ B/2 + C/2 = 1800 – 1350 = 450

⇒ ∠B + ∠C = 900

∴ ∠A = 1800 – (∠B + ∠C) = 900

i.e., ∆ ABC is a right angled.

68. If sin2 (∝ + β/2) is-

a. 1

b. -1

c. 0

d. 0.5

∵ sin2 ∝ + sin2 β = 2

⇒ 1 – cos2 ∝ 1 - cos2 β = 2

⇒ cos2 ∝ cos2 β = 0

⇒ cos ∝ = 0

and cos β = 0

⇒ ∝ = /2 and β = /2

∴ cos (∝ + β/2) = cos [/2 + /2 / 2]

= cos /2 = 0

69. The length of a shadow of a vertical tower is 1/√3 times its height. The angle of elevation of the Sun is-

a. 300

b. 450

c. 600

d. 900

Ans : (Picture)

∵ tan θ = h/1√3 h = √3 = tan 600

70. The graphs of x +2y =3 and 3x-2y = 1 meet the Y-axis at two points having distance-

a. 8/3 units

b. 4/3 units

c. 1 unit

d. 2 units

Ans : When the graphs meet the Y-axis at two points.

Then, [x + 2y = 3] at x = 0 ⇒ [0, y1 =3/2]

[3x - 2y = 1] at x = 0

and i.e., [0, y2 = -1/2]

Required distance = (y1 – y2)

= 3/2 – (- ½) = 2 units

71. If x+1/16x = 1, then the value of 64x3 + 1/64x3 is-

a. 4

b. 52

c. 64

d. 76

∵ x + 1/16x = 1

⇒ 16x2 – 16x + 1 =0

⇒ 16x2 – 16x + 4 = 3

⇒ (4x – 2)2 = 3

⇒ 4x = 2 + √3

⇒ 64x3 = (2±√3)3

=8 + 3√3 + 6√3 (2 + √3)

= 26 + 15√3

∴ 64x3 + 1/64x3 = (26 + 15√3) + 1/ (26 + 15√3)

= (26 + 15√3) + 26 - 15√3/676 -675

=52

72. If a, b, c, are three non-zero real numbers such that a + b + c = 0, and b2 ≠ ca, then the value of a2 + b2 + c2/ b2 –ca is-

a. 3

b. 2

c. 0

d. 1

∵ a + b + c = 0

⇒ a + c = -b

⇒ a2 + c2 = b2 -2ac

⇒ a2 + b2 + c2 = 2b2 – 2ac

∴ a2 + b2 + c2/ b2 ac = 2

73. If a4 + a2 b2 + b4 = 8 and a2 + ab + b2 = 4, then the value of ab is-

a. -1

b. 0

c. 2

d. 1

∵ a4 + a2 b2 + b4/ a2 + ab + b2 = 8/4

⇒ (a2 + b2)2 – (ab) 2/ (a2 + b2 + ab) = 2

⇒ a2 – ab + b2 = 2 ….(1)

and a2 + ab + b2 = 4 …..(2)

⇒ 2ab = 2

⇒ ab = 1

74. If a = 25, b = 15, c = -10, then the value of a3 + b3 + c3 – 3abc/ (a-b)2 + (b-c)2 + (c-a)2 is-

a. 30

b. -15

c. -30

d. 15

∵ a3 + b3 + c3 – 3abc

= (25)3 + (15)3 + (-10)3 – 3*25*15*(-10)

=15625 + 3375 – 1000 + 11250 = 29250

and (a - b)2 + (b – c)2 + (c – a)2

= (10)2 + (25)2 + (-35)2

= (10)2 + 625 + 1225

= 1950

∴ Required value = 29250/1950 =15

75. A, B, C are three points on a circle. The tangent at A meets BC produced at T, ∠BTA = 400, ∠CAT = 440. The angle subtended by BC at the centre of the circle is-

a. 840

b. 920

c. 960

d. 1040

Ans : (Picture)

∠ACB = 400 + 440 = 840

∴ ∠ACO = 900 - 440 = 460 = ∠OAC

⇒ ∠OCB = ∠ACB - ∠ACO

= 840 - 460 = 380 = ∠OBC

∴ ∠BOC = 1800 – (∠OCB + ∠OBC)

= 1800 – (380 + 380) = 1040

76. If the length of a chord of a circle at a distance of 12 cm from the Centre is 10 cm, then the diameter of the circle is-

a. 13 cm

b. 15 cm

c. 26 cm

d. 30 cm

Ans : (Picture)

∵ OA = √OM2 + AM2

= √122 + 52 = 13

∴ Diameter of the circle = 2*OA

= 2*13 = 26cm

77. In ∆ ABC, P and Q are the middle points of the sides AB and AC respectively. R is a point on the segment PQ such that PR : RQ = 1 : 2. If PR = 2 cm, then BC =

a. 4 cm

b. 2 cm

c. 12 cm

d. 6 cm

∵ PR/RQ = ½

But, PR = 2cm

RQ = 2*PR

= 4cm

(Picture)

∴ PQ = PR + RQ

= 2 + 4 = 6 cm

∴ BC = 2*PQ = 12CM

78. If tan θ tan 2θ = 1, then the value of sin2 2θ + tan2 is equal to –

a. ¾

b. 10/3

c. 3 ¾

d. 3

∵ tan θ * tan 2θ = 1

⇒ tan θ * 2 tan θ/1 – tan2 θ = 1

⇒ 2 tan2 θ = 1 - tan2 θ

⇒ 3 tan2 θ = 1

⇒ tan θ = 1/√3 = tan 300

∴ θ = 300

∴ sin2 2θ + tan2 2θ = sin2 600 + tan2 600

= ¾ + 3

=3 ¾

79. The value of cot /20 cot 3/20 cot 5/20 cot 7/20 cot 9/20 is-

a. -1

b. ½

c. 0

d. 1

Ans : Given Exp.

= cot /20. cot 3/20. cot 5/20. cot 7/20. cot 9/20

= cot 90. cot 270. cot 450. cot 630. cot 810

= cot 90. cot 270*1*tan 270. tan 90.

= cot 90. cot 270*1*1/ cot 270*1/ cot 90

=1

80. If sin θ+ cos θ = 17/13, 0<θ<900, then the value of sin θ- cos θ is-

a. 5/17

b. 3/19

c. 7/10

d. 7/13

∵ sin θ + cos θ = 17/13

⇒ sin2 θ + cos2 θ + 2sin θ.cos θ = 289/169

∴ 2sin θ.cos θ = 289 /169 -1

= 289 -169/169 = 120/169

∴ sin2 θ + cos2 θ – 2 sin2 θ.cos θ = 1 – 120/169

= 49/169

⇒ (sin θ – cos θ)2 = (7/13)2

∴ sin θ – cos θ = 7/13

v

1. The product of 2 numbers is 1575 and their quotient is 9/7. Then the sum of the numbers is –

a. 74

b. 78

c. 80

d. 90

Ans : Let the numbers be x and y .

∴ xy = 1575

And x/y = 9/7

∴ xy/x/y = 1575/9/7

∴ y2 = 1225

∴ y = 35 and x = 45

∴ The sum of the numbers = 45+35

= 80

2. The value of (81)3.6 * (9)2.7/ (81)4.2 * (3) is __

a. 3

b. 6

c. 9

d. 8.2

Ans : (81)3.6 * (9)2.7/(81)4.2 * (3) = (3)14.4 * (3)5.4/(3)16.8 * (3)

= 314.4+5.4-16.8-1

= 32

= 9

3. √6+√6+√6+… is equal to –

a. 2

b. 5

c. 4

d. 3

Ans : Let √6+√6+√6+…. be x.

∴ x =√6+x

∴ x2 = 6+x

∴ x2 + x – 6 = 0

∴ (x-3) (x+2) = 0

∴ x = 3

4. The sum of the squares of two natural consecutive odd numbers is 394. The sum of the numbers is –

a. 24

b. 32

c. 40

d. 28

Ans : Let the consecutive odd numbers be x and (x+2)

∵ x2 + (x + 2)2 = 394

∴ x2 + x2 + 4x + 4 = 394

∴ 2 x2 +4x – 390 = 0

∴ x2 + 2x – 195 = 0

∴ (x +15) (x-13) = 0

∴ x = 13

∴ Required sum = 13 +15 =28

5. When (6767 +67) is divided by 68, the remainder is-

a. 1

b. 63

c. 66

d. 67

Ans: (6767+ 67) = 67(6766 + 166)

As 66 is an even number

∴ 6766 is an even number

∴ (6766 + 1) is perfectly divisible by (67 + 1)

i.e. 68

∴ The remainder = 67

6. In a division sum, the divisor is 4 times the quotient and twice the remainder. If a and b are respectively the divisor and the dividend then-

a. 4b-a2 /a =3

b. 4b-2a /a2 =2

c. (a+1)2 = 4b

d. A(a+2)/b = 4

Ans : As divisor is a, and dividend is b.

∴ Quotient = a/4

And Remainder = a/2

∴ b = a * a/4 + a/2

∴ 4b = a2 + 2a

∴ a(a+2)/b = 4

7. If 738 A6A is divisible by 11, then the value of A is-

a. 6

b. 3

c. 9

d. 1

Ans : As 738A6 A is divisible by 11.

∴ A + A + 3 = 6 + 8 + 7

∴ A = 9

8. The east positive integer that should be subtracted from 3011 * 3012 so that the difference is perfect square is-

a. 3009

b. 3010

c. 3011

d. 3012

Ans : ∵ 3011 * 3012 = 3011 (3011 + 1)

= (3011)2 + 3011

∴ Required least number = 3011

9. P, Q, R are employed to do a work for Rs. 5750. P and Q together finished 19/23 of work and Q and R together finished 8/23 of work. Wage of Q, in rupees, is-

a. 2850

b. 3750

c. 2750

d. 1000

Ans : Work done by Q = 19/23 + 8/23 – 1

4/23

∴ Wage of Q = 4/23 * 5750

= Rs. 1000

10. A can do a piece of work in 24 day, B in 32 days and C in 64 days. All begin to do it together, but A leaves after 6 days and B leaves 6 days before the completion of the work. How many days did the work last?

a. 15

b. 20

c. 18

d. 30

Ans : Work done by A = 6/24 = ¼

Work done by B = (x-6)/32

(where x is no. of days in which work is competed)

∵ ¼ + x – 6/32 + x/64 = 1

∴ 16 + 24 – 12 + x /64 = 1

∴ 3x + 4 = 64

∴ x = 60/3 = 20 days

11. The square root of (0.75)3 /1-0.75 + [0.75 + 90.75)2 +1] is-

a. 1

b. 2

c. 3

d. 4

Ans : The square root of

[(0.75)3/1 – 0.75 + {0.75 + (0.75)2 + 1}]

= √1.6875 + 2.3125

=√4

= 2

12. Given that √4096 = 64, the value of √4096 + √40.96 +√0.004096 is-

a. 70.4

b. 70.464

c. 71.104

d. 71.4

Ans : Given Exp. = √4096 + √40.96 +√0.004096

= 64 + 6.4 + 0.064

= 70.464

13. By selling an article at 3/4th of the marked price, there is a gain of 25%. The ratio of the marked price and the cost price is-

a. 5 : 3

b. 3 : 5

c. 3 : 4

d. 4 : 3

Ans : Let of M.P. be Rs. x.

∴ S.P. = Rs.3x/4

and C.P. = 3x/4 * 100/125

= Rs.3x/5

∴ required ratio = x: 3x/5

= 5:3

14. A and B earn in the ratio 2:1. They spend in the ratio 5:3 and save in the ratio 4:1. If the total monthly savings of both A and B are Rs.5,000, the monthly income of B is-

a. Rs.7,000

b. Rs.14,000

c. Rs.5,000

d. Rs.10,000

Ans : Let the monthly income of B be Rs. x.

∴ Monthly income of A = Rs. 2x and

Saving of A =5000 * 4/(4 + 1)

= Rs. 4000

Saving of B = Rs.1000

∵ 2x – 4000/x-1000 = 5/3

⇒ 6x – 12000 = 5x – 5000

∴ x = Rs.7000

15. The ratio of the sum of two numbers and their difference is 5:1. The ratio of the greater number to the smaller number is-

a. 2 : 3

b. 3 : 2

c. 5 : 1

d. 1 : 5

Ans : Let the numbers be x and y.

∵ x + y/ x – y = 5/1

⇒ 5x – 5y = x + y

⇒ 4x = 6y

⇒ x/y = 6/4

x : y = 3 : 2

16. A cistern has 3 pipes A, B and C. A and B can fill it in 3 and 4 hours respectively, and C can empty it in 1 hour. If the pipes are opened at 3 p.m., 4 p.m. and 5 p.m. respectively on the same day, the cistern will be empty at-

a. 7.12 p.m.

b. 7.15 p.m.

c. 7.10 p.m.

d. 7.18 p.m.

Ans : Let the cistern be emptied at x p.m.

∵ x -3/3 + x – 4/4 = x - 5/1

⇒ 4x – 12 + 3x – 12/12 = x – 5/1

⇒ 7x – 24 = 12x – 60

⇒ 5x = 36

∴ x = 36/5

= 7 hr. + 12 min.

= 7.12 p.m.

17. If A works alone, he would take 4 days more to complete the job than if both A and B worked together. If B worked alone, he would take 16 days more to complete the job than if A and B work together. How many days would they take to complete the work if both of them worked together?

a. 10 days

b. 12 days

c. 6 days

d. 8 days

Let x days are taken when they work together.

∴ Time taken by A to complete the work = (x + 4) days

And Time taken by B to complete the work = (x + 16) days

∵ 1/x = 1/x + 4 + 1/x + 16

∴ 1/x = x + 16 + x + 4/(x + 4) (x +16)

∴ 2x2 + 20x = x2 + 20x +64

∴ x2 = 64 (8)2

∴ x = 8 days

18. 250 men can finish a work in 20 days working 5 hours a day. To finish the work within 10 days working 8 hours a day, the minimum number of men required is-

a. 310

b. 300

c. 313

d. 312

Ans : Required no. of men = 250 * 5 * 20/8 * 10

=312.5

313

19. 2 men and 5 women can do a work in 12 days. 5 men 2 women can do that work in 9 days. Only 3 women can finish the same work in-

a. 36 days

b. 21 days

c. 30 days

d. 42 days

∵ (2m + 5w) * 12 = (5m + 2w) * 9

⇒ 24m + 60w = 45m + 18w

⇒ 21m = 42w

⇒ 1m = 2w

∴ 2m + 5w = 9w

∴ required no. of days = 9 * 12/3

= 36

20. While selling, a businessman allows 40% discount on the marked price and there is a loss of 30%. If it is sold at the marked price, profit per cent will be –

a. 10%

b. 20%

c. 16 2/3%

d. 16 1/3%

Ans : Let the M.P. be Rs.100.

∴ S.P = (100-40) = Rs.60

and C. P. = 60 * 100/(100-30) = Rs.600/7

∴ Reqd. % profit = 100-600/7 /600/7 * 100%

= (700 – 600)/7 * 600 * 100 * 7%

= 16 2/3%

21. Successive discount of 10% , 20% and 50% will be equivalent to a single discount of-

a. 36%

b. 64%

c. 80%

d. 56%

Ans : Equivalent to a single discount

= [100 - (100 – 10) (100 – 20) (100 – 50)/100*100]%

= [100 - 90*80*50/10000]%

= [100-36]%

= 64%

22. A retailer offers the following discount schemes for buyers on an article-

I. Two successive discounts of 10%

II. A discount of 12% followed by a discount of 8%.

III. Successive discounts of 15% and 5%

IV. A discount of 20%

The selling price will be minimum under the scheme-

a. I

b. II

c. III

d. IV

Ans : From (i) single discount = [10 + 10 – 10*10/100] % = 19%

From (ii) single discount = [12 + 8 -12*8/100] % = 19.04%

From (iii) single discount = [15 + 5 – 15*5/100] % = 19.25%

From (iv) single discount = 20%

∴ The S.P. will be minimum under the scheme IV.

23. Of three numbers, the second is thrice the first and the third number is three-fourth of the first. If the average of the three numbers is 114, the largest number is –

a. 72

b. 216

c. 354

d. 726

Ans : Let the first number be x.

∴ Second number = 3x and Third number = 3x/4

∵ x + 3x + 3x/4 = 3*114

⇒ 19x/4 = 342

∴ x = 342*4/19 = 72

∴ The largest number = 3*72

= 216

24. A car covers 1/5 of the distance from A to B at the speed of 8 km/hour, 1/10 of the distance at 25 km per hour and the remaining at the speed of 20 km per hour. Find the average speed of the whole journey-

a. 12.625 km/hr

b. 13.625 km/hr

c. 14.625 km/hr

d. 15.625 km/hr

Ans : If the whole journey be x km. The total time taken

= (x/5/8 + x/10/25 + 7x/10/20) hrs

= (x/40 + x/250 + 7x/200) hrs

= 25x + 4x + 35x/1000

= 64x/1000 hrs

∴ Average speed = x/64x/1000

= 15.625 km/hr

25. The average of 3 numbers is 154. The first number is twice the second and the second number is twice the third. The first number is-

a. 264

b. 132

c. 88

d. 66

Ans : Let the first number be x.

∴ Second number = x/2 and Third number = x/4

∵ x + x/2 + x/4 = 3 x 154

⇒ 7x/4 = 462

∴ x = 462*4/7

=264

26. The average salary of all the staff in an office of a corporate house is Rs. 5,000. The average salary of the officers is Rs. 14,000 and that of the rest is Rs. 4,000. If the total number of staff is 500, the number of officers is –

a. 10

b. 15

c. 25

d. 50

Ans : Let the number of officers be x.

∵ 5000*500 = 14000x + 4000(500-x)

∴ 2500000 =14000x + 2000000 – 4000x

∴ x = 500000/10000

= 50

27. The average marks of 40 students in an English exam are 72. Later it is found that three marks 64, 62 and 84 were wrongly entered as 60, 65 and 73. The average after mistakes were rectified is-

a. 70

b. 72

c. 71.9

d. 72.1

Ans : Correct average

= 40*72 + (64 + 62 +84) – 68 – 65 – 73/40

= 2880 + 210 – 206/40 = 2884/40

= 72.1

28. A and B are two alloys of gold and copper prepared by mixing metals in the ratio 7:2 and 7:11 respectively. If equal quantities of the alloys are melted to form a third alloy C, the ratio of gold and copper in C will be-

a. 5 : 7

b. 5 : 9

c. 7 : 5

d. 9 : 5

Ans : Quantity of gold in A = 7/9* wt. of A

Quantity of gold in B = 7/18* wt. of B

If 1 kg of each A and B are mixed to form third alloys C.

Then quantity of gold in 2 kg C = 7/9 + 7/18

= 7/6 kg

And Quantity of copper in 2 kg C = 2 – 7/6

= 5/6 kg

∴ Required ratio = 7/6 : 5/6 = 7 : 5

29. In a laboratory, two bottles contain mixture of acid and water in the ratio 2 : 5 in the first bottle and 7 : 3 in the second. The ratio in which the contents of these two bottles be mixed such that the new mixture has acid and water in the ratio 2 : 3 is-

a. 4 : 15

b. 9 : 8

c. 21 : 8

d. 1 : 2

Ans : Quantity of acid in first bottle = 2/7 x mix.

and Quantity of acid in second bottle = 7/10 x mix.

If x and 1 volumes are taken from I and II bottle respectively to form new mixture.

Then, (2/7 x + 7/10 * 1)/(5x/7 +3/10 * 1) = 2/3

⇒ 6x/7 21/10 = 10x/7 +6/10

⇒ 4x/7 = 15/10

∴ x = 15/10*7/4 = 21/8

∴ Required ratio = x : 1

= 21 : 8

30. A mixture contains 80% acid and rest water. Part of the mixture that should be removed and replaced by same amount of water to make the ratio of acid and water 4 : 3 is-

a. 1/3 rd

b. 3/7 th

c. 2/3 rd

d. 2/7 th

Ans : Let the initial wt. of mixture be 1 kg and x kg of mixture is taken out and replaced by same amount of water.

∵ Amt. of acid/Amt. of water = 0.8 – 0.8x/ (0.2 – 0.2x + x) = 4/3

⇒ 2.4 – 2.4x = 0.8 + 3.2x

⇒ 5.6x = 1.6

∴ x = 1.6/5.6 = 2/7th part

31. An employer reduces the number of his employees in the ratio 9 : 8 and increases their wages in the ratio 14 : 15. If the original wage bill was Rs. 189,900, find the ratio in which the wage bill is decreased-

a. 20 : 21

b. 21 : 20

c. 20 : 19

d. 19 : 21

Ans : Let the initial number of employees be 9x and the employer gives Rs. 14y as wage to each.

∵ 9x * 14y =18900

∴ xy = 150 and The later bill = 8x*15y = 120xy

= 120*150 = 18000

∴ Required ratio = 18000 : 18900

= 20 : 21

32. The batting average for 40 innings of a cricketer is 50 runs. His highest score exceeds his lowest score by 172 runs. If these two innings are excluded, the average of the remaining 38 innings is 48 runs. The highest score of the player is-

a. 165

b. 170

c. 172

d. 174

Ans : Let the max. number of runs be x.

∴ The lowest score = (x-172)

∵ 40*50 = 38*48 + x + (x-172)

⇒ 2000 = 1824 + 2x – 172

∴ x= 174

33. Walking at 3 km per hour, Pintu reaches his school 5 minutes late. If he walks at 4 km per hour he will be 5 minutes early. The distance of Pintu’s school from his house is-

a. 1 ½ Km

b. 2 Km

c. 2 ½ Km

d. 5 Km

Ans : Distance of the school from the house

= 4*3/(4 – 3) * 5 +5/60 km

= 12*1/6

=2 km

34. Nitin bought some oranges at Rs. 40 a dozen and an equal number at Rs.30 a dozen. He sold them at Rs. 45 a dozen and made a profit of Rs.480. The number of oranges, he bought, was-

a. 48 dozen

b. 60 dozen

c. 72 dozen

d. 84 dozen

Ans : Let the number of oranges bought be x.

∴ 45x/12 – 70x/24 = 480

⇒ (45-35)/12 = 480

⇒ x = 480*12/10 = 576

= 48*12 = 48 dozen

35. A man buys two chairs for a total cost of Rs.900. By selling one for 4/5 of its cost and the other for 5/4 of its cost, he makes a profit of Rs.90 on the whole transaction. The cost of the lower priced chair is-

a. Rs.360

b. Rs.400

c. Rs.420

d. Rs.300

Ans : Let the cost price of 1 chair be Rs. x.

∴ C.P. of other chair = Rs. (900-x)

∵ 4/5 x + 5/4 (900-x) = 900 + 90

⇒ 4/5 x +1125 -5x/4 = 990

∴ 9x/20 = 135

∴ x = 135*20/9

= Rs. 300

∴ C.P. of the lower priced chair is Rs. 300.

36. By selling 100 oranges, a vendor gains the selling price of 20 oranges. His gain per cent is-

a. 20

b. 25

c. 30

d. 32

Ans : Let the S.P. of 100 Oranges be Rs. x.

∴ S.P. of 20 oranges = x/100 *20 = Rs. x/5

∴C.P. of 100 oranges = x - x/5

= Rs. 4x/5

∴ Reqd. Profit % = x/5 * 100*5/4x %

= 25%

37. 60% of the cost price of an article is equal to 50% of its selling price. Then the percentage of profit or loss on the cost price is-

a. 20% loss

b. 16 2/3% profit

c. 20% profit

d. 10% loss

Ans : Let the cost price be Rs. 100.

∵ S.P * 50/100 = 100*16/100

∴ S.P = 60*100/50

= Rs. 120

∴ Reqd. % profit = (120 – 100)% = 20%

38. Maninder bought two horses at Rs.40,000 each. He sold one horse at 15% gain, but had to sell the second horse at a loss. If he hand suffered a loss of Rs.3,600 on the whole transaction, then the selling price of the second horse is-

a. Rs.30,000

b. Rs.30,200

c. Rs.30,300

d. Rs.30,400

Ans: C.P. of two horses = 2*40000 = Rs. 80000

and S.P. of two horse = 80000 – 3600 = Rs. 76400

∴ S.P. of the other horse = 76400 – 46000 = 30400

39. A fruit-seller buys x guavas for Rs.y and sells y guavas for Rs. x. If x>y, then he made-

a. x2 – y2 / xy % loss

b. x2 – y2 / xy % gain

c. x2 – y2 / y2 % loss

d. x2 – y2 / y2 * 100% gain

Ans : C.P. of 1 guava = Rs. y/x [x>y]

and S.P. of 1 guava = Rs. x/y

∴ Reqd. Gain% = x/y – y/x/ y/x * 100%

= x2 - y2 /y2 *100%

40. A jar contain 10 red marbles and 30 green ones. How many red marbles must be added to the jar so that 60% of the marbles will be red?

a. 25

b. 30

c. 35

d. 40

Ans : Let after adding x red marbles, the red marbles with be 60% of the total.

∵ (10+x)/ (10 + x) + 30 = 60/100

⇒ 10 + x/40 + x = 3/5

⇒ 50 + 5x = 120 + 3x

x = 70/2 = 35

41. If a number multiplied by 25% of itself gives a number which is 200% more than the number, then the number is –

a. 12

b. 16

c. 35

d. 24

Ans : Let the number be x.

∵ x*25x/100 = x + 200x/100

⇒ x2/4 = 3x

⇒ x2 - 12x = 0

⇒ x – 12 = 0

∴ x = 12

42. The value of an article depreciates every year at the rate of 10% of its value. If the present value of the article is Rs.729, then its worth 3 years ago was-

a. Rs.1250

b. Rs.1000

c. Rs.1125

d. Rs.1200

Ans : Let the worth 3 years ago be Rs. x.

∵ 729 = x (1 – 10/100)3

⇒ 729 = x*9*9*9/10*10*10

∴ x = Rs. 1000

43. The price of onions has been increased by 50%. In order to keep the expenditure on onions the same the percentage of reduction in consumption has to be-

a. 50%

b. 33 1/3%

c. 33%

d. 30%

Ans : Reqd. Percentage of reduction

= 50*100/(100 + 50) %

= 5000/150 %

= 33 1/3%

44. A took two loans altogether of Rs.1200 from B and C. B claimed 14% simple interest per annum, while C claimed 15% per annum. The total interest paid by A in one year was Rs.172. Then, A borrowed-

a. Rs.800 from C

b. Rs.625 from C

c. Rs.400 from B

d. Rs.800 from B

Ans : If A borrowed Rs. x from B. and A borrowed Rs. Rs. (1200 – x) from C.

∵ (1200 – x)*15*1/100 + x*14*1/100

⇒ 18000 – 15x + 14x = 172*100

x = Rs. 800

45. If a regular polygon has each of its angles equal to 3/5 times of two right angles, then the number of side is-

a. 3

b. 5

c. 6

d. 8

Ans : If the number of sides a regular polygon be n.

Then (2n-4)/n = 2*3/5

⇒ (2n – 4)*5 = 6n

∴ n = 5

46. A square is of area 200 sq. m. A new square is formed in such a way that the length of its diagonal is √2 times of the diagonal of the given square. The the area of the new square formed is-

a. 200√2 sq.m

b. 400√2 sq.m

c. 400 sq.m

d. 800 sq.m

Ans : Length of the diagonal of Ist square

= √2*200

= 20 m

∴ Length of the diagonal of new square = 20√2m

∴ Area of the new square = ½*(20.√2)2 = 400 sq. m

47. The heights of a cone, cylinder and hemisphere are equal. If their radii are in the ratio 2 : 3 : 1, then the ratio of their volumes is-

a. 2 : 9 : 2

b. 4 : 9 : 1

c. 4 : 27 : 2

d. 2 : 3 : 1

Ans : Ratio of their volumes [cone : cylinder : hemisphere]

= 1/3(2)2h : (3)2*h : 2/3(1)2.h

= 4/3 : 9 : 2/3 = 4 : 27 : 2

48. A motor-boat can travel at 10 km/hr in still water. It travelled 91 km downstream in a river and then returned to the same place, taking altogether 2 hours. Find the rate of flow of river-

a. 3 km/hr

b. 4 km/hr

c. 2 km/hr

d. 5 km/hr

Ans : Let the rate of flow of river be x km/hr.

∵ 91/(10 + x) + 91/(10 – x) = 20

⇒ 91(10 – x + 10 + x)(10 + x) (10 – x) = 20

⇒ 91*20 = 20(100 – x2)

⇒ x2 = 9 = (3)2

∴ x = 3 km/hr

49. A man driving at 3/4th of his original speed reaches his destination 20 minutes later than the usual time. Then the usual time is-

a. 45 minutes

b. 60 minutes

c. 75 minutes

d. 120 minutes

Ans : Let the original speed be x km/hr and the usual time be y hours.

∵ x * y = ¾ x(y+1/3)

∴ 4y = 3y + 1

∴ y = 1 hr = 60 minutes

50. A motor-boat, travelling at the same speed, can cover 25 km upstream and 39 km downstream in 8 hours. At the same speed, it can travel 35 km upstream and 52 km downstream in 11 hours. The speed of the stream is –

a. 2 km/hr

b. 3 km/hr

c. 4 km/hr

d. 5 km/hr

Ans : Let the speeds of motor boat and the stream be x and y km/hr respectively.

∵ 39/x + y + 25/x – y = 8 …(1)

and 52/x + y + 35/x – y = 11 …(2)

Solving equations (1) and (2), we get-

∴ 100 - 105/x – y = 32 – 33

∴ x – y = 5

and x + y = 13

∴ y = 4 km/hr

51. If a sum of money placed at compound interest, compounded annually, doubles itself in 5 years, then the same amount of money will be 8 times of itself in-

a. 25 years

b. 20 years

c. 15 years

d. 10 years

Ans : Required time = 5log 8/log 2

= 5*3log 2/log 2

= 15 years

52. A person has left an amount of Rs.1,20,000 to be divided between his 2 son aged 14 years and 12 years such that they get equal amounts when each attains 18 years of age. If the amount gets a simple interest of 5% per annum, the younger son’s share at present is-

a. Rs.48,800

b. Rs.57,600

c. Rs.62,400

d. Rs.84,400

Ans : Let the present share of the younger son be Rs. x.

∴ The share of the elder son = Rs. (120000 – x)

∵ x + x*6*5/100

= (120000 – x) + (120000 – x)*4*5/100

⇒ 130x/100 = (120000 –x)*120/100

⇒ 13x = 1440000 – 12x

⇒ 25x = 1440000

∴ x = Rs. 57600

53. If the simple interest on Rs.x at a rate of a% for m years is same as that on Rs. y at a rate of a2% for m2 years, then x : y is equal to-

a. m : a

b. am : 1

c. 1/m : 1/a

d. 1/am : 1

∵ x*a*m/100 = y*a2*m2/100

⇒ x/y = am/1

∴ x : y = am : 1

54. Base of a right prism is an equilateral triangle of side 6 cm. If the volume of the prism is 108√3 cc, its height is-

a. 9 cm

b. 10 cm

c. 11 cm

d. 12 cm

Ans : Height of the prism = 108√3*4/√3*(6)2 = 12cm

55. If a + 1/a + 2 = 0, then the value of (a37 – 1/a100) is-

a. 0

b. -2

c. 1

d. 2

Ans : ∵ a + 1/a + 2 = 0

⇒ a2 + 1 +2a = 0

⇒ (a + 1)2 = 0

⇒ a +1 = 0

∴ a = -1

∴ a37 -1/a100 = (-1) – (1) = -2

56. The value of k for which the graphs of (k-1) x+y-2 = 0 and (2-k) x -3y + 1 = 0 are parallel is-

a. ½

b. -1/2

c. 2

d. -2

Ans : (k – 1) x + y – 2 = 0

∴ y = (1 – k) x + 2 ….(1)

and (2 –k) x – 3y – 1 = 0

3y = (2 – k) x +1

Y = 2 – k/3 x + 1/3 ….(2)

∵ m1 = m2

⇒ 1 – k = 2 – k/3

⇒ 3 – 3k = 2 - k

∴ k = 1/2

57. If a2 + b2 + c2 = 2 (a-b-c) – 3, then the value of (a – b + c) is-

a. -1

b. 3

c. 1

d. -2

Ans : a2 + b2 + c2 = 2(a – b – c) – 3

⇒ a2 - 2a + 1 + b2 + 2b + 1 + c2 + 2c + 1 = 0

⇒ (a – 1)2 + (b + 1)2 + (c + 1)2 = 0

⇒ a – 1 = 0, b + 1 = 0, c + 1 = 0

⇒ a = 1, b = -1, c = -1

∴ a + b – c = 1 – 1 + 1 =1

58. If x2 + 3x + 1 = 0, then the value of x3 + 1/x3 is-

a. -18

b. 18

c. 36

d. -36

Ans : ∵ x2 + 3x + 1 = 0

⇒ x + 3 + 1/x = 0

⇒ x + 1/x = -3

⇒ (x + 1/x)3 = (-3)3

⇒ x3 + 1/x3 + 3 (-3) = -27

∴ x3 + 1/ x3 = -18

59. If xa, xb, xc = 1, then the value of a3 + b3 + c3 is –

a. 9

b. abc

c. a + b + c

d. 3abc

Ans : ∵ xa. xb. xc = 1

⇒ xa + b + c = x0

⇒ a +b + c =0

∴ a3 + b3 + c3 = 3abc

60. Base of a right pyramid is a square, length of diagonal of the base is 24√2 m. If the volume of the pyramid is 1728 cu.m, its height is-

a. 7 m

b. 8 m

c. 9 m

d. 10 m

Ans : Area of the base of the pyramid

= ½ (24√2)2 = 576m2

∴ Height of pyramid + 1728*3/576 = 9m

61. The height of a right circular cone and the radius of its circular base are respectively 9 cm and 3 cm. The cone is cut by a plane parallel to its base so as to divide it into two parts. The volume of the frustum (i.e., the lower part) of the cone is 44 cubic cm. The radius of the upper circular surface of the frustum (taking = 22-7) is-

a. 3√12 cm

b. 3√13 cm

c. 3√6 cm

d. 3√20 cm

Ans : Let the radius of the upper circular part of the frustum be r cm.

(Picture)

Then r/3 = x/9 = AC/AD

∴ x = 3r (where AC = x cm)

⇒ /3h(r12 + r1r2+r22 = 44

⇒ /3(9 – x)*(9 + 3r + r2) = 44

⇒ (9 – x)*(9 + 3r + r2) = 44*3*7/22 = 42

⇒ 81 + 27r + 9r2 – 9x – 3rx - r2x = 42 On putting x = 3r,

⇒ 81 + 27r + 9r2 - 9r2 - 3r3 – 27r = 42

⇒ 3r3 = 39

∴ r = 3√13 cm

62. The ratio of radii of two right circular cylinder is 2 : 3 and their heights are in the ratio 5 : 4. The ratio of their curved surface area is-

a. 5 : 6

b. 3 : 4

c. 4 : 5

d. 2 : 3

Ans : Required ratio = 2*2r*5h/2*3r*4h = 5 : 6

63. A solid cylinder has total surface area of 462 sq.cm. Curved surface area is 1/3rd of its total surface area. The volume of the cylinder is-

a. 530 cm3

b. 536 cm3

c. 539 cm3

d. 545 cm3

Ans : ∵ 2r(r + h) = 462

and 2rh = 1/3*462 = 154

∴ r + h/h = 462/154 =3

∴ r + h = 3h

∴ r = 2h

∴ 2*2h2 = 154

∴ h2 = 154*7/22*4 = 49/4

= (7/2)2

∴ h = 7/2 cm

and r = 7 cm

∴ Volume of the cylinder

= 22/7*49*7/2

= 539 cm3

64. A cylinder and a cone have equal radii of their bases and equal heights. If their curved surface areas are in the ratio 8 : 5, the ratio of their radius and height is-

a. 1 : 2

b. 1 : 3

c. 2 : 3

d. 3 : 4

Ans : Let the radius and height of each are r and h respectively.

∵ 2rh/r√h2 + r2 = 8/5

∴ 10h = 8√r2 + h2

⇒ 100h2 =64r2 + 64h2

∴ h2 = 64 r2/36 = (4/3 r)2

⇒ h = 4/3r

∴ r : h = 3 : 4

65. A solid is hemispherical at the bottom and conical above. If the surface areas of the two parts are equal, then the ratio of radius and height of its conical part is-

a. 1 : 3

b. 1 :1

c. √3 : 1

d. 1 : √3

Ans :

∵ rl = 2r2

⇒ l = 2r

⇒ √h2 + r2 = 2r

⇒ h2 = 3r2

∴ r : h = r/h

= 1 : √3

66. If O is the circumcentre of ∆ ABC and ∠OBC = 350, then the ∠BAC is equal to-

a. 550

b. 1100

c. 700

d. 350

Ans : ∴ ∠BOC = 1800 – (350 + 350) = 1100

∴ ∠BAC = 1/2 * 1100 = 550

67. If I is the incentre of ∆ ABC and ∠ BIC = 1350, then ∆ ABC is-

a. Acute angled

b. Equilateral

c. Right angled

d. Obtuse angled

Ans :

∠BIC = 1350

⇒ B/2 + C/2 = 1800 – 1350 = 450

⇒ ∠B + ∠C = 900

∴ ∠A = 1800 – (∠B + ∠C) = 900

i.e., ∆ ABC is a right angled.

68. If sin2 (∝ + β/2) is-

a. 1

b. -1

c. 0

d. 0.5

∵ sin2 ∝ + sin2 β = 2

⇒ 1 – cos2 ∝ 1 - cos2 β = 2

⇒ cos2 ∝ cos2 β = 0

⇒ cos ∝ = 0

and cos β = 0

⇒ ∝ = /2 and β = /2

∴ cos (∝ + β/2) = cos [/2 + /2 / 2]

= cos /2 = 0

69. The length of a shadow of a vertical tower is 1/√3 times its height. The angle of elevation of the Sun is-

a. 300

b. 450

c. 600

d. 900

Ans :

∵ tan θ = h/1√3 h = √3 = tan 600

70. The graphs of x +2y =3 and 3x-2y = 1 meet the Y-axis at two points having distance-

a. 8/3 units

b. 4/3 units

c. 1 unit

d. 2 units

Ans : When the graphs meet the Y-axis at two points.

Then, [x + 2y = 3] at x = 0 ⇒ [0, y1 =3/2]

[3x - 2y = 1] at x = 0

and i.e., [0, y2 = -1/2]

Required distance = (y1 – y2)

= 3/2 – (- ½) = 2 units

71. If x+1/16x = 1, then the value of 64x3 + 1/64x3 is-

a. 4

b. 52

c. 64

d. 76

∵ x + 1/16x = 1

⇒ 16x2 – 16x + 1 =0

⇒ 16x2 – 16x + 4 = 3

⇒ (4x – 2)2 = 3

⇒ 4x = 2 + √3

⇒ 64x3 = (2±√3)3

=8 + 3√3 + 6√3 (2 + √3)

= 26 + 15√3

∴ 64x3 + 1/64x3 = (26 + 15√3) + 1/ (26 + 15√3)

= (26 + 15√3) + 26 - 15√3/676 -675

=52

72. If a, b, c, are three non-zero real numbers such that a + b + c = 0, and b2 ≠ ca, then the value of a2 + b2 + c2/ b2 –ca is-

a. 3

b. 2

c. 0

d. 1

∵ a + b + c = 0

⇒ a + c = -b

⇒ a2 + c2 = b2 -2ac

⇒ a2 + b2 + c2 = 2b2 – 2ac

∴ a2 + b2 + c2/ b2 ac = 2

73. If a4 + a2 b2 + b4 = 8 and a2 + ab + b2 = 4, then the value of ab is-

a. -1

b. 0

c. 2

d. 1

∵ a4 + a2 b2 + b4/ a2 + ab + b2 = 8/4

⇒ (a2 + b2)2 – (ab) 2/ (a2 + b2 + ab) = 2

⇒ a2 – ab + b2 = 2 ….(1)

and a2 + ab + b2 = 4 …..(2)

⇒ 2ab = 2

⇒ ab = 1

74. If a = 25, b = 15, c = -10, then the value of a3 + b3 + c3 – 3abc/ (a-b)2 + (b-c)2 + (c-a)2 is-

a. 30

b. -15

c. -30

d. 15

∵ a3 + b3 + c3 – 3abc

= (25)3 + (15)3 + (-10)3 – 3*25*15*(-10)

=15625 + 3375 – 1000 + 11250 = 29250

and (a - b)2 + (b – c)2 + (c – a)2

= (10)2 + (25)2 + (-35)2

= (10)2 + 625 + 1225

= 1950

∴ Required value = 29250/1950 =15

75. A, B, C are three points on a circle. The tangent at A meets BC produced at T, ∠BTA = 400, ∠CAT = 440. The angle subtended by BC at the centre of the circle is-

a. 840

b. 920

c. 960

d. 1040

Ans : (Picture)

∠ACB = 400 + 440 = 840

∴ ∠ACO = 900 - 440 = 460 = ∠OAC

⇒ ∠OCB = ∠ACB - ∠ACO

= 840 - 460 = 380 = ∠OBC

∴ ∠BOC = 1800 – (∠OCB + ∠OBC)

= 1800 – (380 + 380) = 1040

76. If the length of a chord of a circle at a distance of 12 cm from the Centre is 10 cm, then the diameter of the circle is-

a. 13 cm

b. 15 cm

c. 26 cm

d. 30 cm

Ans : (Picture)

∵ OA = √OM2 + AM2

= √122 + 52 = 13

∴ Diameter of the circle = 2*OA

= 2*13 = 26cm

77. In ∆ ABC, P and Q are the middle points of the sides AB and AC respectively. R is a point on the segment PQ such that PR : RQ = 1 : 2. If PR = 2 cm, then BC =

a. 4 cm

b. 2 cm

c. 12 cm

d. 6 cm

∵ PR/RQ = ½

But, PR = 2cm

RQ = 2*PR

= 4cm

(Picture)

∴ PQ = PR + RQ

= 2 + 4 = 6 cm

∴ BC = 2*PQ = 12CM

78. If tan θ tan 2θ = 1, then the value of sin2 2θ + tan2 is equal to –

a. ¾

b. 10/3

c. 3 ¾

d. 3

∵ tan θ * tan 2θ = 1

⇒ tan θ * 2 tan θ/1 – tan2 θ = 1

⇒ 2 tan2 θ = 1 - tan2 θ

⇒ 3 tan2 θ = 1

⇒ tan θ = 1/√3 = tan 300

∴ θ = 300

∴ sin2 2θ + tan2 2θ = sin2 600 + tan2 600

= ¾ + 3

=3 ¾

79. The value of cot /20 cot 3/20 cot 5/20 cot 7/20 cot 9/20 is-

a. -1

b. ½

c. 0

d. 1

Ans : Given Exp.

= cot /20. cot 3/20. cot 5/20. cot 7/20. cot 9/20

= cot 90. cot 270. cot 450. cot 630. cot 810

= cot 90. cot 270*1*tan 270. tan 90.

= cot 90. cot 270*1*1/ cot 270*1/ cot 90

=1

80. If sin θ+ cos θ = 17/13, 0<θ<900, then the value of sin θ- cos θ is-

a. 5/17

b. 3/19

c. 7/10

d. 7/13

∵ sin θ + cos θ = 17/13

⇒ sin2 θ + cos2 θ + 2sin θ.cos θ = 289/169

∴ 2sin θ.cos θ = 289 /169 -1

= 289 -169/169 = 120/169

∴ sin2 θ + cos2 θ – 2 sin2 θ.cos θ = 1 – 120/169

= 49/169

⇒ (sin θ – cos θ)2 = (7/13)2

∴ sin θ – cos θ = 7/13

S.S.C. Combined Graduate Level Exam

(Tier-II) 2011

Solved Paper – I

Arithmetical Ability

1. The product of 2 numbers is 1575 and their quotient is 9/7. Then the sum of the numbers is –

a. 74

b. 78

c. 80

d. 90

Ans : Let the numbers be x and y .

∴ xy = 1575

And x/y = 9/7

∴ xy/x/y = 1575/9/7

∴ y2 = 1225

∴ y = 35 and x = 45

∴ The sum of the numbers = 45+35

= 80

2. The value of (81)3.6 * (9)2.7/ (81)4.2 * (3) is __

a. 3

b. 6

c. 9

d. 8.2

Ans : (81)3.6 * (9)2.7/(81)4.2 * (3) = (3)14.4 * (3)5.4/(3)16.8 * (3)

= 314.4+5.4-16.8-1

= 32

= 9

3. √6+√6+√6+… is equal to –

a. 2

b. 5

c. 4

d. 3

Ans : Let √6+√6+√6+…. be x.

∴ x =√6+x

∴ x2 = 6+x

∴ x2 + x – 6 = 0

∴ (x-3) (x+2) = 0

∴ x = 3

4. The sum of the squares of two natural consecutive odd numbers is 394. The sum of the numbers is –

a. 24

b. 32

c. 40

d. 28

Ans : Let the consecutive odd numbers be x and (x+2)

∵ x2 + (x + 2)2 = 394

∴ x2 + x2 + 4x + 4 = 394

∴ 2 x2 +4x – 390 = 0

∴ x2 + 2x – 195 = 0

∴ (x +15) (x-13) = 0

∴ x = 13

∴ Required sum = 13 +15 =28

5. When (6767 +67) is divided by 68, the remainder is-

a. 1

b. 63

c. 66

d. 67

Ans: (6767+ 67) = 67(6766 + 166)

As 66 is an even number

∴ 6766 is an even number

∴ (6766 + 1) is perfectly divisible by (67 + 1)

i.e. 68

∴ The remainder = 67

6. In a division sum, the divisor is 4 times the quotient and twice the remainder. If a and b are respectively the divisor and the dividend then-

a. 4b-a2 /a =3

b. 4b-2a /a2 =2

c. (a+1)2 = 4b

d. A(a+2)/b = 4

Ans : As divisor is a, and dividend is b.

∴ Quotient = a/4

And Remainder = a/2

∴ b = a * a/4 + a/2

∴ 4b = a2 + 2a

∴ a(a+2)/b = 4

7. If 738 A6A is divisible by 11, then the value of A is-

a. 6

b. 3

c. 9

d. 1

Ans : As 738A6 A is divisible by 11.

∴ A + A + 3 = 6 + 8 + 7

∴ A = 9

8. The east positive integer that should be subtracted from 3011 * 3012 so that the difference is perfect square is-

a. 3009

b. 3010

c. 3011

d. 3012

Ans : ∵ 3011 * 3012 = 3011 (3011 + 1)

= (3011)2 + 3011

∴ Required least number = 3011

9. P, Q, R are employed to do a work for Rs. 5750. P and Q together finished 19/23 of work and Q and R together finished 8/23 of work. Wage of Q, in rupees, is-

a. 2850

b. 3750

c. 2750

d. 1000

Ans : Work done by Q = 19/23 + 8/23 – 1

4/23

∴ Wage of Q = 4/23 * 5750

= Rs. 1000

10. A can do a piece of work in 24 day, B in 32 days and C in 64 days. All begin to do it together, but A leaves after 6 days and B leaves 6 days before the completion of the work. How many days did the work last?

a. 15

b. 20

c. 18

d. 30

Ans : Work done by A = 6/24 = ¼

Work done by B = (x-6)/32

(where x is no. of days in which work is competed)

∵ ¼ + x – 6/32 + x/64 = 1

∴ 16 + 24 – 12 + x /64 = 1

∴ 3x + 4 = 64

∴ x = 60/3 = 20 days

11. The square root of (0.75)3 /1-0.75 + [0.75 + 90.75)2 +1] is-

a. 1

b. 2

c. 3

d. 4

Ans : The square root of

[(0.75)3/1 – 0.75 + {0.75 + (0.75)2 + 1}]

= √1.6875 + 2.3125

=√4

= 2

12. Given that √4096 = 64, the value of √4096 + √40.96 +√0.004096 is-

a. 70.4

b. 70.464

c. 71.104

d. 71.4

Ans : Given Exp. = √4096 + √40.96 +√0.004096

= 64 + 6.4 + 0.064

= 70.464

13. By selling an article at 3/4th of the marked price, there is a gain of 25%. The ratio of the marked price and the cost price is-

a. 5 : 3

b. 3 : 5

c. 3 : 4

d. 4 : 3

Ans : Let of M.P. be Rs. x.

∴ S.P. = Rs.3x/4

and C.P. = 3x/4 * 100/125

= Rs.3x/5

∴ required ratio = x: 3x/5

= 5:3

14. A and B earn in the ratio 2:1. They spend in the ratio 5:3 and save in the ratio 4:1. If the total monthly savings of both A and B are Rs.5,000, the monthly income of B is-

a. Rs.7,000

b. Rs.14,000

c. Rs.5,000

d. Rs.10,000

Ans : Let the monthly income of B be Rs. x.

∴ Monthly income of A = Rs. 2x and

Saving of A =5000 * 4/(4 + 1)

= Rs. 4000

Saving of B = Rs.1000

∵ 2x – 4000/x-1000 = 5/3

⇒ 6x – 12000 = 5x – 5000

∴ x = Rs.7000

15. The ratio of the sum of two numbers and their difference is 5:1. The ratio of the greater number to the smaller number is-

a. 2 : 3

b. 3 : 2

c. 5 : 1

d. 1 : 5

Ans : Let the numbers be x and y.

∵ x + y/ x – y = 5/1

⇒ 5x – 5y = x + y

⇒ 4x = 6y

⇒ x/y = 6/4

x : y = 3 : 2

16. A cistern has 3 pipes A, B and C. A and B can fill it in 3 and 4 hours respectively, and C can empty it in 1 hour. If the pipes are opened at 3 p.m., 4 p.m. and 5 p.m. respectively on the same day, the cistern will be empty at-

a. 7.12 p.m.

b. 7.15 p.m.

c. 7.10 p.m.

d. 7.18 p.m.

Ans : Let the cistern be emptied at x p.m.

∵ x -3/3 + x – 4/4 = x - 5/1

⇒ 4x – 12 + 3x – 12/12 = x – 5/1

⇒ 7x – 24 = 12x – 60

⇒ 5x = 36

∴ x = 36/5

= 7 hr. + 12 min.

= 7.12 p.m.

17. If A works alone, he would take 4 days more to complete the job than if both A and B worked together. If B worked alone, he would take 16 days more to complete the job than if A and B work together. How many days would they take to complete the work if both of them worked together?

a. 10 days

b. 12 days

c. 6 days

d. 8 days

Let x days are taken when they work together.

∴ Time taken by A to complete the work = (x + 4) days

And Time taken by B to complete the work = (x + 16) days

∵ 1/x = 1/x + 4 + 1/x + 16

∴ 1/x = x + 16 + x + 4/(x + 4) (x +16)

∴ 2x2 + 20x = x2 + 20x +64

∴ x2 = 64 (8)2

∴ x = 8 days

18. 250 men can finish a work in 20 days working 5 hours a day. To finish the work within 10 days working 8 hours a day, the minimum number of men required is-

a. 310

b. 300

c. 313

d. 312

Ans : Required no. of men = 250 * 5 * 20/8 * 10

=312.5

313

19. 2 men and 5 women can do a work in 12 days. 5 men 2 women can do that work in 9 days. Only 3 women can finish the same work in-

a. 36 days

b. 21 days

c. 30 days

d. 42 days

∵ (2m + 5w) * 12 = (5m + 2w) * 9

⇒ 24m + 60w = 45m + 18w

⇒ 21m = 42w

⇒ 1m = 2w

∴ 2m + 5w = 9w

∴ required no. of days = 9 * 12/3

= 36

20. While selling, a businessman allows 40% discount on the marked price and there is a loss of 30%. If it is sold at the marked price, profit per cent will be –

a. 10%

b. 20%

c. 16 2/3%

d. 16 1/3%

Ans : Let the M.P. be Rs.100.

∴ S.P = (100-40) = Rs.60

and C. P. = 60 * 100/(100-30) = Rs.600/7

∴ Reqd. % profit = 100-600/7 /600/7 * 100%

= (700 – 600)/7 * 600 * 100 * 7%

= 16 2/3%

21. Successive discount of 10% , 20% and 50% will be equivalent to a single discount of-

a. 36%

b. 64%

c. 80%

d. 56%

Ans : Equivalent to a single discount

= [100 - (100 – 10) (100 – 20) (100 – 50)/100*100]%

= [100 - 90*80*50/10000]%

= [100-36]%

= 64%

22. A retailer offers the following discount schemes for buyers on an article-

I. Two successive discounts of 10%

II. A discount of 12% followed by a discount of 8%.

III. Successive discounts of 15% and 5%

IV. A discount of 20%

The selling price will be minimum under the scheme-

a. I

b. II

c. III

d. IV

Ans : From (i) single discount = [10 + 10 – 10*10/100] % = 19%

From (ii) single discount = [12 + 8 -12*8/100] % = 19.04%

From (iii) single discount = [15 + 5 – 15*5/100] % = 19.25%

From (iv) single discount = 20%

∴ The S.P. will be minimum under the scheme IV.

23. Of three numbers, the second is thrice the first and the third number is three-fourth of the first. If the average of the three numbers is 114, the largest number is –

a. 72

b. 216

c. 354

d. 726

Ans : Let the first number be x.

∴ Second number = 3x and Third number = 3x/4

∵ x + 3x + 3x/4 = 3*114

⇒ 19x/4 = 342

∴ x = 342*4/19 = 72

∴ The largest number = 3*72

= 216

24. A car covers 1/5 of the distance from A to B at the speed of 8 km/hour, 1/10 of the distance at 25 km per hour and the remaining at the speed of 20 km per hour. Find the average speed of the whole journey-

a. 12.625 km/hr

b. 13.625 km/hr

c. 14.625 km/hr

d. 15.625 km/hr

Ans : If the whole journey be x km. The total time taken

= (x/5/8 + x/10/25 + 7x/10/20) hrs

= (x/40 + x/250 + 7x/200) hrs

= 25x + 4x + 35x/1000

= 64x/1000 hrs

∴ Average speed = x/64x/1000

= 15.625 km/hr

25. The average of 3 numbers is 154. The first number is twice the second and the second number is twice the third. The first number is-

a. 264

b. 132

c. 88

d. 66

Ans : Let the first number be x.

∴ Second number = x/2 and Third number = x/4

∵ x + x/2 + x/4 = 3 x 154

⇒ 7x/4 = 462

∴ x = 462*4/7

=264

26. The average salary of all the staff in an office of a corporate house is Rs. 5,000. The average salary of the officers is Rs. 14,000 and that of the rest is Rs. 4,000. If the total number of staff is 500, the number of officers is –

a. 10

b. 15

c. 25

d. 50

Ans : Let the number of officers be x.

∵ 5000*500 = 14000x + 4000(500-x)

∴ 2500000 =14000x + 2000000 – 4000x

∴ x = 500000/10000

= 50

27. The average marks of 40 students in an English exam are 72. Later it is found that three marks 64, 62 and 84 were wrongly entered as 60, 65 and 73. The average after mistakes were rectified is-

a. 70

b. 72

c. 71.9

d. 72.1

Ans : Correct average

= 40*72 + (64 + 62 +84) – 68 – 65 – 73/40

= 2880 + 210 – 206/40 = 2884/40

= 72.1

28. A and B are two alloys of gold and copper prepared by mixing metals in the ratio 7:2 and 7:11 respectively. If equal quantities of the alloys are melted to form a third alloy C, the ratio of gold and copper in C will be-

a. 5 : 7

b. 5 : 9

c. 7 : 5

d. 9 : 5

Ans : Quantity of gold in A = 7/9* wt. of A

Quantity of gold in B = 7/18* wt. of B

If 1 kg of each A and B are mixed to form third alloys C.

Then quantity of gold in 2 kg C = 7/9 + 7/18

= 7/6 kg

And Quantity of copper in 2 kg C = 2 – 7/6

= 5/6 kg

∴ Required ratio = 7/6 : 5/6 = 7 : 5

29. In a laboratory, two bottles contain mixture of acid and water in the ratio 2 : 5 in the first bottle and 7 : 3 in the second. The ratio in which the contents of these two bottles be mixed such that the new mixture has acid and water in the ratio 2 : 3 is-

a. 4 : 15

b. 9 : 8

c. 21 : 8

d. 1 : 2

Ans : Quantity of acid in first bottle = 2/7 x mix.

and Quantity of acid in second bottle = 7/10 x mix.

If x and 1 volumes are taken from I and II bottle respectively to form new mixture.

Then, (2/7 x + 7/10 * 1)/(5x/7 +3/10 * 1) = 2/3

⇒ 6x/7 21/10 = 10x/7 +6/10

⇒ 4x/7 = 15/10

∴ x = 15/10*7/4 = 21/8

∴ Required ratio = x : 1

= 21 : 8

30. A mixture contains 80% acid and rest water. Part of the mixture that should be removed and replaced by same amount of water to make the ratio of acid and water 4 : 3 is-

a. 1/3 rd

b. 3/7 th

c. 2/3 rd

d. 2/7 th

Ans : Let the initial wt. of mixture be 1 kg and x kg of mixture is taken out and replaced by same amount of water.

∵ Amt. of acid/Amt. of water = 0.8 – 0.8x/ (0.2 – 0.2x + x) = 4/3

⇒ 2.4 – 2.4x = 0.8 + 3.2x

⇒ 5.6x = 1.6

∴ x = 1.6/5.6 = 2/7th part

31. An employer reduces the number of his employees in the ratio 9 : 8 and increases their wages in the ratio 14 : 15. If the original wage bill was Rs. 189,900, find the ratio in which the wage bill is decreased-

a. 20 : 21

b. 21 : 20

c. 20 : 19

d. 19 : 21

Ans : Let the initial number of employees be 9x and the employer gives Rs. 14y as wage to each.

∵ 9x * 14y =18900

∴ xy = 150 and The later bill = 8x*15y = 120xy

= 120*150 = 18000

∴ Required ratio = 18000 : 18900

= 20 : 21

32. The batting average for 40 innings of a cricketer is 50 runs. His highest score exceeds his lowest score by 172 runs. If these two innings are excluded, the average of the remaining 38 innings is 48 runs. The highest score of the player is-

a. 165

b. 170

c. 172

d. 174

Ans : Let the max. number of runs be x.

∴ The lowest score = (x-172)

∵ 40*50 = 38*48 + x + (x-172)

⇒ 2000 = 1824 + 2x – 172

∴ x= 174

33. Walking at 3 km per hour, Pintu reaches his school 5 minutes late. If he walks at 4 km per hour he will be 5 minutes early. The distance of Pintu’s school from his house is-

a. 1 ½ Km

b. 2 Km

c. 2 ½ Km

d. 5 Km

Ans : Distance of the school from the house

= 4*3/(4 – 3) * 5 +5/60 km

= 12*1/6

=2 km

34. Nitin bought some oranges at Rs. 40 a dozen and an equal number at Rs.30 a dozen. He sold them at Rs. 45 a dozen and made a profit of Rs.480. The number of oranges, he bought, was-

a. 48 dozen

b. 60 dozen

c. 72 dozen

d. 84 dozen

Ans : Let the number of oranges bought be x.

∴ 45x/12 – 70x/24 = 480

⇒ (45-35)/12 = 480

⇒ x = 480*12/10 = 576

= 48*12 = 48 dozen

35. A man buys two chairs for a total cost of Rs.900. By selling one for 4/5 of its cost and the other for 5/4 of its cost, he makes a profit of Rs.90 on the whole transaction. The cost of the lower priced chair is-

a. Rs.360

b. Rs.400

c. Rs.420

d. Rs.300

Ans : Let the cost price of 1 chair be Rs. x.

∴ C.P. of other chair = Rs. (900-x)

∵ 4/5 x + 5/4 (900-x) = 900 + 90

⇒ 4/5 x +1125 -5x/4 = 990

∴ 9x/20 = 135

∴ x = 135*20/9

= Rs. 300

∴ C.P. of the lower priced chair is Rs. 300.

36. By selling 100 oranges, a vendor gains the selling price of 20 oranges. His gain per cent is-

a. 20

b. 25

c. 30

d. 32

Ans : Let the S.P. of 100 Oranges be Rs. x.

∴ S.P. of 20 oranges = x/100 *20 = Rs. x/5

∴C.P. of 100 oranges = x - x/5

= Rs. 4x/5

∴ Reqd. Profit % = x/5 * 100*5/4x %

= 25%

37. 60% of the cost price of an article is equal to 50% of its selling price. Then the percentage of profit or loss on the cost price is-

a. 20% loss

b. 16 2/3% profit

c. 20% profit

d. 10% loss

Ans : Let the cost price be Rs. 100.

∵ S.P * 50/100 = 100*16/100

∴ S.P = 60*100/50

= Rs. 120

∴ Reqd. % profit = (120 – 100)% = 20%

38. Maninder bought two horses at Rs.40,000 each. He sold one horse at 15% gain, but had to sell the second horse at a loss. If he hand suffered a loss of Rs.3,600 on the whole transaction, then the selling price of the second horse is-

a. Rs.30,000

b. Rs.30,200

c. Rs.30,300

d. Rs.30,400

Ans: C.P. of two horses = 2*40000 = Rs. 80000

and S.P. of two horse = 80000 – 3600 = Rs. 76400

∴ S.P. of the other horse = 76400 – 46000 = 30400

39. A fruit-seller buys x guavas for Rs.y and sells y guavas for Rs. x. If x>y, then he made-

a. x2 – y2 / xy % loss

b. x2 – y2 / xy % gain

c. x2 – y2 / y2 % loss

d. x2 – y2 / y2 * 100% gain

Ans : C.P. of 1 guava = Rs. y/x [x>y]

and S.P. of 1 guava = Rs. x/y

∴ Reqd. Gain% = x/y – y/x/ y/x * 100%

= x2 - y2 /y2 *100%

40. A jar contain 10 red marbles and 30 green ones. How many red marbles must be added to the jar so that 60% of the marbles will be red?

a. 25

b. 30

c. 35

d. 40

Ans : Let after adding x red marbles, the red marbles with be 60% of the total.

∵ (10+x)/ (10 + x) + 30 = 60/100

⇒ 10 + x/40 + x = 3/5

⇒ 50 + 5x = 120 + 3x

x = 70/2 = 35

41. If a number multiplied by 25% of itself gives a number which is 200% more than the number, then the number is –

a. 12

b. 16

c. 35

d. 24

Ans : Let the number be x.

∵ x*25x/100 = x + 200x/100

⇒ x2/4 = 3x

⇒ x2 - 12x = 0

⇒ x – 12 = 0

∴ x = 12

42. The value of an article depreciates every year at the rate of 10% of its value. If the present value of the article is Rs.729, then its worth 3 years ago was-

a. Rs.1250

b. Rs.1000

c. Rs.1125

d. Rs.1200

Ans : Let the worth 3 years ago be Rs. x.

∵ 729 = x (1 – 10/100)3

⇒ 729 = x*9*9*9/10*10*10

∴ x = Rs. 1000

43. The price of onions has been increased by 50%. In order to keep the expenditure on onions the same the percentage of reduction in consumption has to be-

a. 50%

b. 33 1/3%

c. 33%

d. 30%

Ans : Reqd. Percentage of reduction

= 50*100/(100 + 50) %

= 5000/150 %

= 33 1/3%

44. A took two loans altogether of Rs.1200 from B and C. B claimed 14% simple interest per annum, while C claimed 15% per annum. The total interest paid by A in one year was Rs.172. Then, A borrowed-

a. Rs.800 from C

b. Rs.625 from C

c. Rs.400 from B

d. Rs.800 from B

Ans : If A borrowed Rs. x from B. and A borrowed Rs. Rs. (1200 – x) from C.

∵ (1200 – x)*15*1/100 + x*14*1/100

⇒ 18000 – 15x + 14x = 172*100

x = Rs. 800

45. If a regular polygon has each of its angles equal to 3/5 times of two right angles, then the number of side is-

a. 3

b. 5

c. 6

d. 8

Ans : If the number of sides a regular polygon be n.

Then (2n-4)/n = 2*3/5

⇒ (2n – 4)*5 = 6n

∴ n = 5

46. A square is of area 200 sq. m. A new square is formed in such a way that the length of its diagonal is √2 times of the diagonal of the given square. The the area of the new square formed is-

a. 200√2 sq.m

b. 400√2 sq.m

c. 400 sq.m

d. 800 sq.m

Ans : Length of the diagonal of Ist square

= √2*200

= 20 m

∴ Length of the diagonal of new square = 20√2m

∴ Area of the new square = ½*(20.√2)2 = 400 sq. m

47. The heights of a cone, cylinder and hemisphere are equal. If their radii are in the ratio 2 : 3 : 1, then the ratio of their volumes is-

a. 2 : 9 : 2

b. 4 : 9 : 1

c. 4 : 27 : 2

d. 2 : 3 : 1

Ans : Ratio of their volumes [cone : cylinder : hemisphere]

= 1/3(2)2h : (3)2*h : 2/3(1)2.h

= 4/3 : 9 : 2/3 = 4 : 27 : 2

48. A motor-boat can travel at 10 km/hr in still water. It travelled 91 km downstream in a river and then returned to the same place, taking altogether 2 hours. Find the rate of flow of river-

a. 3 km/hr

b. 4 km/hr

c. 2 km/hr

d. 5 km/hr

Ans : Let the rate of flow of river be x km/hr.

∵ 91/(10 + x) + 91/(10 – x) = 20

⇒ 91(10 – x + 10 + x)(10 + x) (10 – x) = 20

⇒ 91*20 = 20(100 – x2)

⇒ x2 = 9 = (3)2

∴ x = 3 km/hr

49. A man driving at 3/4th of his original speed reaches his destination 20 minutes later than the usual time. Then the usual time is-

a. 45 minutes

b. 60 minutes

c. 75 minutes

d. 120 minutes

Ans : Let the original speed be x km/hr and the usual time be y hours.

∵ x * y = ¾ x(y+1/3)

∴ 4y = 3y + 1

∴ y = 1 hr = 60 minutes

50. A motor-boat, travelling at the same speed, can cover 25 km upstream and 39 km downstream in 8 hours. At the same speed, it can travel 35 km upstream and 52 km downstream in 11 hours. The speed of the stream is –

a. 2 km/hr

b. 3 km/hr

c. 4 km/hr

d. 5 km/hr

Ans : Let the speeds of motor boat and the stream be x and y km/hr respectively.

∵ 39/x + y + 25/x – y = 8 …(1)

and 52/x + y + 35/x – y = 11 …(2)

Solving equations (1) and (2), we get-

∴ 100 - 105/x – y = 32 – 33

∴ x – y = 5

and x + y = 13

∴ y = 4 km/hr

51. If a sum of money placed at compound interest, compounded annually, doubles itself in 5 years, then the same amount of money will be 8 times of itself in-

a. 25 years

b. 20 years

c. 15 years

d. 10 years

Ans : Required time = 5log 8/log 2

= 5*3log 2/log 2

= 15 years

52. A person has left an amount of Rs.1,20,000 to be divided between his 2 son aged 14 years and 12 years such that they get equal amounts when each attains 18 years of age. If the amount gets a simple interest of 5% per annum, the younger son’s share at present is-

a. Rs.48,800

b. Rs.57,600

c. Rs.62,400

d. Rs.84,400

Ans : Let the present share of the younger son be Rs. x.

∴ The share of the elder son = Rs. (120000 – x)

∵ x + x*6*5/100

= (120000 – x) + (120000 – x)*4*5/100

⇒ 130x/100 = (120000 –x)*120/100

⇒ 13x = 1440000 – 12x

⇒ 25x = 1440000

∴ x = Rs. 57600

53. If the simple interest on Rs.x at a rate of a% for m years is same as that on Rs. y at a rate of a2% for m2 years, then x : y is equal to-

a. m : a

b. am : 1

c. 1/m : 1/a

d. 1/am : 1

∵ x*a*m/100 = y*a2*m2/100

⇒ x/y = am/1

∴ x : y = am : 1

54. Base of a right prism is an equilateral triangle of side 6 cm. If the volume of the prism is 108√3 cc, its height is-

a. 9 cm

b. 10 cm

c. 11 cm

d. 12 cm

Ans : Height of the prism = 108√3*4/√3*(6)2 = 12cm

55. If a + 1/a + 2 = 0, then the value of (a37 – 1/a100) is-

a. 0

b. -2

c. 1

d. 2

Ans : ∵ a + 1/a + 2 = 0

⇒ a2 + 1 +2a = 0

⇒ (a + 1)2 = 0

⇒ a +1 = 0

∴ a = -1

∴ a37 -1/a100 = (-1) – (1) = -2

56. The value of k for which the graphs of (k-1) x+y-2 = 0 and (2-k) x -3y + 1 = 0 are parallel is-

a. ½

b. -1/2

c. 2

d. -2

Ans : (k – 1) x + y – 2 = 0

∴ y = (1 – k) x + 2 ….(1)

and (2 –k) x – 3y – 1 = 0

3y = (2 – k) x +1

Y = 2 – k/3 x + 1/3 ….(2)

∵ m1 = m2

⇒ 1 – k = 2 – k/3

⇒ 3 – 3k = 2 - k

∴ k = 1/2

57. If a2 + b2 + c2 = 2 (a-b-c) – 3, then the value of (a – b + c) is-

a. -1

b. 3

c. 1

d. -2

Ans : a2 + b2 + c2 = 2(a – b – c) – 3

⇒ a2 - 2a + 1 + b2 + 2b + 1 + c2 + 2c + 1 = 0

⇒ (a – 1)2 + (b + 1)2 + (c + 1)2 = 0

⇒ a – 1 = 0, b + 1 = 0, c + 1 = 0

⇒ a = 1, b = -1, c = -1

∴ a + b – c = 1 – 1 + 1 =1

58. If x2 + 3x + 1 = 0, then the value of x3 + 1/x3 is-

a. -18

b. 18

c. 36

d. -36

Ans : ∵ x2 + 3x + 1 = 0

⇒ x + 3 + 1/x = 0

⇒ x + 1/x = -3

⇒ (x + 1/x)3 = (-3)3

⇒ x3 + 1/x3 + 3 (-3) = -27

∴ x3 + 1/ x3 = -18

59. If xa, xb, xc = 1, then the value of a3 + b3 + c3 is –

a. 9

b. abc

c. a + b + c

d. 3abc

Ans : ∵ xa. xb. xc = 1

⇒ xa + b + c = x0

⇒ a +b + c =0

∴ a3 + b3 + c3 = 3abc

60. Base of a right pyramid is a square, length of diagonal of the base is 24√2 m. If the volume of the pyramid is 1728 cu.m, its height is-

a. 7 m

b. 8 m

c. 9 m

d. 10 m

Ans : Area of the base of the pyramid

= ½ (24√2)2 = 576m2

∴ Height of pyramid + 1728*3/576 = 9m

61. The height of a right circular cone and the radius of its circular base are respectively 9 cm and 3 cm. The cone is cut by a plane parallel to its base so as to divide it into two parts. The volume of the frustum (i.e., the lower part) of the cone is 44 cubic cm. The radius of the upper circular surface of the frustum (taking = 22-7) is-

a. 3√12 cm

b. 3√13 cm

c. 3√6 cm

d. 3√20 cm

Ans : Let the radius of the upper circular part of the frustum be r cm.

(Picture)

Then r/3 = x/9 = AC/AD

∴ x = 3r (where AC = x cm)

⇒ /3h(r12 + r1r2+r22 = 44

⇒ /3(9 – x)*(9 + 3r + r2) = 44

⇒ (9 – x)*(9 + 3r + r2) = 44*3*7/22 = 42

⇒ 81 + 27r + 9r2 – 9x – 3rx - r2x = 42 On putting x = 3r,

⇒ 81 + 27r + 9r2 - 9r2 - 3r3 – 27r = 42

⇒ 3r3 = 39

∴ r = 3√13 cm

62. The ratio of radii of two right circular cylinder is 2 : 3 and their heights are in the ratio 5 : 4. The ratio of their curved surface area is-

a. 5 : 6

b. 3 : 4

c. 4 : 5

d. 2 : 3

Ans : Required ratio = 2*2r*5h/2*3r*4h = 5 : 6

63. A solid cylinder has total surface area of 462 sq.cm. Curved surface area is 1/3rd of its total surface area. The volume of the cylinder is-

a. 530 cm3

b. 536 cm3

c. 539 cm3

d. 545 cm3

Ans : ∵ 2r(r + h) = 462

and 2rh = 1/3*462 = 154

∴ r + h/h = 462/154 =3

∴ r + h = 3h

∴ r = 2h

∴ 2*2h2 = 154

∴ h2 = 154*7/22*4 = 49/4

= (7/2)2

∴ h = 7/2 cm

and r = 7 cm

∴ Volume of the cylinder

= 22/7*49*7/2

= 539 cm3

64. A cylinder and a cone have equal radii of their bases and equal heights. If their curved surface areas are in the ratio 8 : 5, the ratio of their radius and height is-

a. 1 : 2

b. 1 : 3

c. 2 : 3

d. 3 : 4

Ans : Let the radius and height of each are r and h respectively.

∵ 2rh/r√h2 + r2 = 8/5

∴ 10h = 8√r2 + h2

⇒ 100h2 =64r2 + 64h2

∴ h2 = 64 r2/36 = (4/3 r)2

⇒ h = 4/3r

∴ r : h = 3 : 4

65. A solid is hemispherical at the bottom and conical above. If the surface areas of the two parts are equal, then the ratio of radius and height of its conical part is-

a. 1 : 3

b. 1 :1

c. √3 : 1

d. 1 : √3

Ans : (Picture)

∵ rl = 2r2

⇒ l = 2r

⇒ √h2 + r2 = 2r

⇒ h2 = 3r2

∴ r : h = r/h

= 1 : √3

66. If O is the circumcentre of ∆ ABC and ∠OBC = 350, then the ∠BAC is equal to-

a. 550

b. 1100

c. 700

d. 350

Ans : ∴ ∠BOC = 1800 – (350 + 350) = 1100

(Picture)

∴ ∠BAC = 1/2 * 1100 = 550

67. If I is the incentre of ∆ ABC and ∠ BIC = 1350, then ∆ ABC is-

a. Acute angled

b. Equilateral

c. Right angled

d. Obtuse angled

Ans : (Picture)

∠BIC = 1350

⇒ B/2 + C/2 = 1800 – 1350 = 450

⇒ ∠B + ∠C = 900

∴ ∠A = 1800 – (∠B + ∠C) = 900

i.e., ∆ ABC is a right angled.

68. If sin2 (∝ + β/2) is-

a. 1

b. -1

c. 0

d. 0.5

∵ sin2 ∝ + sin2 β = 2

⇒ 1 – cos2 ∝ 1 - cos2 β = 2

⇒ cos2 ∝ cos2 β = 0

⇒ cos ∝ = 0

and cos β = 0

⇒ ∝ = /2 and β = /2

∴ cos (∝ + β/2) = cos [/2 + /2 / 2]

= cos /2 = 0

69. The length of a shadow of a vertical tower is 1/√3 times its height. The angle of elevation of the Sun is-

a. 300

b. 450

c. 600

d. 900

Ans : (Picture)

∵ tan θ = h/1√3 h = √3 = tan 600

70. The graphs of x +2y =3 and 3x-2y = 1 meet the Y-axis at two points having distance-

a. 8/3 units

b. 4/3 units

c. 1 unit

d. 2 units

Ans : When the graphs meet the Y-axis at two points.

Then, [x + 2y = 3] at x = 0 ⇒ [0, y1 =3/2]

[3x - 2y = 1] at x = 0

and i.e., [0, y2 = -1/2]

Required distance = (y1 – y2)

= 3/2 – (- ½) = 2 units

71. If x+1/16x = 1, then the value of 64x3 + 1/64x3 is-

a. 4

b. 52

c. 64

d. 76

∵ x + 1/16x = 1

⇒ 16x2 – 16x + 1 =0

⇒ 16x2 – 16x + 4 = 3

⇒ (4x – 2)2 = 3

⇒ 4x = 2 + √3

⇒ 64x3 = (2±√3)3

=8 + 3√3 + 6√3 (2 + √3)

= 26 + 15√3

∴ 64x3 + 1/64x3 = (26 + 15√3) + 1/ (26 + 15√3)

= (26 + 15√3) + 26 - 15√3/676 -675

=52

72. If a, b, c, are three non-zero real numbers such that a + b + c = 0, and b2 ≠ ca, then the value of a2 + b2 + c2/ b2 –ca is-

a. 3

b. 2

c. 0

d. 1

∵ a + b + c = 0

⇒ a + c = -b

⇒ a2 + c2 = b2 -2ac

⇒ a2 + b2 + c2 = 2b2 – 2ac

∴ a2 + b2 + c2/ b2 ac = 2

73. If a4 + a2 b2 + b4 = 8 and a2 + ab + b2 = 4, then the value of ab is-

a. -1

b. 0

c. 2

d. 1

∵ a4 + a2 b2 + b4/ a2 + ab + b2 = 8/4

⇒ (a2 + b2)2 – (ab) 2/ (a2 + b2 + ab) = 2

⇒ a2 – ab + b2 = 2 ….(1)

and a2 + ab + b2 = 4 …..(2)

⇒ 2ab = 2

⇒ ab = 1

74. If a = 25, b = 15, c = -10, then the value of a3 + b3 + c3 – 3abc/ (a-b)2 + (b-c)2 + (c-a)2 is-

a. 30

b. -15

c. -30

d. 15

∵ a3 + b3 + c3 – 3abc

= (25)3 + (15)3 + (-10)3 – 3*25*15*(-10)

=15625 + 3375 – 1000 + 11250 = 29250

and (a - b)2 + (b – c)2 + (c – a)2

= (10)2 + (25)2 + (-35)2

= (10)2 + 625 + 1225

= 1950

∴ Required value = 29250/1950 =15

75. A, B, C are three points on a circle. The tangent at A meets BC produced at T, ∠BTA = 400, ∠CAT = 440. The angle subtended by BC at the centre of the circle is-

a. 840

b. 920

c. 960

d. 1040

Ans : (Picture)

∠ACB = 400 + 440 = 840

∴ ∠ACO = 900 - 440 = 460 = ∠OAC

⇒ ∠OCB = ∠ACB - ∠ACO

= 840 - 460 = 380 = ∠OBC

∴ ∠BOC = 1800 – (∠OCB + ∠OBC)

= 1800 – (380 + 380) = 1040

76. If the length of a chord of a circle at a distance of 12 cm from the Centre is 10 cm, then the diameter of the circle is-

a. 13 cm

b. 15 cm

c. 26 cm

d. 30 cm

Ans : (Picture)

∵ OA = √OM2 + AM2

= √122 + 52 = 13

∴ Diameter of the circle = 2*OA

= 2*13 = 26cm

77. In ∆ ABC, P and Q are the middle points of the sides AB and AC respectively. R is a point on the segment PQ such that PR : RQ = 1 : 2. If PR = 2 cm, then BC =

a. 4 cm

b. 2 cm

c. 12 cm

d. 6 cm

∵ PR/RQ = ½

But, PR = 2cm

RQ = 2*PR

= 4cm

(Picture)

∴ PQ = PR + RQ

= 2 + 4 = 6 cm

∴ BC = 2*PQ = 12CM

78. If tan θ tan 2θ = 1, then the value of sin2 2θ + tan2 is equal to –

a. ¾

b. 10/3

c. 3 ¾

d. 3

∵ tan θ * tan 2θ = 1

⇒ tan θ * 2 tan θ/1 – tan2 θ = 1

⇒ 2 tan2 θ = 1 - tan2 θ

⇒ 3 tan2 θ = 1

⇒ tan θ = 1/√3 = tan 300

∴ θ = 300

∴ sin2 2θ + tan2 2θ = sin2 600 + tan2 600

= ¾ + 3

=3 ¾

79. The value of cot /20 cot 3/20 cot 5/20 cot 7/20 cot 9/20 is-

a. -1

b. ½

c. 0

d. 1

Ans : Given Exp.

= cot /20. cot 3/20. cot 5/20. cot 7/20. cot 9/20

= cot 90. cot 270. cot 450. cot 630. cot 810

= cot 90. cot 270*1*tan 270. tan 90.

= cot 90. cot 270*1*1/ cot 270*1/ cot 90

=1

80. If sin θ+ cos θ = 17/13, 0<θ<900, then the value of sin θ- cos θ is-

a. 5/17

b. 3/19

c. 7/10

d. 7/13

∵ sin θ + cos θ = 17/13

⇒ sin2 θ + cos2 θ + 2sin θ.cos θ = 289/169

∴ 2sin θ.cos θ = 289 /169 -1

= 289 -169/169 = 120/169

∴ sin2 θ + cos2 θ – 2 sin2 θ.cos θ = 1 – 120/169

= 49/169

⇒ (sin θ – cos θ)2 = (7/13)2

∴ sin θ – cos θ = 7/13

S.S.C. Combined Graduate Level Exam

(Tier-II) 2011

Solved Paper – I

Arithmetical Ability

1. The product of 2 numbers is 1575 and their quotient is 9/7. Then the sum of the numbers is –

a. 74

b. 78

c. 80

d. 90

Ans : Let the numbers be x and y .

∴ xy = 1575

And x/y = 9/7

∴ xy/x/y = 1575/9/7

∴ y2 = 1225

∴ y = 35 and x = 45

∴ The sum of the numbers = 45+35

= 80

2. The value of (81)3.6 * (9)2.7/ (81)4.2 * (3) is __

a. 3

b. 6

c. 9

d. 8.2

Ans : (81)3.6 * (9)2.7/(81)4.2 * (3) = (3)14.4 * (3)5.4/(3)16.8 * (3)

= 314.4+5.4-16.8-1

= 32

= 9

3. √6+√6+√6+… is equal to –

a. 2

b. 5

c. 4

d. 3

Ans : Let √6+√6+√6+…. be x.

∴ x =√6+x

∴ x2 = 6+x

∴ x2 + x – 6 = 0

∴ (x-3) (x+2) = 0

∴ x = 3

4. The sum of the squares of two natural consecutive odd numbers is 394. The sum of the numbers is –

a. 24

b. 32

c. 40

d. 28

Ans : Let the consecutive odd numbers be x and (x+2)

∵ x2 + (x + 2)2 = 394

∴ x2 + x2 + 4x + 4 = 394

∴ 2 x2 +4x – 390 = 0

∴ x2 + 2x – 195 = 0

∴ (x +15) (x-13) = 0

∴ x = 13

∴ Required sum = 13 +15 =28

5. When (6767 +67) is divided by 68, the remainder is-

a. 1

b. 63

c. 66

d. 67

Ans: (6767+ 67) = 67(6766 + 166)

As 66 is an even number

∴ 6766 is an even number

∴ (6766 + 1) is perfectly divisible by (67 + 1)

i.e. 68

∴ The remainder = 67

6. In a division sum, the divisor is 4 times the quotient and twice the remainder. If a and b are respectively the divisor and the dividend then-

a. 4b-a2 /a =3

b. 4b-2a /a2 =2

c. (a+1)2 = 4b

d. A(a+2)/b = 4

Ans : As divisor is a, and dividend is b.

∴ Quotient = a/4

And Remainder = a/2

∴ b = a * a/4 + a/2

∴ 4b = a2 + 2a

∴ a(a+2)/b = 4

7. If 738 A6A is divisible by 11, then the value of A is-

a. 6

b. 3

c. 9

d. 1

Ans : As 738A6 A is divisible by 11.

∴ A + A + 3 = 6 + 8 + 7

∴ A = 9

8. The east positive integer that should be subtracted from 3011 * 3012 so that the difference is perfect square is-

a. 3009

b. 3010

c. 3011

d. 3012

Ans : ∵ 3011 * 3012 = 3011 (3011 + 1)

= (3011)2 + 3011

∴ Required least number = 3011

9. P, Q, R are employed to do a work for Rs. 5750. P and Q together finished 19/23 of work and Q and R together finished 8/23 of work. Wage of Q, in rupees, is-

a. 2850

b. 3750

c. 2750

d. 1000

Ans : Work done by Q = 19/23 + 8/23 – 1

4/23

∴ Wage of Q = 4/23 * 5750

= Rs. 1000

10. A can do a piece of work in 24 day, B in 32 days and C in 64 days. All begin to do it together, but A leaves after 6 days and B leaves 6 days before the completion of the work. How many days did the work last?

a. 15

b. 20

c. 18

d. 30

Ans : Work done by A = 6/24 = ¼

Work done by B = (x-6)/32

(where x is no. of days in which work is competed)

∵ ¼ + x – 6/32 + x/64 = 1

∴ 16 + 24 – 12 + x /64 = 1

∴ 3x + 4 = 64

∴ x = 60/3 = 20 days

11. The square root of (0.75)3 /1-0.75 + [0.75 + 90.75)2 +1] is-

a. 1

b. 2

c. 3

d. 4

Ans : The square root of

[(0.75)3/1 – 0.75 + {0.75 + (0.75)2 + 1}]

= √1.6875 + 2.3125

=√4

= 2

12. Given that √4096 = 64, the value of √4096 + √40.96 +√0.004096 is-

a. 70.4

b. 70.464

c. 71.104

d. 71.4

Ans : Given Exp. = √4096 + √40.96 +√0.004096

= 64 + 6.4 + 0.064

= 70.464

13. By selling an article at 3/4th of the marked price, there is a gain of 25%. The ratio of the marked price and the cost price is-

a. 5 : 3

b. 3 : 5

c. 3 : 4

d. 4 : 3

Ans : Let of M.P. be Rs. x.

∴ S.P. = Rs.3x/4

and C.P. = 3x/4 * 100/125

= Rs.3x/5

∴ required ratio = x: 3x/5

= 5:3

14. A and B earn in the ratio 2:1. They spend in the ratio 5:3 and save in the ratio 4:1. If the total monthly savings of both A and B are Rs.5,000, the monthly income of B is-

a. Rs.7,000

b. Rs.14,000

c. Rs.5,000

d. Rs.10,000

Ans : Let the monthly income of B be Rs. x.

∴ Monthly income of A = Rs. 2x and

Saving of A =5000 * 4/(4 + 1)

= Rs. 4000

Saving of B = Rs.1000

∵ 2x – 4000/x-1000 = 5/3

⇒ 6x – 12000 = 5x – 5000

∴ x = Rs.7000

15. The ratio of the sum of two numbers and their difference is 5:1. The ratio of the greater number to the smaller number is-

a. 2 : 3

b. 3 : 2

c. 5 : 1

d. 1 : 5

Ans : Let the numbers be x and y.

∵ x + y/ x – y = 5/1

⇒ 5x – 5y = x + y

⇒ 4x = 6y

⇒ x/y = 6/4

x : y = 3 : 2

16. A cistern has 3 pipes A, B and C. A and B can fill it in 3 and 4 hours respectively, and C can empty it in 1 hour. If the pipes are opened at 3 p.m., 4 p.m. and 5 p.m. respectively on the same day, the cistern will be empty at-

a. 7.12 p.m.

b. 7.15 p.m.

c. 7.10 p.m.

d. 7.18 p.m.

Ans : Let the cistern be emptied at x p.m.

∵ x -3/3 + x – 4/4 = x - 5/1

⇒ 4x – 12 + 3x – 12/12 = x – 5/1

⇒ 7x – 24 = 12x – 60

⇒ 5x = 36

∴ x = 36/5

= 7 hr. + 12 min.

= 7.12 p.m.

17. If A works alone, he would take 4 days more to complete the job than if both A and B worked together. If B worked alone, he would take 16 days more to complete the job than if A and B work together. How many days would they take to complete the work if both of them worked together?

a. 10 days

b. 12 days

c. 6 days

d. 8 days

Let x days are taken when they work together.

∴ Time taken by A to complete the work = (x + 4) days

And Time taken by B to complete the work = (x + 16) days

∵ 1/x = 1/x + 4 + 1/x + 16

∴ 1/x = x + 16 + x + 4/(x + 4) (x +16)

∴ 2x2 + 20x = x2 + 20x +64

∴ x2 = 64 (8)2

∴ x = 8 days

18. 250 men can finish a work in 20 days working 5 hours a day. To finish the work within 10 days working 8 hours a day, the minimum number of men required is-

a. 310

b. 300

c. 313

d. 312

Ans : Required no. of men = 250 * 5 * 20/8 * 10

=312.5

313

19. 2 men and 5 women can do a work in 12 days. 5 men 2 women can do that work in 9 days. Only 3 women can finish the same work in-

a. 36 days

b. 21 days

c. 30 days

d. 42 days

∵ (2m + 5w) * 12 = (5m + 2w) * 9

⇒ 24m + 60w = 45m + 18w

⇒ 21m = 42w

⇒ 1m = 2w

∴ 2m + 5w = 9w

∴ required no. of days = 9 * 12/3

= 36

20. While selling, a businessman allows 40% discount on the marked price and there is a loss of 30%. If it is sold at the marked price, profit per cent will be –

a. 10%

b. 20%

c. 16 2/3%

d. 16 1/3%

Ans : Let the M.P. be Rs.100.

∴ S.P = (100-40) = Rs.60

and C. P. = 60 * 100/(100-30) = Rs.600/7

∴ Reqd. % profit = 100-600/7 /600/7 * 100%

= (700 – 600)/7 * 600 * 100 * 7%

= 16 2/3%

21. Successive discount of 10% , 20% and 50% will be equivalent to a single discount of-

a. 36%

b. 64%

c. 80%

d. 56%

Ans : Equivalent to a single discount

= [100 - (100 – 10) (100 – 20) (100 – 50)/100*100]%

= [100 - 90*80*50/10000]%

= [100-36]%

= 64%

22. A retailer offers the following discount schemes for buyers on an article-

I. Two successive discounts of 10%

II. A discount of 12% followed by a discount of 8%.

III. Successive discounts of 15% and 5%

IV. A discount of 20%

The selling price will be minimum under the scheme-

a. I

b. II

c. III

d. IV

Ans : From (i) single discount = [10 + 10 – 10*10/100] % = 19%

From (ii) single discount = [12 + 8 -12*8/100] % = 19.04%

From (iii) single discount = [15 + 5 – 15*5/100] % = 19.25%

From (iv) single discount = 20%

∴ The S.P. will be minimum under the scheme IV.

23. Of three numbers, the second is thrice the first and the third number is three-fourth of the first. If the average of the three numbers is 114, the largest number is –

a. 72

b. 216

c. 354

d. 726

Ans : Let the first number be x.

∴ Second number = 3x and Third number = 3x/4

∵ x + 3x + 3x/4 = 3*114

⇒ 19x/4 = 342

∴ x = 342*4/19 = 72

∴ The largest number = 3*72

= 216

24. A car covers 1/5 of the distance from A to B at the speed of 8 km/hour, 1/10 of the distance at 25 km per hour and the remaining at the speed of 20 km per hour. Find the average speed of the whole journey-

a. 12.625 km/hr

b. 13.625 km/hr

c. 14.625 km/hr

d. 15.625 km/hr

Ans : If the whole journey be x km. The total time taken

= (x/5/8 + x/10/25 + 7x/10/20) hrs

= (x/40 + x/250 + 7x/200) hrs

= 25x + 4x + 35x/1000

= 64x/1000 hrs

∴ Average speed = x/64x/1000

= 15.625 km/hr

25. The average of 3 numbers is 154. The first number is twice the second and the second number is twice the third. The first number is-

a. 264

b. 132

c. 88

d. 66

Ans : Let the first number be x.

∴ Second number = x/2 and Third number = x/4

∵ x + x/2 + x/4 = 3 x 154

⇒ 7x/4 = 462

∴ x = 462*4/7

=264

26. The average salary of all the staff in an office of a corporate house is Rs. 5,000. The average salary of the officers is Rs. 14,000 and that of the rest is Rs. 4,000. If the total number of staff is 500, the number of officers is –

a. 10

b. 15

c. 25

d. 50

Ans : Let the number of officers be x.

∵ 5000*500 = 14000x + 4000(500-x)

∴ 2500000 =14000x + 2000000 – 4000x

∴ x = 500000/10000

= 50

27. The average marks of 40 students in an English exam are 72. Later it is found that three marks 64, 62 and 84 were wrongly entered as 60, 65 and 73. The average after mistakes were rectified is-

a. 70

b. 72

c. 71.9

d. 72.1

Ans : Correct average

= 40*72 + (64 + 62 +84) – 68 – 65 – 73/40

= 2880 + 210 – 206/40 = 2884/40

= 72.1

28. A and B are two alloys of gold and copper prepared by mixing metals in the ratio 7:2 and 7:11 respectively. If equal quantities of the alloys are melted to form a third alloy C, the ratio of gold and copper in C will be-

a. 5 : 7

b. 5 : 9

c. 7 : 5

d. 9 : 5

Ans : Quantity of gold in A = 7/9* wt. of A

Quantity of gold in B = 7/18* wt. of B

If 1 kg of each A and B are mixed to form third alloys C.

Then quantity of gold in 2 kg C = 7/9 + 7/18

= 7/6 kg

And Quantity of copper in 2 kg C = 2 – 7/6

= 5/6 kg

∴ Required ratio = 7/6 : 5/6 = 7 : 5

29. In a laboratory, two bottles contain mixture of acid and water in the ratio 2 : 5 in the first bottle and 7 : 3 in the second. The ratio in which the contents of these two bottles be mixed such that the new mixture has acid and water in the ratio 2 : 3 is-

a. 4 : 15

b. 9 : 8

c. 21 : 8

d. 1 : 2

Ans : Quantity of acid in first bottle = 2/7 x mix.

and Quantity of acid in second bottle = 7/10 x mix.

If x and 1 volumes are taken from I and II bottle respectively to form new mixture.

Then, (2/7 x + 7/10 * 1)/(5x/7 +3/10 * 1) = 2/3

⇒ 6x/7 21/10 = 10x/7 +6/10

⇒ 4x/7 = 15/10

∴ x = 15/10*7/4 = 21/8

∴ Required ratio = x : 1

= 21 : 8

30. A mixture contains 80% acid and rest water. Part of the mixture that should be removed and replaced by same amount of water to make the ratio of acid and water 4 : 3 is-

a. 1/3 rd

b. 3/7 th

c. 2/3 rd

d. 2/7 th

Ans : Let the initial wt. of mixture be 1 kg and x kg of mixture is taken out and replaced by same amount of water.

∵ Amt. of acid/Amt. of water = 0.8 – 0.8x/ (0.2 – 0.2x + x) = 4/3

⇒ 2.4 – 2.4x = 0.8 + 3.2x

⇒ 5.6x = 1.6

∴ x = 1.6/5.6 = 2/7th part

31. An employer reduces the number of his employees in the ratio 9 : 8 and increases their wages in the ratio 14 : 15. If the original wage bill was Rs. 189,900, find the ratio in which the wage bill is decreased-

a. 20 : 21

b. 21 : 20

c. 20 : 19

d. 19 : 21

Ans : Let the initial number of employees be 9x and the employer gives Rs. 14y as wage to each.

∵ 9x * 14y =18900

∴ xy = 150 and The later bill = 8x*15y = 120xy

= 120*150 = 18000

∴ Required ratio = 18000 : 18900

= 20 : 21

32. The batting average for 40 innings of a cricketer is 50 runs. His highest score exceeds his lowest score by 172 runs. If these two innings are excluded, the average of the remaining 38 innings is 48 runs. The highest score of the player is-

a. 165

b. 170

c. 172

d. 174

Ans : Let the max. number of runs be x.

∴ The lowest score = (x-172)

∵ 40*50 = 38*48 + x + (x-172)

⇒ 2000 = 1824 + 2x – 172

∴ x= 174

33. Walking at 3 km per hour, Pintu reaches his school 5 minutes late. If he walks at 4 km per hour he will be 5 minutes early. The distance of Pintu’s school from his house is-

a. 1 ½ Km

b. 2 Km

c. 2 ½ Km

d. 5 Km

Ans : Distance of the school from the house

= 4*3/(4 – 3) * 5 +5/60 km

= 12*1/6

=2 km

34. Nitin bought some oranges at Rs. 40 a dozen and an equal number at Rs.30 a dozen. He sold them at Rs. 45 a dozen and made a profit of Rs.480. The number of oranges, he bought, was-

a. 48 dozen

b. 60 dozen

c. 72 dozen

d. 84 dozen

Ans : Let the number of oranges bought be x.

∴ 45x/12 – 70x/24 = 480

⇒ (45-35)/12 = 480

⇒ x = 480*12/10 = 576

= 48*12 = 48 dozen

35. A man buys two chairs for a total cost of Rs.900. By selling one for 4/5 of its cost and the other for 5/4 of its cost, he makes a profit of Rs.90 on the whole transaction. The cost of the lower priced chair is-

a. Rs.360

b. Rs.400

c. Rs.420

d. Rs.300

Ans : Let the cost price of 1 chair be Rs. x.

∴ C.P. of other chair = Rs. (900-x)

∵ 4/5 x + 5/4 (900-x) = 900 + 90

⇒ 4/5 x +1125 -5x/4 = 990

∴ 9x/20 = 135

∴ x = 135*20/9

= Rs. 300

∴ C.P. of the lower priced chair is Rs. 300.

36. By selling 100 oranges, a vendor gains the selling price of 20 oranges. His gain per cent is-

a. 20

b. 25

c. 30

d. 32

Ans : Let the S.P. of 100 Oranges be Rs. x.

∴ S.P. of 20 oranges = x/100 *20 = Rs. x/5

∴C.P. of 100 oranges = x - x/5

= Rs. 4x/5

∴ Reqd. Profit % = x/5 * 100*5/4x %

= 25%

37. 60% of the cost price of an article is equal to 50% of its selling price. Then the percentage of profit or loss on the cost price is-

a. 20% loss

b. 16 2/3% profit

c. 20% profit

d. 10% loss

Ans : Let the cost price be Rs. 100.

∵ S.P * 50/100 = 100*16/100

∴ S.P = 60*100/50

= Rs. 120

∴ Reqd. % profit = (120 – 100)% = 20%

38. Maninder bought two horses at Rs.40,000 each. He sold one horse at 15% gain, but had to sell the second horse at a loss. If he hand suffered a loss of Rs.3,600 on the whole transaction, then the selling price of the second horse is-

a. Rs.30,000

b. Rs.30,200

c. Rs.30,300

d. Rs.30,400

Ans: C.P. of two horses = 2*40000 = Rs. 80000

and S.P. of two horse = 80000 – 3600 = Rs. 76400

∴ S.P. of the other horse = 76400 – 46000 = 30400

39. A fruit-seller buys x guavas for Rs.y and sells y guavas for Rs. x. If x>y, then he made-

a. x2 – y2 / xy % loss

b. x2 – y2 / xy % gain

c. x2 – y2 / y2 % loss

d. x2 – y2 / y2 * 100% gain

Ans : C.P. of 1 guava = Rs. y/x [x>y]

and S.P. of 1 guava = Rs. x/y

∴ Reqd. Gain% = x/y – y/x/ y/x * 100%

= x2 - y2 /y2 *100%

40. A jar contain 10 red marbles and 30 green ones. How many red marbles must be added to the jar so that 60% of the marbles will be red?

a. 25

b. 30

c. 35

d. 40

Ans : Let after adding x red marbles, the red marbles with be 60% of the total.

∵ (10+x)/ (10 + x) + 30 = 60/100

⇒ 10 + x/40 + x = 3/5

⇒ 50 + 5x = 120 + 3x

x = 70/2 = 35

41. If a number multiplied by 25% of itself gives a number which is 200% more than the number, then the number is –

a. 12

b. 16

c. 35

d. 24

Ans : Let the number be x.

∵ x*25x/100 = x + 200x/100

⇒ x2/4 = 3x

⇒ x2 - 12x = 0

⇒ x – 12 = 0

∴ x = 12

42. The value of an article depreciates every year at the rate of 10% of its value. If the present value of the article is Rs.729, then its worth 3 years ago was-

a. Rs.1250

b. Rs.1000

c. Rs.1125

d. Rs.1200

Ans : Let the worth 3 years ago be Rs. x.

∵ 729 = x (1 – 10/100)3

⇒ 729 = x*9*9*9/10*10*10

∴ x = Rs. 1000

43. The price of onions has been increased by 50%. In order to keep the expenditure on onions the same the percentage of reduction in consumption has to be-

a. 50%

b. 33 1/3%

c. 33%

d. 30%

Ans : Reqd. Percentage of reduction

= 50*100/(100 + 50) %

= 5000/150 %

= 33 1/3%

44. A took two loans altogether of Rs.1200 from B and C. B claimed 14% simple interest per annum, while C claimed 15% per annum. The total interest paid by A in one year was Rs.172. Then, A borrowed-

a. Rs.800 from C

b. Rs.625 from C

c. Rs.400 from B

d. Rs.800 from B

Ans : If A borrowed Rs. x from B. and A borrowed Rs. Rs. (1200 – x) from C.

∵ (1200 – x)*15*1/100 + x*14*1/100

⇒ 18000 – 15x + 14x = 172*100

x = Rs. 800

45. If a regular polygon has each of its angles equal to 3/5 times of two right angles, then the number of side is-

a. 3

b. 5

c. 6

d. 8

Ans : If the number of sides a regular polygon be n.

Then (2n-4)/n = 2*3/5

⇒ (2n – 4)*5 = 6n

∴ n = 5

46. A square is of area 200 sq. m. A new square is formed in such a way that the length of its diagonal is √2 times of the diagonal of the given square. The the area of the new square formed is-

a. 200√2 sq.m

b. 400√2 sq.m

c. 400 sq.m

d. 800 sq.m

Ans : Length of the diagonal of Ist square

= √2*200

= 20 m

∴ Length of the diagonal of new square = 20√2m

∴ Area of the new square = ½*(20.√2)2 = 400 sq. m

47. The heights of a cone, cylinder and hemisphere are equal. If their radii are in the ratio 2 : 3 : 1, then the ratio of their volumes is-

a. 2 : 9 : 2

b. 4 : 9 : 1

c. 4 : 27 : 2

d. 2 : 3 : 1

Ans : Ratio of their volumes [cone : cylinder : hemisphere]

= 1/3(2)2h : (3)2*h : 2/3(1)2.h

= 4/3 : 9 : 2/3 = 4 : 27 : 2

48. A motor-boat can travel at 10 km/hr in still water. It travelled 91 km downstream in a river and then returned to the same place, taking altogether 2 hours. Find the rate of flow of river-

a. 3 km/hr

b. 4 km/hr

c. 2 km/hr

d. 5 km/hr

Ans : Let the rate of flow of river be x km/hr.

∵ 91/(10 + x) + 91/(10 – x) = 20

⇒ 91(10 – x + 10 + x)(10 + x) (10 – x) = 20

⇒ 91*20 = 20(100 – x2)

⇒ x2 = 9 = (3)2

∴ x = 3 km/hr

49. A man driving at 3/4th of his original speed reaches his destination 20 minutes later than the usual time. Then the usual time is-

a. 45 minutes

b. 60 minutes

c. 75 minutes

d. 120 minutes

Ans : Let the original speed be x km/hr and the usual time be y hours.

∵ x * y = ¾ x(y+1/3)

∴ 4y = 3y + 1

∴ y = 1 hr = 60 minutes

50. A motor-boat, travelling at the same speed, can cover 25 km upstream and 39 km downstream in 8 hours. At the same speed, it can travel 35 km upstream and 52 km downstream in 11 hours. The speed of the stream is –

a. 2 km/hr

b. 3 km/hr

c. 4 km/hr

d. 5 km/hr

Ans : Let the speeds of motor boat and the stream be x and y km/hr respectively.

∵ 39/x + y + 25/x – y = 8 …(1)

and 52/x + y + 35/x – y = 11 …(2)

Solving equations (1) and (2), we get-

∴ 100 - 105/x – y = 32 – 33

∴ x – y = 5

and x + y = 13

∴ y = 4 km/hr

51. If a sum of money placed at compound interest, compounded annually, doubles itself in 5 years, then the same amount of money will be 8 times of itself in-

a. 25 years

b. 20 years

c. 15 years

d. 10 years

Ans : Required time = 5log 8/log 2

= 5*3log 2/log 2

= 15 years

52. A person has left an amount of Rs.1,20,000 to be divided between his 2 son aged 14 years and 12 years such that they get equal amounts when each attains 18 years of age. If the amount gets a simple interest of 5% per annum, the younger son’s share at present is-

a. Rs.48,800

b. Rs.57,600

c. Rs.62,400

d. Rs.84,400

Ans : Let the present share of the younger son be Rs. x.

∴ The share of the elder son = Rs. (120000 – x)

∵ x + x*6*5/100

= (120000 – x) + (120000 – x)*4*5/100

⇒ 130x/100 = (120000 –x)*120/100

⇒ 13x = 1440000 – 12x

⇒ 25x = 1440000

∴ x = Rs. 57600

53. If the simple interest on Rs.x at a rate of a% for m years is same as that on Rs. y at a rate of a2% for m2 years, then x : y is equal to-

a. m : a

b. am : 1

c. 1/m : 1/a

d. 1/am : 1

∵ x*a*m/100 = y*a2*m2/100

⇒ x/y = am/1

∴ x : y = am : 1

54. Base of a right prism is an equilateral triangle of side 6 cm. If the volume of the prism is 108√3 cc, its height is-

a. 9 cm

b. 10 cm

c. 11 cm

d. 12 cm

Ans : Height of the prism = 108√3*4/√3*(6)2 = 12cm

55. If a + 1/a + 2 = 0, then the value of (a37 – 1/a100) is-

a. 0

b. -2

c. 1

d. 2

Ans : ∵ a + 1/a + 2 = 0

⇒ a2 + 1 +2a = 0

⇒ (a + 1)2 = 0

⇒ a +1 = 0

∴ a = -1

∴ a37 -1/a100 = (-1) – (1) = -2

56. The value of k for which the graphs of (k-1) x+y-2 = 0 and (2-k) x -3y + 1 = 0 are parallel is-

a. ½

b. -1/2

c. 2

d. -2

Ans : (k – 1) x + y – 2 = 0

∴ y = (1 – k) x + 2 ….(1)

and (2 –k) x – 3y – 1 = 0

3y = (2 – k) x +1

Y = 2 – k/3 x + 1/3 ….(2)

∵ m1 = m2

⇒ 1 – k = 2 – k/3

⇒ 3 – 3k = 2 - k

∴ k = 1/2

57. If a2 + b2 + c2 = 2 (a-b-c) – 3, then the value of (a – b + c) is-

a. -1

b. 3

c. 1

d. -2

Ans : a2 + b2 + c2 = 2(a – b – c) – 3

⇒ a2 - 2a + 1 + b2 + 2b + 1 + c2 + 2c + 1 = 0

⇒ (a – 1)2 + (b + 1)2 + (c + 1)2 = 0

⇒ a – 1 = 0, b + 1 = 0, c + 1 = 0

⇒ a = 1, b = -1, c = -1

∴ a + b – c = 1 – 1 + 1 =1

58. If x2 + 3x + 1 = 0, then the value of x3 + 1/x3 is-

a. -18

b. 18

c. 36

d. -36

Ans : ∵ x2 + 3x + 1 = 0

⇒ x + 3 + 1/x = 0

⇒ x + 1/x = -3

⇒ (x + 1/x)3 = (-3)3

⇒ x3 + 1/x3 + 3 (-3) = -27

∴ x3 + 1/ x3 = -18

59. If xa, xb, xc = 1, then the value of a3 + b3 + c3 is –

a. 9

b. abc

c. a + b + c

d. 3abc

Ans : ∵ xa. xb. xc = 1

⇒ xa + b + c = x0

⇒ a +b + c =0

∴ a3 + b3 + c3 = 3abc

60. Base of a right pyramid is a square, length of diagonal of the base is 24√2 m. If the volume of the pyramid is 1728 cu.m, its height is-

a. 7 m

b. 8 m

c. 9 m

d. 10 m

Ans : Area of the base of the pyramid

= ½ (24√2)2 = 576m2

∴ Height of pyramid + 1728*3/576 = 9m

61. The height of a right circular cone and the radius of its circular base are respectively 9 cm and 3 cm. The cone is cut by a plane parallel to its base so as to divide it into two parts. The volume of the frustum (i.e., the lower part) of the cone is 44 cubic cm. The radius of the upper circular surface of the frustum (taking = 22-7) is-

a. 3√12 cm

b. 3√13 cm

c. 3√6 cm

d. 3√20 cm

Ans : Let the radius of the upper circular part of the frustum be r cm.

(Picture)

Then r/3 = x/9 = AC/AD

∴ x = 3r (where AC = x cm)

⇒ /3h(r12 + r1r2+r22 = 44

⇒ /3(9 – x)*(9 + 3r + r2) = 44

⇒ (9 – x)*(9 + 3r + r2) = 44*3*7/22 = 42

⇒ 81 + 27r + 9r2 – 9x – 3rx - r2x = 42 On putting x = 3r,

⇒ 81 + 27r + 9r2 - 9r2 - 3r3 – 27r = 42

⇒ 3r3 = 39

∴ r = 3√13 cm

62. The ratio of radii of two right circular cylinder is 2 : 3 and their heights are in the ratio 5 : 4. The ratio of their curved surface area is-

a. 5 : 6

b. 3 : 4

c. 4 : 5

d. 2 : 3

Ans : Required ratio = 2*2r*5h/2*3r*4h = 5 : 6

63. A solid cylinder has total surface area of 462 sq.cm. Curved surface area is 1/3rd of its total surface area. The volume of the cylinder is-

a. 530 cm3

b. 536 cm3

c. 539 cm3

d. 545 cm3

Ans : ∵ 2r(r + h) = 462

and 2rh = 1/3*462 = 154

∴ r + h/h = 462/154 =3

∴ r + h = 3h

∴ r = 2h

∴ 2*2h2 = 154

∴ h2 = 154*7/22*4 = 49/4

= (7/2)2

∴ h = 7/2 cm

and r = 7 cm

∴ Volume of the cylinder

= 22/7*49*7/2

= 539 cm3

64. A cylinder and a cone have equal radii of their bases and equal heights. If their curved surface areas are in the ratio 8 : 5, the ratio of their radius and height is-

a. 1 : 2

b. 1 : 3

c. 2 : 3

d. 3 : 4

Ans : Let the radius and height of each are r and h respectively.

∵ 2rh/r√h2 + r2 = 8/5

∴ 10h = 8√r2 + h2

⇒ 100h2 =64r2 + 64h2

∴ h2 = 64 r2/36 = (4/3 r)2

⇒ h = 4/3r

∴ r : h = 3 : 4

65. A solid is hemispherical at the bottom and conical above. If the surface areas of the two parts are equal, then the ratio of radius and height of its conical part is-

a. 1 : 3

b. 1 :1

c. √3 : 1

d. 1 : √3

Ans : (Picture)

∵ rl = 2r2

⇒ l = 2r

⇒ √h2 + r2 = 2r

⇒ h2 = 3r2

∴ r : h = r/h

= 1 : √3

66. If O is the circumcentre of ∆ ABC and ∠OBC = 350, then the ∠BAC is equal to-

a. 550

b. 1100

c. 700

d. 350

Ans : ∴ ∠BOC = 1800 – (350 + 350) = 1100

(Picture)

∴ ∠BAC = 1/2 * 1100 = 550

67. If I is the incentre of ∆ ABC and ∠ BIC = 1350, then ∆ ABC is-

a. Acute angled

b. Equilateral

c. Right angled

d. Obtuse angled

Ans : (Picture)

∠BIC = 1350

⇒ B/2 + C/2 = 1800 – 1350 = 450

⇒ ∠B + ∠C = 900

∴ ∠A = 1800 – (∠B + ∠C) = 900

i.e., ∆ ABC is a right angled.

68. If sin2 (∝ + β/2) is-

a. 1

b. -1

c. 0

d. 0.5

∵ sin2 ∝ + sin2 β = 2

⇒ 1 – cos2 ∝ 1 - cos2 β = 2

⇒ cos2 ∝ cos2 β = 0

⇒ cos ∝ = 0

and cos β = 0

⇒ ∝ = /2 and β = /2

∴ cos (∝ + β/2) = cos [/2 + /2 / 2]

= cos /2 = 0

69. The length of a shadow of a vertical tower is 1/√3 times its height. The angle of elevation of the Sun is-

a. 300

b. 450

c. 600

d. 900

Ans : (Picture)

∵ tan θ = h/1√3 h = √3 = tan 600

70. The graphs of x +2y =3 and 3x-2y = 1 meet the Y-axis at two points having distance-

a. 8/3 units

b. 4/3 units

c. 1 unit

d. 2 units

Ans : When the graphs meet the Y-axis at two points.

Then, [x + 2y = 3] at x = 0 ⇒ [0, y1 =3/2]

[3x - 2y = 1] at x = 0

and i.e., [0, y2 = -1/2]

Required distance = (y1 – y2)

= 3/2 – (- ½) = 2 units

71. If x+1/16x = 1, then the value of 64x3 + 1/64x3 is-

a. 4

b. 52

c. 64

d. 76

∵ x + 1/16x = 1

⇒ 16x2 – 16x + 1 =0

⇒ 16x2 – 16x + 4 = 3

⇒ (4x – 2)2 = 3

⇒ 4x = 2 + √3

⇒ 64x3 = (2±√3)3

=8 + 3√3 + 6√3 (2 + √3)

= 26 + 15√3

∴ 64x3 + 1/64x3 = (26 + 15√3) + 1/ (26 + 15√3)

= (26 + 15√3) + 26 - 15√3/676 -675

=52

72. If a, b, c, are three non-zero real numbers such that a + b + c = 0, and b2 ≠ ca, then the value of a2 + b2 + c2/ b2 –ca is-

a. 3

b. 2

c. 0

d. 1

∵ a + b + c = 0

⇒ a + c = -b

⇒ a2 + c2 = b2 -2ac

⇒ a2 + b2 + c2 = 2b2 – 2ac

∴ a2 + b2 + c2/ b2 ac = 2

73. If a4 + a2 b2 + b4 = 8 and a2 + ab + b2 = 4, then the value of ab is-

a. -1

b. 0

c. 2

d. 1

∵ a4 + a2 b2 + b4/ a2 + ab + b2 = 8/4

⇒ (a2 + b2)2 – (ab) 2/ (a2 + b2 + ab) = 2

⇒ a2 – ab + b2 = 2 ….(1)

and a2 + ab + b2 = 4 …..(2)

⇒ 2ab = 2

⇒ ab = 1

74. If a = 25, b = 15, c = -10, then the value of a3 + b3 + c3 – 3abc/ (a-b)2 + (b-c)2 + (c-a)2 is-

a. 30

b. -15

c. -30

d. 15

∵ a3 + b3 + c3 – 3abc

= (25)3 + (15)3 + (-10)3 – 3*25*15*(-10)

=15625 + 3375 – 1000 + 11250 = 29250

and (a - b)2 + (b – c)2 + (c – a)2

= (10)2 + (25)2 + (-35)2

= (10)2 + 625 + 1225

= 1950

∴ Required value = 29250/1950 =15

75. A, B, C are three points on a circle. The tangent at A meets BC produced at T, ∠BTA = 400, ∠CAT = 440. The angle subtended by BC at the centre of the circle is-

a. 840

b. 920

c. 960

d. 1040

Ans : (Picture)

∠ACB = 400 + 440 = 840

∴ ∠ACO = 900 - 440 = 460 = ∠OAC

⇒ ∠OCB = ∠ACB - ∠ACO

= 840 - 460 = 380 = ∠OBC

∴ ∠BOC = 1800 – (∠OCB + ∠OBC)

= 1800 – (380 + 380) = 1040

76. If the length of a chord of a circle at a distance of 12 cm from the Centre is 10 cm, then the diameter of the circle is-

a. 13 cm

b. 15 cm

c. 26 cm

d. 30 cm

Ans : (Picture)

∵ OA = √OM2 + AM2

= √122 + 52 = 13

∴ Diameter of the circle = 2*OA

= 2*13 = 26cm

77. In ∆ ABC, P and Q are the middle points of the sides AB and AC respectively. R is a point on the segment PQ such that PR : RQ = 1 : 2. If PR = 2 cm, then BC =

a. 4 cm

b. 2 cm

c. 12 cm

d. 6 cm

∵ PR/RQ = ½

But, PR = 2cm

RQ = 2*PR

= 4cm

(Picture)

∴ PQ = PR + RQ

= 2 + 4 = 6 cm

∴ BC = 2*PQ = 12CM

78. If tan θ tan 2θ = 1, then the value of sin2 2θ + tan2 is equal to –

a. ¾

b. 10/3

c. 3 ¾

d. 3

∵ tan θ * tan 2θ = 1

⇒ tan θ * 2 tan θ/1 – tan2 θ = 1

⇒ 2 tan2 θ = 1 - tan2 θ

⇒ 3 tan2 θ = 1

⇒ tan θ = 1/√3 = tan 300

∴ θ = 300

∴ sin2 2θ + tan2 2θ = sin2 600 + tan2 600

= ¾ + 3

=3 ¾

79. The value of cot /20 cot 3/20 cot 5/20 cot 7/20 cot 9/20 is-

a. -1

b. ½

c. 0

d. 1

Ans : Given Exp.

= cot /20. cot 3/20. cot 5/20. cot 7/20. cot 9/20

= cot 90. cot 270. cot 450. cot 630. cot 810

= cot 90. cot 270*1*tan 270. tan 90.

= cot 90. cot 270*1*1/ cot 270*1/ cot 90

=1

80. If sin θ+ cos θ = 17/13, 0<θ<900, then the value of sin θ- cos θ is-

a. 5/17

b. 3/19

c. 7/10

d. 7/13

∵ sin θ + cos θ = 17/13

⇒ sin2 θ + cos2 θ + 2sin θ.cos θ = 289/169

∴ 2sin θ.cos θ = 289 /169 -1

= 289 -169/169 = 120/169

∴ sin2 θ + cos2 θ – 2 sin2 θ.cos θ = 1 – 120/169

= 49/169

⇒ (sin θ – cos θ)2 = (7/13)2

∴ sin θ – cos θ = 7/13

S.S.C. Combined Graduate Level Exam

(Tier-II) 2011

Solved Paper – I

Arithmetical Ability

1. The product of 2 numbers is 1575 and their quotient is 9/7. Then the sum of the numbers is –

a. 74

b. 78

c. 80

d. 90

Ans : Let the numbers be x and y .

∴ xy = 1575

And x/y = 9/7

∴ xy/x/y = 1575/9/7

∴ y2 = 1225

∴ y = 35 and x = 45

∴ The sum of the numbers = 45+35

= 80

2. The value of (81)3.6 * (9)2.7/ (81)4.2 * (3) is __

a. 3

b. 6

c. 9

d. 8.2

Ans : (81)3.6 * (9)2.7/(81)4.2 * (3) = (3)14.4 * (3)5.4/(3)16.8 * (3)

= 314.4+5.4-16.8-1

= 32

= 9

3. √6+√6+√6+… is equal to –

a. 2

b. 5

c. 4

d. 3

Ans : Let √6+√6+√6+…. be x.

∴ x =√6+x

∴ x2 = 6+x

∴ x2 + x – 6 = 0

∴ (x-3) (x+2) = 0

∴ x = 3

4. The sum of the squares of two natural consecutive odd numbers is 394. The sum of the numbers is –

a. 24

b. 32

c. 40

d. 28

Ans : Let the consecutive odd numbers be x and (x+2)

∵ x2 + (x + 2)2 = 394

∴ x2 + x2 + 4x + 4 = 394

∴ 2 x2 +4x – 390 = 0

∴ x2 + 2x – 195 = 0

∴ (x +15) (x-13) = 0

∴ x = 13

∴ Required sum = 13 +15 =28

5. When (6767 +67) is divided by 68, the remainder is-

a. 1

b. 63

c. 66

d. 67

Ans: (6767+ 67) = 67(6766 + 166)

As 66 is an even number

∴ 6766 is an even number

∴ (6766 + 1) is perfectly divisible by (67 + 1)

i.e. 68

∴ The remainder = 67

6. In a division sum, the divisor is 4 times the quotient and twice the remainder. If a and b are respectively the divisor and the dividend then-

a. 4b-a2 /a =3

b. 4b-2a /a2 =2

c. (a+1)2 = 4b

d. A(a+2)/b = 4

Ans : As divisor is a, and dividend is b.

∴ Quotient = a/4

And Remainder = a/2

∴ b = a * a/4 + a/2

∴ 4b = a2 + 2a

∴ a(a+2)/b = 4

7. If 738 A6A is divisible by 11, then the value of A is-

a. 6

b. 3

c. 9

d. 1

Ans : As 738A6 A is divisible by 11.

∴ A + A + 3 = 6 + 8 + 7

∴ A = 9

8. The east positive integer that should be subtracted from 3011 * 3012 so that the difference is perfect square is-

a. 3009

b. 3010

c. 3011

d. 3012

Ans : ∵ 3011 * 3012 = 3011 (3011 + 1)

= (3011)2 + 3011

∴ Required least number = 3011

9. P, Q, R are employed to do a work for Rs. 5750. P and Q together finished 19/23 of work and Q and R together finished 8/23 of work. Wage of Q, in rupees, is-

a. 2850

b. 3750

c. 2750

d. 1000

Ans : Work done by Q = 19/23 + 8/23 – 1

4/23

∴ Wage of Q = 4/23 * 5750

= Rs. 1000

10. A can do a piece of work in 24 day, B in 32 days and C in 64 days. All begin to do it together, but A leaves after 6 days and B leaves 6 days before the completion of the work. How many days did the work last?

a. 15

b. 20

c. 18

d. 30

Ans : Work done by A = 6/24 = ¼

Work done by B = (x-6)/32

(where x is no. of days in which work is competed)

∵ ¼ + x – 6/32 + x/64 = 1

∴ 16 + 24 – 12 + x /64 = 1

∴ 3x + 4 = 64

∴ x = 60/3 = 20 days

11. The square root of (0.75)3 /1-0.75 + [0.75 + 90.75)2 +1] is-

a. 1

b. 2

c. 3

d. 4

Ans : The square root of

[(0.75)3/1 – 0.75 + {0.75 + (0.75)2 + 1}]

= √1.6875 + 2.3125

=√4

= 2

12. Given that √4096 = 64, the value of √4096 + √40.96 +√0.004096 is-

a. 70.4

b. 70.464

c. 71.104

d. 71.4

Ans : Given Exp. = √4096 + √40.96 +√0.004096

= 64 + 6.4 + 0.064

= 70.464

13. By selling an article at 3/4th of the marked price, there is a gain of 25%. The ratio of the marked price and the cost price is-

a. 5 : 3

b. 3 : 5

c. 3 : 4

d. 4 : 3

Ans : Let of M.P. be Rs. x.

∴ S.P. = Rs.3x/4

and C.P. = 3x/4 * 100/125

= Rs.3x/5

∴ required ratio = x: 3x/5

= 5:3

14. A and B earn in the ratio 2:1. They spend in the ratio 5:3 and save in the ratio 4:1. If the total monthly savings of both A and B are Rs.5,000, the monthly income of B is-

a. Rs.7,000

b. Rs.14,000

c. Rs.5,000

d. Rs.10,000

Ans : Let the monthly income of B be Rs. x.

∴ Monthly income of A = Rs. 2x and

Saving of A =5000 * 4/(4 + 1)

= Rs. 4000

Saving of B = Rs.1000

∵ 2x – 4000/x-1000 = 5/3

⇒ 6x – 12000 = 5x – 5000

∴ x = Rs.7000

15. The ratio of the sum of two numbers and their difference is 5:1. The ratio of the greater number to the smaller number is-

a. 2 : 3

b. 3 : 2

c. 5 : 1

d. 1 : 5

Ans : Let the numbers be x and y.

∵ x + y/ x – y = 5/1

⇒ 5x – 5y = x + y

⇒ 4x = 6y

⇒ x/y = 6/4

x : y = 3 : 2

16. A cistern has 3 pipes A, B and C. A and B can fill it in 3 and 4 hours respectively, and C can empty it in 1 hour. If the pipes are opened at 3 p.m., 4 p.m. and 5 p.m. respectively on the same day, the cistern will be empty at-

a. 7.12 p.m.

b. 7.15 p.m.

c. 7.10 p.m.

d. 7.18 p.m.

Ans : Let the cistern be emptied at x p.m.

∵ x -3/3 + x – 4/4 = x - 5/1

⇒ 4x – 12 + 3x – 12/12 = x – 5/1

⇒ 7x – 24 = 12x – 60

⇒ 5x = 36

∴ x = 36/5

= 7 hr. + 12 min.

= 7.12 p.m.

17. If A works alone, he would take 4 days more to complete the job than if both A and B worked together. If B worked alone, he would take 16 days more to complete the job than if A and B work together. How many days would they take to complete the work if both of them worked together?

a. 10 days

b. 12 days

c. 6 days

d. 8 days

Let x days are taken when they work together.

∴ Time taken by A to complete the work = (x + 4) days

And Time taken by B to complete the work = (x + 16) days

∵ 1/x = 1/x + 4 + 1/x + 16

∴ 1/x = x + 16 + x + 4/(x + 4) (x +16)

∴ 2x2 + 20x = x2 + 20x +64

∴ x2 = 64 (8)2

∴ x = 8 days

18. 250 men can finish a work in 20 days working 5 hours a day. To finish the work within 10 days working 8 hours a day, the minimum number of men required is-

a. 310

b. 300

c. 313

d. 312

Ans : Required no. of men = 250 * 5 * 20/8 * 10

=312.5

313

19. 2 men and 5 women can do a work in 12 days. 5 men 2 women can do that work in 9 days. Only 3 women can finish the same work in-

a. 36 days

b. 21 days

c. 30 days

d. 42 days

∵ (2m + 5w) * 12 = (5m + 2w) * 9

⇒ 24m + 60w = 45m + 18w

⇒ 21m = 42w

⇒ 1m = 2w

∴ 2m + 5w = 9w

∴ required no. of days = 9 * 12/3

= 36

20. While selling, a businessman allows 40% discount on the marked price and there is a loss of 30%. If it is sold at the marked price, profit per cent will be –

a. 10%

b. 20%

c. 16 2/3%

d. 16 1/3%

Ans : Let the M.P. be Rs.100.

∴ S.P = (100-40) = Rs.60

and C. P. = 60 * 100/(100-30) = Rs.600/7

∴ Reqd. % profit = 100-600/7 /600/7 * 100%

= (700 – 600)/7 * 600 * 100 * 7%

= 16 2/3%

21. Successive discount of 10% , 20% and 50% will be equivalent to a single discount of-

a. 36%

b. 64%

c. 80%

d. 56%

Ans : Equivalent to a single discount

= [100 - (100 – 10) (100 – 20) (100 – 50)/100*100]%

= [100 - 90*80*50/10000]%

= [100-36]%

= 64%

22. A retailer offers the following discount schemes for buyers on an article-

I. Two successive discounts of 10%

II. A discount of 12% followed by a discount of 8%.

III. Successive discounts of 15% and 5%

IV. A discount of 20%

The selling price will be minimum under the scheme-

a. I

b. II

c. III

d. IV

Ans : From (i) single discount = [10 + 10 – 10*10/100] % = 19%

From (ii) single discount = [12 + 8 -12*8/100] % = 19.04%

From (iii) single discount = [15 + 5 – 15*5/100] % = 19.25%

From (iv) single discount = 20%

∴ The S.P. will be minimum under the scheme IV.

23. Of three numbers, the second is thrice the first and the third number is three-fourth of the first. If the average of the three numbers is 114, the largest number is –

a. 72

b. 216

c. 354

d. 726

Ans : Let the first number be x.

∴ Second number = 3x and Third number = 3x/4

∵ x + 3x + 3x/4 = 3*114

⇒ 19x/4 = 342

∴ x = 342*4/19 = 72

∴ The largest number = 3*72

= 216

24. A car covers 1/5 of the distance from A to B at the speed of 8 km/hour, 1/10 of the distance at 25 km per hour and the remaining at the speed of 20 km per hour. Find the average speed of the whole journey-

a. 12.625 km/hr

b. 13.625 km/hr

c. 14.625 km/hr

d. 15.625 km/hr

Ans : If the whole journey be x km. The total time taken

= (x/5/8 + x/10/25 + 7x/10/20) hrs

= (x/40 + x/250 + 7x/200) hrs

= 25x + 4x + 35x/1000

= 64x/1000 hrs

∴ Average speed = x/64x/1000

= 15.625 km/hr

25. The average of 3 numbers is 154. The first number is twice the second and the second number is twice the third. The first number is-

a. 264

b. 132

c. 88

d. 66

Ans : Let the first number be x.

∴ Second number = x/2 and Third number = x/4

∵ x + x/2 + x/4 = 3 x 154

⇒ 7x/4 = 462

∴ x = 462*4/7

=264

26. The average salary of all the staff in an office of a corporate house is Rs. 5,000. The average salary of the officers is Rs. 14,000 and that of the rest is Rs. 4,000. If the total number of staff is 500, the number of officers is –

a. 10

b. 15

c. 25

d. 50

Ans : Let the number of officers be x.

∵ 5000*500 = 14000x + 4000(500-x)

∴ 2500000 =14000x + 2000000 – 4000x

∴ x = 500000/10000

= 50

27. The average marks of 40 students in an English exam are 72. Later it is found that three marks 64, 62 and 84 were wrongly entered as 60, 65 and 73. The average after mistakes were rectified is-

a. 70

b. 72

c. 71.9

d. 72.1

Ans : Correct average

= 40*72 + (64 + 62 +84) – 68 – 65 – 73/40

= 2880 + 210 – 206/40 = 2884/40

= 72.1

28. A and B are two alloys of gold and copper prepared by mixing metals in the ratio 7:2 and 7:11 respectively. If equal quantities of the alloys are melted to form a third alloy C, the ratio of gold and copper in C will be-

a. 5 : 7

b. 5 : 9

c. 7 : 5

d. 9 : 5

Ans : Quantity of gold in A = 7/9* wt. of A

Quantity of gold in B = 7/18* wt. of B

If 1 kg of each A and B are mixed to form third alloys C.

Then quantity of gold in 2 kg C = 7/9 + 7/18

= 7/6 kg

And Quantity of copper in 2 kg C = 2 – 7/6

= 5/6 kg

∴ Required ratio = 7/6 : 5/6 = 7 : 5

29. In a laboratory, two bottles contain mixture of acid and water in the ratio 2 : 5 in the first bottle and 7 : 3 in the second. The ratio in which the contents of these two bottles be mixed such that the new mixture has acid and water in the ratio 2 : 3 is-

a. 4 : 15

b. 9 : 8

c. 21 : 8

d. 1 : 2

Ans : Quantity of acid in first bottle = 2/7 x mix.

and Quantity of acid in second bottle = 7/10 x mix.

If x and 1 volumes are taken from I and II bottle respectively to form new mixture.

Then, (2/7 x + 7/10 * 1)/(5x/7 +3/10 * 1) = 2/3

⇒ 6x/7 21/10 = 10x/7 +6/10

⇒ 4x/7 = 15/10

∴ x = 15/10*7/4 = 21/8

∴ Required ratio = x : 1

= 21 : 8

30. A mixture contains 80% acid and rest water. Part of the mixture that should be removed and replaced by same amount of water to make the ratio of acid and water 4 : 3 is-

a. 1/3 rd

b. 3/7 th

c. 2/3 rd

d. 2/7 th

Ans : Let the initial wt. of mixture be 1 kg and x kg of mixture is taken out and replaced by same amount of water.

∵ Amt. of acid/Amt. of water = 0.8 – 0.8x/ (0.2 – 0.2x + x) = 4/3

⇒ 2.4 – 2.4x = 0.8 + 3.2x

⇒ 5.6x = 1.6

∴ x = 1.6/5.6 = 2/7th part

31. An employer reduces the number of his employees in the ratio 9 : 8 and increases their wages in the ratio 14 : 15. If the original wage bill was Rs. 189,900, find the ratio in which the wage bill is decreased-

a. 20 : 21

b. 21 : 20

c. 20 : 19

d. 19 : 21

Ans : Let the initial number of employees be 9x and the employer gives Rs. 14y as wage to each.

∵ 9x * 14y =18900

∴ xy = 150 and The later bill = 8x*15y = 120xy

= 120*150 = 18000

∴ Required ratio = 18000 : 18900

= 20 : 21

32. The batting average for 40 innings of a cricketer is 50 runs. His highest score exceeds his lowest score by 172 runs. If these two innings are excluded, the average of the remaining 38 innings is 48 runs. The highest score of the player is-

a. 165

b. 170

c. 172

d. 174

Ans : Let the max. number of runs be x.

∴ The lowest score = (x-172)

∵ 40*50 = 38*48 + x + (x-172)

⇒ 2000 = 1824 + 2x – 172

∴ x= 174

33. Walking at 3 km per hour, Pintu reaches his school 5 minutes late. If he walks at 4 km per hour he will be 5 minutes early. The distance of Pintu’s school from his house is-

a. 1 ½ Km

b. 2 Km

c. 2 ½ Km

d. 5 Km

Ans : Distance of the school from the house

= 4*3/(4 – 3) * 5 +5/60 km

= 12*1/6

=2 km

34. Nitin bought some oranges at Rs. 40 a dozen and an equal number at Rs.30 a dozen. He sold them at Rs. 45 a dozen and made a profit of Rs.480. The number of oranges, he bought, was-

a. 48 dozen

b. 60 dozen

c. 72 dozen

d. 84 dozen

Ans : Let the number of oranges bought be x.

∴ 45x/12 – 70x/24 = 480

⇒ (45-35)/12 = 480

⇒ x = 480*12/10 = 576

= 48*12 = 48 dozen

35. A man buys two chairs for a total cost of Rs.900. By selling one for 4/5 of its cost and the other for 5/4 of its cost, he makes a profit of Rs.90 on the whole transaction. The cost of the lower priced chair is-

a. Rs.360

b. Rs.400

c. Rs.420

d. Rs.300

Ans : Let the cost price of 1 chair be Rs. x.

∴ C.P. of other chair = Rs. (900-x)

∵ 4/5 x + 5/4 (900-x) = 900 + 90

⇒ 4/5 x +1125 -5x/4 = 990

∴ 9x/20 = 135

∴ x = 135*20/9

= Rs. 300

∴ C.P. of the lower priced chair is Rs. 300.

36. By selling 100 oranges, a vendor gains the selling price of 20 oranges. His gain per cent is-

a. 20

b. 25

c. 30

d. 32

Ans : Let the S.P. of 100 Oranges be Rs. x.

∴ S.P. of 20 oranges = x/100 *20 = Rs. x/5

∴C.P. of 100 oranges = x - x/5

= Rs. 4x/5

∴ Reqd. Profit % = x/5 * 100*5/4x %

= 25%

37. 60% of the cost price of an article is equal to 50% of its selling price. Then the percentage of profit or loss on the cost price is-

a. 20% loss

b. 16 2/3% profit

c. 20% profit

d. 10% loss

Ans : Let the cost price be Rs. 100.

∵ S.P * 50/100 = 100*16/100

∴ S.P = 60*100/50

= Rs. 120

∴ Reqd. % profit = (120 – 100)% = 20%

38. Maninder bought two horses at Rs.40,000 each. He sold one horse at 15% gain, but had to sell the second horse at a loss. If he hand suffered a loss of Rs.3,600 on the whole transaction, then the selling price of the second horse is-

a. Rs.30,000

b. Rs.30,200

c. Rs.30,300

d. Rs.30,400

Ans: C.P. of two horses = 2*40000 = Rs. 80000

and S.P. of two horse = 80000 – 3600 = Rs. 76400

∴ S.P. of the other horse = 76400 – 46000 = 30400

39. A fruit-seller buys x guavas for Rs.y and sells y guavas for Rs. x. If x>y, then he made-

a. x2 – y2 / xy % loss

b. x2 – y2 / xy % gain

c. x2 – y2 / y2 % loss

d. x2 – y2 / y2 * 100% gain

Ans : C.P. of 1 guava = Rs. y/x [x>y]

and S.P. of 1 guava = Rs. x/y

∴ Reqd. Gain% = x/y – y/x/ y/x * 100%

= x2 - y2 /y2 *100%

40. A jar contain 10 red marbles and 30 green ones. How many red marbles must be added to the jar so that 60% of the marbles will be red?

a. 25

b. 30

c. 35

d. 40

Ans : Let after adding x red marbles, the red marbles with be 60% of the total.

∵ (10+x)/ (10 + x) + 30 = 60/100

⇒ 10 + x/40 + x = 3/5

⇒ 50 + 5x = 120 + 3x

x = 70/2 = 35

41. If a number multiplied by 25% of itself gives a number which is 200% more than the number, then the number is –

a. 12

b. 16

c. 35

d. 24

Ans : Let the number be x.

∵ x*25x/100 = x + 200x/100

⇒ x2/4 = 3x

⇒ x2 - 12x = 0

⇒ x – 12 = 0

∴ x = 12

42. The value of an article depreciates every year at the rate of 10% of its value. If the present value of the article is Rs.729, then its worth 3 years ago was-

a. Rs.1250

b. Rs.1000

c. Rs.1125

d. Rs.1200

Ans : Let the worth 3 years ago be Rs. x.

∵ 729 = x (1 – 10/100)3

⇒ 729 = x*9*9*9/10*10*10

∴ x = Rs. 1000

43. The price of onions has been increased by 50%. In order to keep the expenditure on onions the same the percentage of reduction in consumption has to be-

a. 50%

b. 33 1/3%

c. 33%

d. 30%

Ans : Reqd. Percentage of reduction

= 50*100/(100 + 50) %

= 5000/150 %

= 33 1/3%

44. A took two loans altogether of Rs.1200 from B and C. B claimed 14% simple interest per annum, while C claimed 15% per annum. The total interest paid by A in one year was Rs.172. Then, A borrowed-

a. Rs.800 from C

b. Rs.625 from C

c. Rs.400 from B

d. Rs.800 from B

Ans : If A borrowed Rs. x from B. and A borrowed Rs. Rs. (1200 – x) from C.

∵ (1200 – x)*15*1/100 + x*14*1/100

⇒ 18000 – 15x + 14x = 172*100

x = Rs. 800

45. If a regular polygon has each of its angles equal to 3/5 times of two right angles, then the number of side is-

a. 3

b. 5

c. 6

d. 8

Ans : If the number of sides a regular polygon be n.

Then (2n-4)/n = 2*3/5

⇒ (2n – 4)*5 = 6n

∴ n = 5

46. A square is of area 200 sq. m. A new square is formed in such a way that the length of its diagonal is √2 times of the diagonal of the given square. The the area of the new square formed is-

a. 200√2 sq.m

b. 400√2 sq.m

c. 400 sq.m

d. 800 sq.m

Ans : Length of the diagonal of Ist square

= √2*200

= 20 m

∴ Length of the diagonal of new square = 20√2m

∴ Area of the new square = ½*(20.√2)2 = 400 sq. m

47. The heights of a cone, cylinder and hemisphere are equal. If their radii are in the ratio 2 : 3 : 1, then the ratio of their volumes is-

a. 2 : 9 : 2

b. 4 : 9 : 1

c. 4 : 27 : 2

d. 2 : 3 : 1

Ans : Ratio of their volumes [cone : cylinder : hemisphere]

= 1/3(2)2h : (3)2*h : 2/3(1)2.h

= 4/3 : 9 : 2/3 = 4 : 27 : 2

48. A motor-boat can travel at 10 km/hr in still water. It travelled 91 km downstream in a river and then returned to the same place, taking altogether 2 hours. Find the rate of flow of river-

a. 3 km/hr

b. 4 km/hr

c. 2 km/hr

d. 5 km/hr

Ans : Let the rate of flow of river be x km/hr.

∵ 91/(10 + x) + 91/(10 – x) = 20

⇒ 91(10 – x + 10 + x)(10 + x) (10 – x) = 20

⇒ 91*20 = 20(100 – x2)

⇒ x2 = 9 = (3)2

∴ x = 3 km/hr

49. A man driving at 3/4th of his original speed reaches his destination 20 minutes later than the usual time. Then the usual time is-

a. 45 minutes

b. 60 minutes

c. 75 minutes

d. 120 minutes

Ans : Let the original speed be x km/hr and the usual time be y hours.

∵ x * y = ¾ x(y+1/3)

∴ 4y = 3y + 1

∴ y = 1 hr = 60 minutes

50. A motor-boat, travelling at the same speed, can cover 25 km upstream and 39 km downstream in 8 hours. At the same speed, it can travel 35 km upstream and 52 km downstream in 11 hours. The speed of the stream is –

a. 2 km/hr

b. 3 km/hr

c. 4 km/hr

d. 5 km/hr

Ans : Let the speeds of motor boat and the stream be x and y km/hr respectively.

∵ 39/x + y + 25/x – y = 8 …(1)

and 52/x + y + 35/x – y = 11 …(2)

Solving equations (1) and (2), we get-

∴ 100 - 105/x – y = 32 – 33

∴ x – y = 5

and x + y = 13

∴ y = 4 km/hr

51. If a sum of money placed at compound interest, compounded annually, doubles itself in 5 years, then the same amount of money will be 8 times of itself in-

a. 25 years

b. 20 years

c. 15 years

d. 10 years

Ans : Required time = 5log 8/log 2

= 5*3log 2/log 2

= 15 years

52. A person has left an amount of Rs.1,20,000 to be divided between his 2 son aged 14 years and 12 years such that they get equal amounts when each attains 18 years of age. If the amount gets a simple interest of 5% per annum, the younger son’s share at present is-

a. Rs.48,800

b. Rs.57,600

c. Rs.62,400

d. Rs.84,400

Ans : Let the present share of the younger son be Rs. x.

∴ The share of the elder son = Rs. (120000 – x)

∵ x + x*6*5/100

= (120000 – x) + (120000 – x)*4*5/100

⇒ 130x/100 = (120000 –x)*120/100

⇒ 13x = 1440000 – 12x

⇒ 25x = 1440000

∴ x = Rs. 57600

53. If the simple interest on Rs.x at a rate of a% for m years is same as that on Rs. y at a rate of a2% for m2 years, then x : y is equal to-

a. m : a

b. am : 1

c. 1/m : 1/a

d. 1/am : 1

∵ x*a*m/100 = y*a2*m2/100

⇒ x/y = am/1

∴ x : y = am : 1

54. Base of a right prism is an equilateral triangle of side 6 cm. If the volume of the prism is 108√3 cc, its height is-

a. 9 cm

b. 10 cm

c. 11 cm

d. 12 cm

Ans : Height of the prism = 108√3*4/√3*(6)2 = 12cm

55. If a + 1/a + 2 = 0, then the value of (a37 – 1/a100) is-

a. 0

b. -2

c. 1

d. 2

Ans : ∵ a + 1/a + 2 = 0

⇒ a2 + 1 +2a = 0

⇒ (a + 1)2 = 0

⇒ a +1 = 0

∴ a = -1

∴ a37 -1/a100 = (-1) – (1) = -2

56. The value of k for which the graphs of (k-1) x+y-2 = 0 and (2-k) x -3y + 1 = 0 are parallel is-

a. ½

b. -1/2

c. 2

d. -2

Ans : (k – 1) x + y – 2 = 0

∴ y = (1 – k) x + 2 ….(1)

and (2 –k) x – 3y – 1 = 0

3y = (2 – k) x +1

Y = 2 – k/3 x + 1/3 ….(2)

∵ m1 = m2

⇒ 1 – k = 2 – k/3

⇒ 3 – 3k = 2 - k

∴ k = 1/2

57. If a2 + b2 + c2 = 2 (a-b-c) – 3, then the value of (a – b + c) is-

a. -1

b. 3

c. 1

d. -2

Ans : a2 + b2 + c2 = 2(a – b – c) – 3

⇒ a2 - 2a + 1 + b2 + 2b + 1 + c2 + 2c + 1 = 0

⇒ (a – 1)2 + (b + 1)2 + (c + 1)2 = 0

⇒ a – 1 = 0, b + 1 = 0, c + 1 = 0

⇒ a = 1, b = -1, c = -1

∴ a + b – c = 1 – 1 + 1 =1

58. If x2 + 3x + 1 = 0, then the value of x3 + 1/x3 is-

a. -18

b. 18

c. 36

d. -36

Ans : ∵ x2 + 3x + 1 = 0

⇒ x + 3 + 1/x = 0

⇒ x + 1/x = -3

⇒ (x + 1/x)3 = (-3)3

⇒ x3 + 1/x3 + 3 (-3) = -27

∴ x3 + 1/ x3 = -18

59. If xa, xb, xc = 1, then the value of a3 + b3 + c3 is –

a. 9

b. abc

c. a + b + c

d. 3abc

Ans : ∵ xa. xb. xc = 1

⇒ xa + b + c = x0

⇒ a +b + c =0

∴ a3 + b3 + c3 = 3abc

60. Base of a right pyramid is a square, length of diagonal of the base is 24√2 m. If the volume of the pyramid is 1728 cu.m, its height is-

a. 7 m

b. 8 m

c. 9 m

d. 10 m

Ans : Area of the base of the pyramid

= ½ (24√2)2 = 576m2

∴ Height of pyramid + 1728*3/576 = 9m

61. The height of a right circular cone and the radius of its circular base are respectively 9 cm and 3 cm. The cone is cut by a plane parallel to its base so as to divide it into two parts. The volume of the frustum (i.e., the lower part) of the cone is 44 cubic cm. The radius of the upper circular surface of the frustum (taking = 22-7) is-

a. 3√12 cm

b. 3√13 cm

c. 3√6 cm

d. 3√20 cm

Ans : Let the radius of the upper circular part of the frustum be r cm.

(Picture)

Then r/3 = x/9 = AC/AD

∴ x = 3r (where AC = x cm)

⇒ /3h(r12 + r1r2+r22 = 44

⇒ /3(9 – x)*(9 + 3r + r2) = 44

⇒ (9 – x)*(9 + 3r + r2) = 44*3*7/22 = 42

⇒ 81 + 27r + 9r2 – 9x – 3rx - r2x = 42 On putting x = 3r,

⇒ 81 + 27r + 9r2 - 9r2 - 3r3 – 27r = 42

⇒ 3r3 = 39

∴ r = 3√13 cm

62. The ratio of radii of two right circular cylinder is 2 : 3 and their heights are in the ratio 5 : 4. The ratio of their curved surface area is-

a. 5 : 6

b. 3 : 4

c. 4 : 5

d. 2 : 3

Ans : Required ratio = 2*2r*5h/2*3r*4h = 5 : 6

63. A solid cylinder has total surface area of 462 sq.cm. Curved surface area is 1/3rd of its total surface area. The volume of the cylinder is-

a. 530 cm3

b. 536 cm3

c. 539 cm3

d. 545 cm3

Ans : ∵ 2r(r + h) = 462

and 2rh = 1/3*462 = 154

∴ r + h/h = 462/154 =3

∴ r + h = 3h

∴ r = 2h

∴ 2*2h2 = 154

∴ h2 = 154*7/22*4 = 49/4

= (7/2)2

∴ h = 7/2 cm

and r = 7 cm

∴ Volume of the cylinder

= 22/7*49*7/2

= 539 cm3

64. A cylinder and a cone have equal radii of their bases and equal heights. If their curved surface areas are in the ratio 8 : 5, the ratio of their radius and height is-

a. 1 : 2

b. 1 : 3

c. 2 : 3

d. 3 : 4

Ans : Let the radius and height of each are r and h respectively.

∵ 2rh/r√h2 + r2 = 8/5

∴ 10h = 8√r2 + h2

⇒ 100h2 =64r2 + 64h2

∴ h2 = 64 r2/36 = (4/3 r)2

⇒ h = 4/3r

∴ r : h = 3 : 4

65. A solid is hemispherical at the bottom and conical above. If the surface areas of the two parts are equal, then the ratio of radius and height of its conical part is-

a. 1 : 3

b. 1 :1

c. √3 : 1

d. 1 : √3

Ans : (Picture)

∵ rl = 2r2

⇒ l = 2r

⇒ √h2 + r2 = 2r

⇒ h2 = 3r2

∴ r : h = r/h

= 1 : √3

66. If O is the circumcentre of ∆ ABC and ∠OBC = 350, then the ∠BAC is equal to-

a. 550

b. 1100

c. 700

d. 350

Ans : ∴ ∠BOC = 1800 – (350 + 350) = 1100

(Picture)

∴ ∠BAC = 1/2 * 1100 = 550

67. If I is the incentre of ∆ ABC and ∠ BIC = 1350, then ∆ ABC is-

a. Acute angled

b. Equilateral

c. Right angled

d. Obtuse angled

Ans : (Picture)

∠BIC = 1350

⇒ B/2 + C/2 = 1800 – 1350 = 450

⇒ ∠B + ∠C = 900

∴ ∠A = 1800 – (∠B + ∠C) = 900

i.e., ∆ ABC is a right angled.

68. If sin2 (∝ + β/2) is-

a. 1

b. -1

c. 0

d. 0.5

∵ sin2 ∝ + sin2 β = 2

⇒ 1 – cos2 ∝ 1 - cos2 β = 2

⇒ cos2 ∝ cos2 β = 0

⇒ cos ∝ = 0

and cos β = 0

⇒ ∝ = /2 and β = /2

∴ cos (∝ + β/2) = cos [/2 + /2 / 2]

= cos /2 = 0

69. The length of a shadow of a vertical tower is 1/√3 times its height. The angle of elevation of the Sun is-

a. 300

b. 450

c. 600

d. 900

Ans : (Picture)

∵ tan θ = h/1√3 h = √3 = tan 600

70. The graphs of x +2y =3 and 3x-2y = 1 meet the Y-axis at two points having distance-

a. 8/3 units

b. 4/3 units

c. 1 unit

d. 2 units

Ans : When the graphs meet the Y-axis at two points.

Then, [x + 2y = 3] at x = 0 ⇒ [0, y1 =3/2]

[3x - 2y = 1] at x = 0

and i.e., [0, y2 = -1/2]

Required distance = (y1 – y2)

= 3/2 – (- ½) = 2 units

71. If x+1/16x = 1, then the value of 64x3 + 1/64x3 is-

a. 4

b. 52

c. 64

d. 76

∵ x + 1/16x = 1

⇒ 16x2 – 16x + 1 =0

⇒ 16x2 – 16x + 4 = 3

⇒ (4x – 2)2 = 3

⇒ 4x = 2 + √3

⇒ 64x3 = (2±√3)3

=8 + 3√3 + 6√3 (2 + √3)

= 26 + 15√3

∴ 64x3 + 1/64x3 = (26 + 15√3) + 1/ (26 + 15√3)

= (26 + 15√3) + 26 - 15√3/676 -675

=52

72. If a, b, c, are three non-zero real numbers such that a + b + c = 0, and b2 ≠ ca, then the value of a2 + b2 + c2/ b2 –ca is-

a. 3

b. 2

c. 0

d. 1

∵ a + b + c = 0

⇒ a + c = -b

⇒ a2 + c2 = b2 -2ac

⇒ a2 + b2 + c2 = 2b2 – 2ac

∴ a2 + b2 + c2/ b2 ac = 2

73. If a4 + a2 b2 + b4 = 8 and a2 + ab + b2 = 4, then the value of ab is-

a. -1

b. 0

c. 2

d. 1

∵ a4 + a2 b2 + b4/ a2 + ab + b2 = 8/4

⇒ (a2 + b2)2 – (ab) 2/ (a2 + b2 + ab) = 2

⇒ a2 – ab + b2 = 2 ….(1)

and a2 + ab + b2 = 4 …..(2)

⇒ 2ab = 2

⇒ ab = 1

74. If a = 25, b = 15, c = -10, then the value of a3 + b3 + c3 – 3abc/ (a-b)2 + (b-c)2 + (c-a)2 is-

a. 30

b. -15

c. -30

d. 15

∵ a3 + b3 + c3 – 3abc

= (25)3 + (15)3 + (-10)3 – 3*25*15*(-10)

=15625 + 3375 – 1000 + 11250 = 29250

and (a - b)2 + (b – c)2 + (c – a)2

= (10)2 + (25)2 + (-35)2

= (10)2 + 625 + 1225

= 1950

∴ Required value = 29250/1950 =15

75. A, B, C are three points on a circle. The tangent at A meets BC produced at T, ∠BTA = 400, ∠CAT = 440. The angle subtended by BC at the centre of the circle is-

a. 840

b. 920

c. 960

d. 1040

Ans : (Picture)

∠ACB = 400 + 440 = 840

∴ ∠ACO = 900 - 440 = 460 = ∠OAC

⇒ ∠OCB = ∠ACB - ∠ACO

= 840 - 460 = 380 = ∠OBC

∴ ∠BOC = 1800 – (∠OCB + ∠OBC)

= 1800 – (380 + 380) = 1040

76. If the length of a chord of a circle at a distance of 12 cm from the Centre is 10 cm, then the diameter of the circle is-

a. 13 cm

b. 15 cm

c. 26 cm

d. 30 cm

Ans : (Picture)

∵ OA = √OM2 + AM2

= √122 + 52 = 13

∴ Diameter of the circle = 2*OA

= 2*13 = 26cm

77. In ∆ ABC, P and Q are the middle points of the sides AB and AC respectively. R is a point on the segment PQ such that PR : RQ = 1 : 2. If PR = 2 cm, then BC =

a. 4 cm

b. 2 cm

c. 12 cm

d. 6 cm

∵ PR/RQ = ½

But, PR = 2cm

RQ = 2*PR

= 4cm

(Picture)

∴ PQ = PR + RQ

= 2 + 4 = 6 cm

∴ BC = 2*PQ = 12CM

78. If tan θ tan 2θ = 1, then the value of sin2 2θ + tan2 is equal to –

a. ¾

b. 10/3

c. 3 ¾

d. 3

∵ tan θ * tan 2θ = 1

⇒ tan θ * 2 tan θ/1 – tan2 θ = 1

⇒ 2 tan2 θ = 1 - tan2 θ

⇒ 3 tan2 θ = 1

⇒ tan θ = 1/√3 = tan 300

∴ θ = 300

∴ sin2 2θ + tan2 2θ = sin2 600 + tan2 600

= ¾ + 3

=3 ¾

79. The value of cot /20 cot 3/20 cot 5/20 cot 7/20 cot 9/20 is-

a. -1

b. ½

c. 0

d. 1

Ans : Given Exp.

= cot /20. cot 3/20. cot 5/20. cot 7/20. cot 9/20

= cot 90. cot 270. cot 450. cot 630. cot 810

= cot 90. cot 270*1*tan 270. tan 90.

= cot 90. cot 270*1*1/ cot 270*1/ cot 90

=1

80. If sin θ+ cos θ = 17/13, 0<θ<900, then the value of sin θ- cos θ is-

a. 5/17

b. 3/19

c. 7/10

d. 7/13

∵ sin θ + cos θ = 17/13

⇒ sin2 θ + cos2 θ + 2sin θ.cos θ = 289/169

∴ 2sin θ.cos θ = 289 /169 -1

= 289 -169/169 = 120/169

∴ sin2 θ + cos2 θ – 2 sin2 θ.cos θ = 1 – 120/169

= 49/169

⇒ (sin θ – cos θ)2 = (7/13)2

∴ sin θ – cos θ = 7/13

v

SSC Placement Paper : Bangalore ,25 October 2008

SUBEX LTD PAPER ON 25TH OCTOBER AT BANGALORE

Company profile

Subex Limited

Adarsh Tech Park

Outer Ring Road, Devearabisannalli

Bangalore - 560,037, India

www.subexworld.com

hii...I,m Krishnakumar Pisharody from Kerala...25oct2008 I attended one aptitude test for the post of (Engineer -Technical support in Subex Ltd,Outer ring road, Banglore.

First round was aptitude test , contains 25 questions, (Verbal, Analytical reasoning& brain teasers ).40 minutes was the test duration...It was overall easy...Here I describe some questions..

Verbal: They give some passages and after that some objective type questions..

Analytical Reasoning:

Q1.In one company there are 3 men, tom,jack,karan and 3 women.Eliza ,anni,meri.Company allots 9 cottages in one row. For one person one room will be allot. The staffs give some conditions to arrange their own cottages, that are some body dont wanna get room at end points. Ladies said that they cant live with adjacent cottages ,their next cottage must be vacant. somebody wanna live next to the end cottages. somebody wanna live along with their friends....these are the conditions .Then questions are based on this description like , How many live next to vacant cottages?, Anni lives in between who &who?,etc.....

Q2)Give some menu card for a hotel ,(monday holiday) some items are get only particular days, some items available only on one day per week, some items twice or thrice per week, and question based on this description...

BRAIN TEASERS:-

Q1) A and B do a work with 5 days ,A and C in 6 days ,B and C in 20 days.. If a do alone how many days it will take?

Q2) Three Orange, Six banana, Eight apple worth Rs.30. 7 Orange ,4 Banana ,2 apple worth Rs.20 , then 5Orange,5Banana,5Apple can get with how much ?

Q3)How many zeros will display in the product of number between 1-100?

Company profile

Subex Limited

Adarsh Tech Park

Outer Ring Road, Devearabisannalli

Bangalore - 560,037, India

www.subexworld.com

hii...I,m Krishnakumar Pisharody from Kerala...25oct2008 I attended one aptitude test for the post of (Engineer -Technical support in Subex Ltd,Outer ring road, Banglore.

First round was aptitude test , contains 25 questions, (Verbal, Analytical reasoning& brain teasers ).40 minutes was the test duration...It was overall easy...Here I describe some questions..

Verbal: They give some passages and after that some objective type questions..

Analytical Reasoning:

Q1.In one company there are 3 men, tom,jack,karan and 3 women.Eliza ,anni,meri.Company allots 9 cottages in one row. For one person one room will be allot. The staffs give some conditions to arrange their own cottages, that are some body dont wanna get room at end points. Ladies said that they cant live with adjacent cottages ,their next cottage must be vacant. somebody wanna live next to the end cottages. somebody wanna live along with their friends....these are the conditions .Then questions are based on this description like , How many live next to vacant cottages?, Anni lives in between who &who?,etc.....

Q2)Give some menu card for a hotel ,(monday holiday) some items are get only particular days, some items available only on one day per week, some items twice or thrice per week, and question based on this description...

BRAIN TEASERS:-

Q1) A and B do a work with 5 days ,A and C in 6 days ,B and C in 20 days.. If a do alone how many days it will take?

Q2) Three Orange, Six banana, Eight apple worth Rs.30. 7 Orange ,4 Banana ,2 apple worth Rs.20 , then 5Orange,5Banana,5Apple can get with how much ?

Q3)How many zeros will display in the product of number between 1-100?

SSC Placement Paper : Berhampur ,12 March 2008

1.Written: (1 hour)

* 20 objective question on c, c++.

* 2 programs out which only 1 has to written.

-Objective questions is quite difficult it requires deep knowledge of C.

-mainly pointer.

-Most probably they give String ,File Handling & Linklist programs.

2.Technical Interview:(1h 30 min)

* Question based on C. (pointer, Memory Allocation, String, RAM, Struture, file etc)

* They gave me program to write ie Double link list program using C. All inserting a node ,deleting a node, traversing of list.

3.Tchnical Interview:(1hour)

* Question on C, C++ and various concepts of C++, electronics and various branch subjects.

* Here also they gave me to write a single program which include all the concepts of OOPs (like polymorphyism, overloading, abstraction, encapsulation, dynamic binding etc)

4.HR Interview:(30 min)

* They asked general HR Questions.for this be confident, gentle and effective.

5.Telephonic Interview:(10 min)

* For our batch we had a telephonic round for this they have informed me one day before, the time for the telephonic interview.

* They asked me Questions from 5th Semester.ie form Microprcessor, Advance Electronic Circuit etc

some question from flip flops and digital circuits.

it was product based company.4.5 package.3000 stipend for 6 months.I realy feel glade that i was selected for subex World.

I am really thankful to my parents, friends ,teachers, Collage for their great support.

* 20 objective question on c, c++.

* 2 programs out which only 1 has to written.

-Objective questions is quite difficult it requires deep knowledge of C.

-mainly pointer.

-Most probably they give String ,File Handling & Linklist programs.

2.Technical Interview:(1h 30 min)

* Question based on C. (pointer, Memory Allocation, String, RAM, Struture, file etc)

* They gave me program to write ie Double link list program using C. All inserting a node ,deleting a node, traversing of list.

3.Tchnical Interview:(1hour)

* Question on C, C++ and various concepts of C++, electronics and various branch subjects.

* Here also they gave me to write a single program which include all the concepts of OOPs (like polymorphyism, overloading, abstraction, encapsulation, dynamic binding etc)

4.HR Interview:(30 min)

* They asked general HR Questions.for this be confident, gentle and effective.

5.Telephonic Interview:(10 min)

* For our batch we had a telephonic round for this they have informed me one day before, the time for the telephonic interview.

* They asked me Questions from 5th Semester.ie form Microprcessor, Advance Electronic Circuit etc

some question from flip flops and digital circuits.

it was product based company.4.5 package.3000 stipend for 6 months.I realy feel glade that i was selected for subex World.

I am really thankful to my parents, friends ,teachers, Collage for their great support.

SSC Placement Paper : Kolkota ,20 February 2011

Directions: In questions no. 76 to 80, a part of the sentence is underlined. Below are given alternatives to the underlined part at (A), (B) and (C) which may lmprove the sentence. Choose the correct alternative. In case no improvement is needed your answer is (D).

76. If I was you I would not sign the document.

(A) If I have been you

(B) If I were you

(C) If I had been you

(D) No improvement

77. They were astonished through his failure in the examination.

(A) from

(B) at

(C) with

(D) No improvement

The increasing sale of luxuries IS an index of the country,s prosperity.

(A) appendix

(B) pointer

(C) mark

(D) No improvement

When are you starting to write to your d 86. frien ?

(A) wanting

(B) going

(C) thinking

(D) No improvement

I prefer to ride than to walk.

(A) ride to walk

(B) riding than walking

(C) riding to walking

(D) No improvement

Directions: In questions no. 81 to 85, out of the four alternatives choose the one which can be substituted for the given words / sentence.

81. To give one,s authority to another

(A) Assign (B) Delegate (C) Represent (D) Designate

82. One who intervenes between two or more parties to settle differences

(A) Neutral (B) Intermediary

(C) Judge (D) Connoisseur

(A) Rebellion

(C) Revolt

(B) Mutiny

(D) Anarchy

(A) Unheard

(C) Audible

High sea waves earthquake

(A) Tsunami

(C) Hurricane

(B) Faint

(D) Inaudible

(B) Tornado

(D) Cyclone

Directions : In questions no. 86 to 90, groups of four words are given. In each group, one word is correctly spelt. Find the correctly spelt word and mark your answer in the Answer Sheet.

(A) budgetery

(C) budgetary

(A) occurence

(C) ocurence

(A) pidistrian

(C) pedestrian

(A) seperately

(C) seperatley

(A) embarrassed

(C) embarrased

(B) bugetary

(D) budge try

(B) occurrence

(D) occurance

(B) pedistrian

(D) pidestrian

(B) separately

(D) separatly

(B) embarassed

(D) embarased

Directions: In the following passage (Questions no. 91 to 100), some of the words have been left out. First read the passage over and try to understand what it is about. Then fill in the blanks with the help of the alternatives given. Mark your answer in the Answer Sheet.

When Anil passed his final university examination and got his 91 he decided to 92 and invited all his friends to a party to be 93 the following Sunday. He spent most of that Saturday 94 things ready and at 7.40 the first guest arrived. After that a steady 95 of people 96 and Anil was busy chatting to people and introducing them to one another. Anil had a wide 97 of friends and not everyone at the party knew everyone ~. ,A party is always a good way to break the 99 and get people talking,, Anil thought. The party soon got going and there was a feeling of relief at the 100 that

91. (A) licence

(C) degree

93. (A) planned

(C) offered

94. (A) preparing

(C) getting

95. (A) stream

(C) river

(B) bachelorhood

(D) diploma

(B) feast

(D) commemorate

(B) celebrated

(D) held

(B) putting

(D) doing

(B) current

(D) movement

96. (A) turned out

(C) turned in

97. (A) number

(C) circle

98. (A) else

(C) different

99. (A) silence

(C) monotony

100. (A) subject

(C) fact

(B) turned up

(D) turned down

(B) group

(D) round

(B) other

(D) person

(B) atmosphere

(D) ice

(B) point

(D) matter

PART-IV

GENERAL AWARENESS

(A) West Bengal

(C) Kerala

(B) Tamil Nadu

(D) Orissa

102. Which of the following countries connected by the Palk Strait ?

(A) India and Sri Lanka

(B) North Korea and South Korea

(C) Pakistan and China

(D) Britain and France

103. Match the following:

1. Hazaribagh a. Coal

2. Neyveli b. Iron

3. Jharia c. Lignite

4. Rourkela d. Mica

(A) lc, 2d, 3a, 4b

(B) ld, 2c, 3a, 4b

(C) la, 2b, 3c, 4d

(D) ld, 2c, 3b, 4a

What is the minimum percentage of votes a political party must get to acquire the status of a registered party?

(A) 1%

(C) 3%

(B) 2%

(D) 4%

106. Economic development depends on

(A) Natural resources

(B) Capital formation

(C) Size of the market

(D) All of the above

107. National Income is generated from

(A) any money-making activity

(B) any laborious activity

(C) any profit-making activity

(D) any productive activity

108. Money supply is governed by the

(A) Planning Commission

(B) Finance Commission

(C) Reserve Bank of India

(D) Commercial Banks

109. The headquarters of WTO is at

(A) New York (B) Doha

(C) Uruguay (D) Geneva

110. Which state is called the ,Rice Bowl, of India?

(A)

(B)

(C)

(D)

Andhra Pradesh

Tamil Nadu

105. Who estimated the National Income for 111. The highest waterfall of India is the first time in India ?

(A)

(B)

(C)

(D)

Mahalanobis

Dadabhai Naoroji

V.K.R.V. Rao

Sardar Patel

(A)

(B)

(C)

(D)

Shimsha falls

Hogenakkal falls

Courtallam falls

cqm - IV

~1¥i10:t4 JiHCf5I-ft

101/~-~ ~ ~ ~ ~ ~ ~ ? 1~ ~ ~ CfiT 31Tffiq; mm fc!ttT ,R f.:rR ~ ?

(A) ~ aPm1 (B) ~ ~ (A) >l1~fuCfl W~

(C) ~ (D) 3TIfum (B) ~ ~

102. f~"if"lRsid ~ ,B ~ ~ ,~ -Rz, ~ ¥ (C) ~ CfiT 31IcfiR

/, ~ ~ ? ". (D) ~ ~

(A) ~ ~ m~ 1~ ~ 3W1 ~ ~ QTill ~ ?

(B) ~ CfiTW:rr ~ ~ CfiTW:rr (A) ~ ~-3c:(C) QlfCfl«1H ~ ~ (B) ~ ~

"1"1fufcIT~~

(D) ~ ~ m (C) . ~ ~ "1"1fufcIT~~

(D) ~ 3c:~-~ ~ ~ f.14Gld cn1 \iITill ~ ?

(A) ~ 3WWT

(B) fcffi 3WWT

(C) ~ rorcf ~

(D) 04IQIRCfl ~

~it.3TI. CfiT ~&1IW1~ ~ ?

(A) ~ <.IT(C) ~ (D) ~

,lWd ~ ~ ~ "CflT ,~ CfiT Cfiiro, (~

~) CfiQT ~ ~ ?

(A) amr ~

(B) ~ ~

(C) ~

(D) Cfl,""lf2Cfl

111. ,lWd ~ ~ ~ ,JIi1>lQld "Cflf.:rm ~ ?

(A) ful:mr >rm1

(B) {il ~ ~ Cfr,1ffi

(C) cfi) dR1li >r,1ffi

(D) itrr >r,1ffi

f.1"i f"l f@d CfiT fiR;rr;r ~ :

1. ~

2. ~

3. ,lfU ~

(f",P~I$2)

(A) Ie, 2d, 3a, 4b 109.

(B) ld, 2e, 3a, 4b ~

(C) la, 2b, 3e, 4d

(D) ld, 2e, 3b, 4a

7 110. {1"Io1Riq; GR trR it ~ em-,B-em ~ ~ If(f ~?

(A) 1%

(C) 3%

(B) 2%

(D) 4%

1~ ~ ~ ~ 3W1 CfiT ~ ~ ~

~ fcn<:rr ~ ?

(A) liQ(1(B) ~~

(C) cft.it.3iR.cft. ucr

(D) ~ ~

112. Which one of the following is known as 118. Arrange the following Magadhan the ,immovable property, in the cell ? dynasties in chronological order :

(A) Carbohydrate (B) Fat

(C) Protein (D) Nucleic acid

113. Water from soil enters into the root

hairs owing to

(A) Atmospheric pressure

(B) Capillary pressure

(C) Root pressure

(D) Osmotic pressure

114. What is meant by a ,pir, III the Sufi tradition?

(A) The Supreme God

(B) The Guru of the Sufis

(C) The greatest of all Sufi saints

(D) The orthodox teacher who contests the Sufi beliefs

115. Khalsa Panth was created by Guru Gobind Singh in which year?

(A) 1599

(C) 1707

(B) 1699

(D) 1657

116. Who propounded the Panchsheel Principles?

(A) Mahatma Gandhi

(B) Lord Buddha

(C) Pandit Jawahar Lal Nehru

(D) Swami Dayanand Saraswati

117. On April 12, 1944 Subhash Chandra Bose hoisted the INA Flag in a town. In which StatelUnion Territory is that town now?

(A) Andaman and Nicobar Islands

(B) Tripura

(C) Manipur

(D) Mizoram

II. Sisunagas

III. Mauryas

IV. Haryankas

(A) IV, II, III and I

(B) II, I, IV and III

(C) IV, II, I and III

(D) III, I, IV and II

119. The term of office of the Comptroller and Auditor General of India is

(A) 3 years

(C) 5 years

(B) 4 years

(D) 6 years

120. Who was the first Chief Election Commissioner of India ?

(A) G.V. Mavlankar

(B) T. Swaminathan

(C) K.V.K. Sundaram

(D) Sukumar Sen

121. What IS the retirement age for a

Supreme Court Judge?

(A) 62 years

(C) 68 years

(B) 65 years

(D) 70 years

122. Name the ,Political Guru, of Mahatma Gandhi.

(A) Gopalakrishna Gokhale

(B) Bal Gangadhar Tilak

(C) Aurobindo Ghosh

(D) Lala Lajpat Rai

11y f.ilOOif~Rsl(j "# -B fcnOO M ch1f.(A) ~~

(B) ~ CflT ~

(C) ri ~ ~ it m~

(D) ~ cn1 ~3TI it ft;m: ~ CfTffi

q£q,(IClI<] ~

1/15. ~ ~ fuQ "[RT ~ tf~ cn1~ ~

Cfff it cn17f{ m ?

(A) 1599

(C) 1707

(B) 1699

(D) 1657

cjiI~jh.1it ~ CflT 51fd IClCfl "Cflf.:r 2lT ?

(A) ~ ,1T~

(B) ~ ~

(C) ~~~~-

(D) ~ ~ *H~ffi

117. 12~, 1944 q;) ~ ~ W ~~;pr:

/ "# ,~~ WIT, CflT ~ ~ 2lT I

~ ";f1R ~ WlGr fcru ~/~ ~ ~

it ~ ?

(A) ~ ~ f.ich1cill,( &T1-~

(B) ~

(C) ~

(D) floilJflV!

171f.i8lO. Oif~Rsl(j l1,l1:l it ,(1,Jlcj~n q;) CflI0ful1lj*1l,(

~:

I. -;:R:

II. f.< I~III. 11TIV. ~

(A) IV; II, III ~ I

(B) II, I, IV ~ III

(C) IV, II, I ~ III

(D) III, I, IV ~ II

1/19. ,liffif it ~ ~ l1QI~@qD~ CflT Cfll4Cfll0

fcmR ~ CflT mcrr ~ ?

(A) 3 Cfff (B) 4 Cfff

(C) 5 Cfff (D) 6 Cfff

1/ ,liffif cnr wm lJ&1 ¥fCl, ~ "Cflf.:r 2lT ?

(A) ~.cIT. 111Cl8Cfl,(

(B) it. ~141(C) it.cIT.it. ~

(D) ~ ~

121/ ~ o:<:tllll0ll it ~~ ~ B,iIf.i~R1 Ch1

( ~ fcf;(r;ft mm ~ ? .

(A) 62 Cfff (B) 65 Cfff

(C) 68 Cfff (D) 70 Cfff

1~ ~ ,1T~CflT ,,(I,Jl,i1R1Cfl :r" "Cflt.1 2lT ?

(A) JnqI0~t,IJ1 ~

(B) ~ ~ fuw:fi

(C) ~ ~

(D) ~ 01,Jlq(,j U<:f

123. The average life span of red blood 130. Which of the following is not a corpuscles is about computer network?

(A) 100 - 200 days

(B) 100 - 120 days

(C) 160 - 180 days

(D) 150 - 200 days

(A) Wide area network

(B) Local area network

(C) Personal network

(D) Metropolitan area network

124. Dormancy period of animals during 131. Winter season is called

(A) Aestivation

(C) Hibernation

(B) Regeneration

(D) Mutation

125. The angle in which a cricket ball should be hit to travel maximum horizontal distance is

(A) 60° with horizontal

(B) 45° with horizontal

(C) 30° with horizontal

(D) 15° with horizontal

126. The minimum number of geostationary satellites needed for uninterrupted global coverage is

(A) 3

(C) 2

(B) 4

(D) 1

127. The best conductor of electricity among the following is

(A) Copper (B) Iron

(C) Aluminium (D) Silver

128. Breeding and management of bees IS known as

(A) Sericulture

(C) Pisciculture

(B) Silviculture

(D) Apiculture

129. The vitamin necessary for coagulation

of blood is

(A) Vitamin B

(C) Vitamin K

(B) Vitamin C

(D) Vitamin E

When a group of computers IS connected together III a small area without the help of telephone lines, it is called

(A) Remote Communication Network

(RCN)

(B) Local Area Network (LAN)

(C) Wide Area Network (WAN)

(D) Value Added Network (VAN)

132. Which one of the following elements is used in the manufacture of fertilizers?

(A) Fluorine

(C) Lead

(B) Potassium

(D) Aluminium

(A) Isoprene

(C) Butadiene

(B) Styrene

(D) Ethylene

134. In addition to hydrogen, the other abundant element present on Sun,s surface is

(A) Helium

(C) Argon

(B) Neon

(D) Oxygen

135. Which of the following is the major constituent of LPG?

(A) Methane (B) Ethane

(C) Propane (D) Butane

136. Flight Recorder is technically called

(A) Dark box (B) Blind box

(C) Black box (D) Altitude meter

123. ~ ~-~,* q)f ~ ~-~ ,Wfl1lT

/ fcncH ~ q)f m-m ~ 7

(A) 100- 200 ~

(B) 100-120 ~

(C) 160- 180 ~

(D) 150- 200 ~

124. ~ ~ if ~,*ih ~-~ ~ cp;rr ~

/7

(A) ~{"CI~~H

(C) QI~(B) ;(lJlrt{~H

(D) R{2~H

125; ~ Ch1 ~ ~ ~ cituT ~ 11m \jfRf

I ~,~ CfQ 31f~ ~ ~ Mi~

~7

(A) ~ ~ 60° q)f cituT

(B) ~ -B 45° q)f cituT

(C) ~ -B 30° q)f cituT

(D) ~ ~ 15° q)f cituT

126. ~ ~ ~ m ~ Cfi11--B-Cfll1 133 7fcncH 8l~CflIct1 ~ ~it ~ 7

(A) 3 (B) 4

(C) 2 (D) 1

127. r;p:.,f~f~d if -B ~ ~ ~-~

~ 7

(A) "dT(C) ~128. ~~fcR94,i ih ~ ~ ~~ ~ cp;rr ~

/ i 7

(A) ~;(lCfl0,H (B) f~C,4lCflfil{

(C) fq;ffi) Cflfil{ (D) Q>!{l Cflfil{

Rk1f~fujd if ~ ~ CflW{c:{~ ;rit.~7

(A) ~ l$r ~

(B) ~ l$r ~

(C) ~4Rt;Cfl ~

(D) ~QHJI;(l4 l$r ~

~ ~ CflW{dl ~ ~ mz -B l$r if ~

2ct1!"h1,i ih dTU ih, m ~ ~ \ffi(IT ~,

m~cp;rr~i 7

(A) ~ ~ ~ (RCN)

(B) ~ l$r ~ (LAN)

(C) ~ l$r ~ (WAN)

(D) ~ ~ ~ (VAN)

132. ~ ih f.1l1fuT if Rk1f~f~d if -B ~ ~ m if ~ 7jfffiT ~ 7

(A) lRj3ffiR (B) c{j2F~14~

(C) ,BTm (D) ~5lICflfdCfl~ ~ ~ ~ 7

(A) 3i1~«J>fl,i (B) ~r~;(l,i

(C) ilX[c:l:SI{,i (D) "Qf~

134. ~ Ch1 ~ en: QI~~I,JI,i ih 31ffiCIT ~

/ ~ ~ (A) Q1f~4~ (B) f.:r3lf;r

(C) 3fi135. Rk1f~fujd if -B ~ ~.-qr.~. q)f ~ /~~7

(A) it~ (B)~?R .

(C) ~ (D) ~

129. ~ ih ~ ~ ~ FClGlfil,i 3ilcH4Cfl ~? 136. ~-3if~ q)f dCfl"?!Chl-.:w1 cp;rr ~ ?

I (A) MGlfil,i B (B) MGlfil,i C (A) ,ffiU ,was prepared by

(A) J.H. Gibbon

(B) Jonas E. Salk

(C) Robert Edwards

(D) James Simpson

(A) Mercury

(C) Cadmium

(B) Lead

(D) Zinc

(A) Troposphere

(C) Stratosphere

(B) Ionosphere

(D) Exosphere

140. Who discovered sea route to India via 147. The National Integration Council (NIC)

The ,Cape of Good Hope, ? is chaired by the

(A) Vasco da Gama

(B) Amundsen

(C) Christopher Columbus

(D) John Cabot

141. The 16-year-old school-boy, Arjun Vajpai who became the youngest Indian to successfully climb the world,s highest peak, Mount Everest, on May 22, 2010, is from which state ?

(A) Uttar Pradesh

(B) Madhya Pradesh

(C) National Capital Territory of Delhi

(D) Uttarakhand

142. Which one of the following is an ,Air-to-Air, missile?

(A) Prithvi

(C) Akash

(B) Agni

(D) Astra

(A) Prime Minister

(B) Finance Minister

(C) Home Minister

(D) President of India

(A) Nagasaki

(C) Tokyo

(B) Hiroshima

(D) Hong Kong

149. Which of the following countries has entered into an agreement with India for the supply of Advanced Jet Trainer (HAWK)?

(A) Russia (B) U.S.A.

(C) England (D) France

150. The name of the ,Cargo Ship, which sank off Mumbai coast recently, causing oil spill in the Arabian Sea, was

(A) Sagar Jyoti

(C) MSC Chitra

(B) Venus

(D) Golden Eagle

1~~31H ~ ~ cn1~ ~ ?

(A) ~ (B) T.fR

(C) ~ (D) lpn«1If1111

,"

~ ~ ~ it we, ~3TI 1/44. ~·71m.~. IDU ~ BCfl2I£l«i ~ Ch1

~ q.;T if qJfI1ld ~ Tf<1T~ ?

(A) ~ ern if (B) cff:q ern if

(C) ~ q.;T if (D):qr: q.;T if

14y lOil,ill1ldl Wr ~ CfiRUTit ~31T ~ ?

.( (A) -qro . (B) mm

(C) et:sfiilll1 (D) ~

138. q:;~ mm I~?

(A) ~

(C) ~

(B) Cfl,if2Cfl

(D) ffil:R;r ;:rrs-.:>

~ cn1Ucn2Wl ih fc;m:~ w:rreIT ~

~ q;rr{ 2TI ?

(A) ~.~. ~

(B) ~ t ~

(C) ~ q::sqg*i

"

(D) ~ f~J:qfl,i

140/~ 3W:n:rs ~, ih ~ ~ nCfi it ~

(~cn1~~cn12TI? .

(A) ~ fu ,lT11T

(B) ~

(C) fs:flf2.ILfl{ chl<1illfl

(D) 7jjf.:r ~

141:1 16-crff(~~"il1it~~,R~

22~, 2010 cn1 flLflZ1clllicfCfl ~ Cffffi ~

Cfil1 ~ CfiTCX1fu; ~, fcnB ~ CfiT~ Cffffi

~ ?

(A) ~~

(B) 11m ~

(C) ~ mfucf ~, ~

(D) 3f114. R~f( 1f@d if it ~ ~ ,~-it-~,

Cffffi ~ ~ ?

(A) ",2~ (B) 3lfT.:r

(C) ~ (D) ~

146. ~ ,R(f q:;Ql ~ ~ ~ ? 7(A) ~ (B) 3"l11l,i4:sZ1

(C) WldIQ4:sZ1 (D) illf%J01,:sZ1

147. ~ ~ ~ (~.~.m.) CfiT ~ /~~?

(A) *lR "4it

(B) fcffi"4it

(C) ~ "4it

(D) ~ ih ~

14 . ~"illR ~ (A) ,iIJllfliChl (B) f%U~ll1l

(C) itcFIT (D) ~rJICflrJI

1~ if it ~ ~ it ~ CfiT3?Rf ~

~ (~.~.~it.) it ~ cnm ~3TI

~ ?

(A) ~ (B) ~~~

(C) . ~ (D) >om

15~ ,l1lZ1ql~Cfl ~, CfiT ;wr ~, \ilT QR1

if ~ it fl~S::d2 ,R ~ Tf<1T ~ am ~

~mrRif~~Tf<1T~ I

(A) mrR ~ (B) cft;m

(C) ~.~.m. ~ (D) ~ ~

MANNER IN WHICH ANSWERS ARE TO BE GIVEN

StR ~ ~ fafU

Directions: Each question or incomplete statement is followed by four alternative suggested answers or completions. In each case, you are required to select the one that correctly answers the question or completes the statement and blacken l-l appropriate rectangle A, B, C

or D by HB pencil against the question concerned in the Answer-Sheet. (For V.". candidates corresponding rectangle will be

blackened by the scribe.)

The following example illustrates the manner in which the questions are required to be answered.

Example:

Question No. ,Q, -

Out of the four words given below, three are alike in some way and one is different. Find the ODD word ;

(A) Girl

(B) Boy

(C) Woman

(D) Chair

Explanation : In the above example, the correct answer is ,Chair, and this answer has been suggested at ,D,. Accordingly, the answer is to be indicated by blackening l-l the rectangle by HB pencil in column ,D, against Question

No. ,Q, in the manner indicated below;

Question No. ,Q,

There is only one correct answer to each question. You should~lacken 1-1 the rectangle of the appropriate column, viz., A, B, C or D. If you blacken l-l more than one rectangle against anyone question the answer will be treated as wrong. If you wish to cancel any answer, you should completely erase that black mark in the rectangle in the Answer-Sheet, and then blacken the rectangle of revised response. You are NOT required to mark your answers in this Booklet. All answers must be indicated in

the Answer-Sheet only.

76. If I was you I would not sign the document.

(A) If I have been you

(B) If I were you

(C) If I had been you

(D) No improvement

77. They were astonished through his failure in the examination.

(A) from

(B) at

(C) with

(D) No improvement

The increasing sale of luxuries IS an index of the country,s prosperity.

(A) appendix

(B) pointer

(C) mark

(D) No improvement

When are you starting to write to your d 86. frien ?

(A) wanting

(B) going

(C) thinking

(D) No improvement

I prefer to ride than to walk.

(A) ride to walk

(B) riding than walking

(C) riding to walking

(D) No improvement

Directions: In questions no. 81 to 85, out of the four alternatives choose the one which can be substituted for the given words / sentence.

81. To give one,s authority to another

(A) Assign (B) Delegate (C) Represent (D) Designate

82. One who intervenes between two or more parties to settle differences

(A) Neutral (B) Intermediary

(C) Judge (D) Connoisseur

(A) Rebellion

(C) Revolt

(B) Mutiny

(D) Anarchy

(A) Unheard

(C) Audible

High sea waves earthquake

(A) Tsunami

(C) Hurricane

(B) Faint

(D) Inaudible

(B) Tornado

(D) Cyclone

Directions : In questions no. 86 to 90, groups of four words are given. In each group, one word is correctly spelt. Find the correctly spelt word and mark your answer in the Answer Sheet.

(A) budgetery

(C) budgetary

(A) occurence

(C) ocurence

(A) pidistrian

(C) pedestrian

(A) seperately

(C) seperatley

(A) embarrassed

(C) embarrased

(B) bugetary

(D) budge try

(B) occurrence

(D) occurance

(B) pedistrian

(D) pidestrian

(B) separately

(D) separatly

(B) embarassed

(D) embarased

Directions: In the following passage (Questions no. 91 to 100), some of the words have been left out. First read the passage over and try to understand what it is about. Then fill in the blanks with the help of the alternatives given. Mark your answer in the Answer Sheet.

When Anil passed his final university examination and got his 91 he decided to 92 and invited all his friends to a party to be 93 the following Sunday. He spent most of that Saturday 94 things ready and at 7.40 the first guest arrived. After that a steady 95 of people 96 and Anil was busy chatting to people and introducing them to one another. Anil had a wide 97 of friends and not everyone at the party knew everyone ~. ,A party is always a good way to break the 99 and get people talking,, Anil thought. The party soon got going and there was a feeling of relief at the 100 that

91. (A) licence

(C) degree

93. (A) planned

(C) offered

94. (A) preparing

(C) getting

95. (A) stream

(C) river

(B) bachelorhood

(D) diploma

(B) feast

(D) commemorate

(B) celebrated

(D) held

(B) putting

(D) doing

(B) current

(D) movement

96. (A) turned out

(C) turned in

97. (A) number

(C) circle

98. (A) else

(C) different

99. (A) silence

(C) monotony

100. (A) subject

(C) fact

(B) turned up

(D) turned down

(B) group

(D) round

(B) other

(D) person

(B) atmosphere

(D) ice

(B) point

(D) matter

PART-IV

GENERAL AWARENESS

(A) West Bengal

(C) Kerala

(B) Tamil Nadu

(D) Orissa

102. Which of the following countries connected by the Palk Strait ?

(A) India and Sri Lanka

(B) North Korea and South Korea

(C) Pakistan and China

(D) Britain and France

103. Match the following:

1. Hazaribagh a. Coal

2. Neyveli b. Iron

3. Jharia c. Lignite

4. Rourkela d. Mica

(A) lc, 2d, 3a, 4b

(B) ld, 2c, 3a, 4b

(C) la, 2b, 3c, 4d

(D) ld, 2c, 3b, 4a

What is the minimum percentage of votes a political party must get to acquire the status of a registered party?

(A) 1%

(C) 3%

(B) 2%

(D) 4%

106. Economic development depends on

(A) Natural resources

(B) Capital formation

(C) Size of the market

(D) All of the above

107. National Income is generated from

(A) any money-making activity

(B) any laborious activity

(C) any profit-making activity

(D) any productive activity

108. Money supply is governed by the

(A) Planning Commission

(B) Finance Commission

(C) Reserve Bank of India

(D) Commercial Banks

109. The headquarters of WTO is at

(A) New York (B) Doha

(C) Uruguay (D) Geneva

110. Which state is called the ,Rice Bowl, of India?

(A)

(B)

(C)

(D)

Andhra Pradesh

Tamil Nadu

105. Who estimated the National Income for 111. The highest waterfall of India is the first time in India ?

(A)

(B)

(C)

(D)

Mahalanobis

Dadabhai Naoroji

V.K.R.V. Rao

Sardar Patel

(A)

(B)

(C)

(D)

Shimsha falls

Hogenakkal falls

Courtallam falls

cqm - IV

~1¥i10:t4 JiHCf5I-ft

101/~-~ ~ ~ ~ ~ ~ ~ ? 1~ ~ ~ CfiT 31Tffiq; mm fc!ttT ,R f.:rR ~ ?

(A) ~ aPm1 (B) ~ ~ (A) >l1~fuCfl W~

(C) ~ (D) 3TIfum (B) ~ ~

102. f~"if"lRsid ~ ,B ~ ~ ,~ -Rz, ~ ¥ (C) ~ CfiT 31IcfiR

/, ~ ~ ? ". (D) ~ ~

(A) ~ ~ m~ 1~ ~ 3W1 ~ ~ QTill ~ ?

(B) ~ CfiTW:rr ~ ~ CfiTW:rr (A) ~ ~-3c:(C) QlfCfl«1H ~ ~ (B) ~ ~

"1"1fufcIT~~

(D) ~ ~ m (C) . ~ ~ "1"1fufcIT~~

(D) ~ 3c:~-~ ~ ~ f.14Gld cn1 \iITill ~ ?

(A) ~ 3WWT

(B) fcffi 3WWT

(C) ~ rorcf ~

(D) 04IQIRCfl ~

~it.3TI. CfiT ~&1IW1~ ~ ?

(A) ~ <.IT(C) ~ (D) ~

,lWd ~ ~ ~ "CflT ,~ CfiT Cfiiro, (~

~) CfiQT ~ ~ ?

(A) amr ~

(B) ~ ~

(C) ~

(D) Cfl,""lf2Cfl

111. ,lWd ~ ~ ~ ,JIi1>lQld "Cflf.:rm ~ ?

(A) ful:mr >rm1

(B) {il ~ ~ Cfr,1ffi

(C) cfi) dR1li >r,1ffi

(D) itrr >r,1ffi

f.1"i f"l f@d CfiT fiR;rr;r ~ :

1. ~

2. ~

3. ,lfU ~

(f",P~I$2)

(A) Ie, 2d, 3a, 4b 109.

(B) ld, 2e, 3a, 4b ~

(C) la, 2b, 3e, 4d

(D) ld, 2e, 3b, 4a

7 110. {1"Io1Riq; GR trR it ~ em-,B-em ~ ~ If(f ~?

(A) 1%

(C) 3%

(B) 2%

(D) 4%

1~ ~ ~ ~ 3W1 CfiT ~ ~ ~

~ fcn<:rr ~ ?

(A) liQ(1(B) ~~

(C) cft.it.3iR.cft. ucr

(D) ~ ~

112. Which one of the following is known as 118. Arrange the following Magadhan the ,immovable property, in the cell ? dynasties in chronological order :

(A) Carbohydrate (B) Fat

(C) Protein (D) Nucleic acid

113. Water from soil enters into the root

hairs owing to

(A) Atmospheric pressure

(B) Capillary pressure

(C) Root pressure

(D) Osmotic pressure

114. What is meant by a ,pir, III the Sufi tradition?

(A) The Supreme God

(B) The Guru of the Sufis

(C) The greatest of all Sufi saints

(D) The orthodox teacher who contests the Sufi beliefs

115. Khalsa Panth was created by Guru Gobind Singh in which year?

(A) 1599

(C) 1707

(B) 1699

(D) 1657

116. Who propounded the Panchsheel Principles?

(A) Mahatma Gandhi

(B) Lord Buddha

(C) Pandit Jawahar Lal Nehru

(D) Swami Dayanand Saraswati

117. On April 12, 1944 Subhash Chandra Bose hoisted the INA Flag in a town. In which StatelUnion Territory is that town now?

(A) Andaman and Nicobar Islands

(B) Tripura

(C) Manipur

(D) Mizoram

II. Sisunagas

III. Mauryas

IV. Haryankas

(A) IV, II, III and I

(B) II, I, IV and III

(C) IV, II, I and III

(D) III, I, IV and II

119. The term of office of the Comptroller and Auditor General of India is

(A) 3 years

(C) 5 years

(B) 4 years

(D) 6 years

120. Who was the first Chief Election Commissioner of India ?

(A) G.V. Mavlankar

(B) T. Swaminathan

(C) K.V.K. Sundaram

(D) Sukumar Sen

121. What IS the retirement age for a

Supreme Court Judge?

(A) 62 years

(C) 68 years

(B) 65 years

(D) 70 years

122. Name the ,Political Guru, of Mahatma Gandhi.

(A) Gopalakrishna Gokhale

(B) Bal Gangadhar Tilak

(C) Aurobindo Ghosh

(D) Lala Lajpat Rai

11y f.ilOOif~Rsl(j "# -B fcnOO M ch1f.(A) ~~

(B) ~ CflT ~

(C) ri ~ ~ it m~

(D) ~ cn1 ~3TI it ft;m: ~ CfTffi

q£q,(IClI<] ~

1/15. ~ ~ fuQ "[RT ~ tf~ cn1~ ~

Cfff it cn17f{ m ?

(A) 1599

(C) 1707

(B) 1699

(D) 1657

cjiI~jh.1it ~ CflT 51fd IClCfl "Cflf.:r 2lT ?

(A) ~ ,1T~

(B) ~ ~

(C) ~~~~-

(D) ~ ~ *H~ffi

117. 12~, 1944 q;) ~ ~ W ~~;pr:

/ "# ,~~ WIT, CflT ~ ~ 2lT I

~ ";f1R ~ WlGr fcru ~/~ ~ ~

it ~ ?

(A) ~ ~ f.ich1cill,( &T1-~

(B) ~

(C) ~

(D) floilJflV!

171f.i8lO. Oif~Rsl(j l1,l1:l it ,(1,Jlcj~n q;) CflI0ful1lj*1l,(

~:

I. -;:R:

II. f.< I~III. 11TIV. ~

(A) IV; II, III ~ I

(B) II, I, IV ~ III

(C) IV, II, I ~ III

(D) III, I, IV ~ II

1/19. ,liffif it ~ ~ l1QI~@qD~ CflT Cfll4Cfll0

fcmR ~ CflT mcrr ~ ?

(A) 3 Cfff (B) 4 Cfff

(C) 5 Cfff (D) 6 Cfff

1/ ,liffif cnr wm lJ&1 ¥fCl, ~ "Cflf.:r 2lT ?

(A) ~.cIT. 111Cl8Cfl,(

(B) it. ~141(C) it.cIT.it. ~

(D) ~ ~

121/ ~ o:<:tllll0ll it ~~ ~ B,iIf.i~R1 Ch1

( ~ fcf;(r;ft mm ~ ? .

(A) 62 Cfff (B) 65 Cfff

(C) 68 Cfff (D) 70 Cfff

1~ ~ ,1T~CflT ,,(I,Jl,i1R1Cfl :r" "Cflt.1 2lT ?

(A) JnqI0~t,IJ1 ~

(B) ~ ~ fuw:fi

(C) ~ ~

(D) ~ 01,Jlq(,j U<:f

123. The average life span of red blood 130. Which of the following is not a corpuscles is about computer network?

(A) 100 - 200 days

(B) 100 - 120 days

(C) 160 - 180 days

(D) 150 - 200 days

(A) Wide area network

(B) Local area network

(C) Personal network

(D) Metropolitan area network

124. Dormancy period of animals during 131. Winter season is called

(A) Aestivation

(C) Hibernation

(B) Regeneration

(D) Mutation

125. The angle in which a cricket ball should be hit to travel maximum horizontal distance is

(A) 60° with horizontal

(B) 45° with horizontal

(C) 30° with horizontal

(D) 15° with horizontal

126. The minimum number of geostationary satellites needed for uninterrupted global coverage is

(A) 3

(C) 2

(B) 4

(D) 1

127. The best conductor of electricity among the following is

(A) Copper (B) Iron

(C) Aluminium (D) Silver

128. Breeding and management of bees IS known as

(A) Sericulture

(C) Pisciculture

(B) Silviculture

(D) Apiculture

129. The vitamin necessary for coagulation

of blood is

(A) Vitamin B

(C) Vitamin K

(B) Vitamin C

(D) Vitamin E

When a group of computers IS connected together III a small area without the help of telephone lines, it is called

(A) Remote Communication Network

(RCN)

(B) Local Area Network (LAN)

(C) Wide Area Network (WAN)

(D) Value Added Network (VAN)

132. Which one of the following elements is used in the manufacture of fertilizers?

(A) Fluorine

(C) Lead

(B) Potassium

(D) Aluminium

(A) Isoprene

(C) Butadiene

(B) Styrene

(D) Ethylene

134. In addition to hydrogen, the other abundant element present on Sun,s surface is

(A) Helium

(C) Argon

(B) Neon

(D) Oxygen

135. Which of the following is the major constituent of LPG?

(A) Methane (B) Ethane

(C) Propane (D) Butane

136. Flight Recorder is technically called

(A) Dark box (B) Blind box

(C) Black box (D) Altitude meter

123. ~ ~-~,* q)f ~ ~-~ ,Wfl1lT

/ fcncH ~ q)f m-m ~ 7

(A) 100- 200 ~

(B) 100-120 ~

(C) 160- 180 ~

(D) 150- 200 ~

124. ~ ~ if ~,*ih ~-~ ~ cp;rr ~

/7

(A) ~{"CI~~H

(C) QI~(B) ;(lJlrt{~H

(D) R{2~H

125; ~ Ch1 ~ ~ ~ cituT ~ 11m \jfRf

I ~,~ CfQ 31f~ ~ ~ Mi~

~7

(A) ~ ~ 60° q)f cituT

(B) ~ -B 45° q)f cituT

(C) ~ -B 30° q)f cituT

(D) ~ ~ 15° q)f cituT

126. ~ ~ ~ m ~ Cfi11--B-Cfll1 133 7fcncH 8l~CflIct1 ~ ~it ~ 7

(A) 3 (B) 4

(C) 2 (D) 1

127. r;p:.,f~f~d if -B ~ ~ ~-~

~ 7

(A) "dT(C) ~128. ~~fcR94,i ih ~ ~ ~~ ~ cp;rr ~

/ i 7

(A) ~;(lCfl0,H (B) f~C,4lCflfil{

(C) fq;ffi) Cflfil{ (D) Q>!{l Cflfil{

Rk1f~fujd if ~ ~ CflW{c:{~ ;rit.~7

(A) ~ l$r ~

(B) ~ l$r ~

(C) ~4Rt;Cfl ~

(D) ~QHJI;(l4 l$r ~

~ ~ CflW{dl ~ ~ mz -B l$r if ~

2ct1!"h1,i ih dTU ih, m ~ ~ \ffi(IT ~,

m~cp;rr~i 7

(A) ~ ~ ~ (RCN)

(B) ~ l$r ~ (LAN)

(C) ~ l$r ~ (WAN)

(D) ~ ~ ~ (VAN)

132. ~ ih f.1l1fuT if Rk1f~f~d if -B ~ ~ m if ~ 7jfffiT ~ 7

(A) lRj3ffiR (B) c{j2F~14~

(C) ,BTm (D) ~5lICflfdCfl~ ~ ~ ~ 7

(A) 3i1~«J>fl,i (B) ~r~;(l,i

(C) ilX[c:l:SI{,i (D) "Qf~

134. ~ Ch1 ~ en: QI~~I,JI,i ih 31ffiCIT ~

/ ~ ~ (A) Q1f~4~ (B) f.:r3lf;r

(C) 3fi135. Rk1f~fujd if -B ~ ~.-qr.~. q)f ~ /~~7

(A) it~ (B)~?R .

(C) ~ (D) ~

129. ~ ih ~ ~ ~ FClGlfil,i 3ilcH4Cfl ~? 136. ~-3if~ q)f dCfl"?!Chl-.:w1 cp;rr ~ ?

I (A) MGlfil,i B (B) MGlfil,i C (A) ,ffiU ,was prepared by

(A) J.H. Gibbon

(B) Jonas E. Salk

(C) Robert Edwards

(D) James Simpson

(A) Mercury

(C) Cadmium

(B) Lead

(D) Zinc

(A) Troposphere

(C) Stratosphere

(B) Ionosphere

(D) Exosphere

140. Who discovered sea route to India via 147. The National Integration Council (NIC)

The ,Cape of Good Hope, ? is chaired by the

(A) Vasco da Gama

(B) Amundsen

(C) Christopher Columbus

(D) John Cabot

141. The 16-year-old school-boy, Arjun Vajpai who became the youngest Indian to successfully climb the world,s highest peak, Mount Everest, on May 22, 2010, is from which state ?

(A) Uttar Pradesh

(B) Madhya Pradesh

(C) National Capital Territory of Delhi

(D) Uttarakhand

142. Which one of the following is an ,Air-to-Air, missile?

(A) Prithvi

(C) Akash

(B) Agni

(D) Astra

(A) Prime Minister

(B) Finance Minister

(C) Home Minister

(D) President of India

(A) Nagasaki

(C) Tokyo

(B) Hiroshima

(D) Hong Kong

149. Which of the following countries has entered into an agreement with India for the supply of Advanced Jet Trainer (HAWK)?

(A) Russia (B) U.S.A.

(C) England (D) France

150. The name of the ,Cargo Ship, which sank off Mumbai coast recently, causing oil spill in the Arabian Sea, was

(A) Sagar Jyoti

(C) MSC Chitra

(B) Venus

(D) Golden Eagle

1~~31H ~ ~ cn1~ ~ ?

(A) ~ (B) T.fR

(C) ~ (D) lpn«1If1111

,"

~ ~ ~ it we, ~3TI 1/44. ~·71m.~. IDU ~ BCfl2I£l«i ~ Ch1

~ q.;T if qJfI1ld ~ Tf<1T~ ?

(A) ~ ern if (B) cff:q ern if

(C) ~ q.;T if (D):qr: q.;T if

14y lOil,ill1ldl Wr ~ CfiRUTit ~31T ~ ?

.( (A) -qro . (B) mm

(C) et:sfiilll1 (D) ~

138. q:;~ mm I~?

(A) ~

(C) ~

(B) Cfl,if2Cfl

(D) ffil:R;r ;:rrs-.:>

~ cn1Ucn2Wl ih fc;m:~ w:rreIT ~

~ q;rr{ 2TI ?

(A) ~.~. ~

(B) ~ t ~

(C) ~ q::sqg*i

"

(D) ~ f~J:qfl,i

140/~ 3W:n:rs ~, ih ~ ~ nCfi it ~

(~cn1~~cn12TI? .

(A) ~ fu ,lT11T

(B) ~

(C) fs:flf2.ILfl{ chl<1illfl

(D) 7jjf.:r ~

141:1 16-crff(~~"il1it~~,R~

22~, 2010 cn1 flLflZ1clllicfCfl ~ Cffffi ~

Cfil1 ~ CfiTCX1fu; ~, fcnB ~ CfiT~ Cffffi

~ ?

(A) ~~

(B) 11m ~

(C) ~ mfucf ~, ~

(D) 3f114. R~f( 1f@d if it ~ ~ ,~-it-~,

Cffffi ~ ~ ?

(A) ",2~ (B) 3lfT.:r

(C) ~ (D) ~

146. ~ ,R(f q:;Ql ~ ~ ~ ? 7(A) ~ (B) 3"l11l,i4:sZ1

(C) WldIQ4:sZ1 (D) illf%J01,:sZ1

147. ~ ~ ~ (~.~.m.) CfiT ~ /~~?

(A) *lR "4it

(B) fcffi"4it

(C) ~ "4it

(D) ~ ih ~

14 . ~"illR ~ (A) ,iIJllfliChl (B) f%U~ll1l

(C) itcFIT (D) ~rJICflrJI

1~ if it ~ ~ it ~ CfiT3?Rf ~

~ (~.~.~it.) it ~ cnm ~3TI

~ ?

(A) ~ (B) ~~~

(C) . ~ (D) >om

15~ ,l1lZ1ql~Cfl ~, CfiT ;wr ~, \ilT QR1

if ~ it fl~S::d2 ,R ~ Tf<1T ~ am ~

~mrRif~~Tf<1T~ I

(A) mrR ~ (B) cft;m

(C) ~.~.m. ~ (D) ~ ~

MANNER IN WHICH ANSWERS ARE TO BE GIVEN

StR ~ ~ fafU

Directions: Each question or incomplete statement is followed by four alternative suggested answers or completions. In each case, you are required to select the one that correctly answers the question or completes the statement and blacken l-l appropriate rectangle A, B, C

or D by HB pencil against the question concerned in the Answer-Sheet. (For V.". candidates corresponding rectangle will be

blackened by the scribe.)

The following example illustrates the manner in which the questions are required to be answered.

Example:

Question No. ,Q, -

Out of the four words given below, three are alike in some way and one is different. Find the ODD word ;

(A) Girl

(B) Boy

(C) Woman

(D) Chair

Explanation : In the above example, the correct answer is ,Chair, and this answer has been suggested at ,D,. Accordingly, the answer is to be indicated by blackening l-l the rectangle by HB pencil in column ,D, against Question

No. ,Q, in the manner indicated below;

Question No. ,Q,

There is only one correct answer to each question. You should~lacken 1-1 the rectangle of the appropriate column, viz., A, B, C or D. If you blacken l-l more than one rectangle against anyone question the answer will be treated as wrong. If you wish to cancel any answer, you should completely erase that black mark in the rectangle in the Answer-Sheet, and then blacken the rectangle of revised response. You are NOT required to mark your answers in this Booklet. All answers must be indicated in

the Answer-Sheet only.