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AMI Placement Paper : AMI Whole Testpaper

AMI PAPER ON 10th JULY

Hi friendsÖ. AMI came to my college on 10th July and selected only two members (Including me). Actually their recruitment process is very tough.

 

We had the interview for 2 days in first day; they come to my college and conduct only 2 rounds. They conduct remaining rounds in the company on next day itself. AMI is in kumaran nagar near sathyabama deemed university Chennai.

 

FIRST ROUND

Actually three sections in first round.

  1. C.
  2. C++.
  3. Logical reasoning.

In the first section they ask only 10 questions. Itís fully based on pointer concepts and basics in c. Each question carries one mark.
In the second section also they ask only 10 questions. Itís fully based on oops concepts. Each question carries one mark.
In the third section they ask 10 questions. Its easy only better study logical reasoning in sakunthala devi book. Each question carries different marks.
After that they will shortlist.

 

SECOND ROUND
The second round is not an elimination round. In this they conduct again written test. In this test we select either Assembly Language Program or Java Programming. Only 30 minutes time we have to attend all questions. They ask only 10 questions in ALP. But in java they ask 20 questions itís quite tough. Better take ALP it is plus point to yours. I took ALP only. Better go through the previous AMI question papers. I got many questions like this only. For example ďswap two members without using the temporary variablesĒ

 

THIRD ROUND

This is technical HR round. In this they check our technical skill fully. Normally itís for 45 minutes if your performance is good. Otherwise they will ask even for one and half an hour. Better concentrate on C and ALP. Once u knows the ALP itís a very good chance to got place in AMI. But you have to very good knowledge in both languages. For me they ask only for 40 minutes. Not only technical, they ask some logical reasoning questions also.

After this round there is a huge elimination. For us we will attend 9 members. But they short listed only 3 members.

 

FOURTH ROUND

Fourth round is GD. We donít have GD because only 3 members got cleared the technical. But they said fourth round is GD at time of pre-placement talk.

 

FIFTH ROUND

Fifth round is personal HR. This is quite easy. They mainly ask about the hobbies and what are all mentioned in the resume. Before going to this interview something know about your hobbies and what are all mentioned in the resume. Surely they will ask about the positive and negative points of your hobbies. Be bold but donít ask about the BOND. Normally they ask bond for 3 years. If they ask about the bond to you, simply tell I agree for the bond.

 

AMI Placement Paper : AMI Whole Testpaper

AMERICAN MEGATRENDS Inc.(AMI) BIOS- 24th FEB 2004 (KOLKATHA)

Logical Reasoning:

1. 3 types of chickens: Baby chickens cost 5 cents, hen chickens cost 3 dollares, rooster chickens cost 5 dollars. Buy 100 chickens for 100 dollars.How many will you have from each?

2. Calculate : (x-a)(x-b)(x-c)....(x-z)?

3. What is the day after 4 days after 2 days before the day before tomorrow?

4. 2 trains each of length 1 mile enter into 2 tunnels of lenth 1 mileand 2 miles apart. speed if trains are 5 and 10 miles/hr. Give the postion as viewed by a helicopter when the trains meet?

5. A says I,m not guilty
B says C is not guilty
C says A is not guilty
If all above are true.(i) Can we say anyone is guilty/innocent?
(ii)Who is most likely to be guilty?

6. 18 story building - Accountant,s office is 5 times the lawyer,s office. Architects office 2 floors up then he is halfway between Dentist,s & Accountant,s office. Architects office halfway down means betweenDentist,s and Lawyer,s office. Give the order

7. How many tennis matches are to be arranged to conduct a tournament with 213 players?

8. In a family 7 children don,t eat spinach, 6 don,t eat carrot, 5 don,t eat beans, 4 don,t eat spinach & carrots, 3 don,t eat carrot & beans, 2 don,t eat beans & spinach. 1 doesn,t eat all 3. Find the no. of children.

C/C++
-----
a. void main()
{
char a[6] = "INDIA";
while(*a)
{
printf("%c",*a);
a++;
}
}

1. india  2. I N D I A  3. no output  4. error  5.Null

b. protected derivation

c. int m = 10;
main()
{
int m = 20;
{int m = 30;}
printf("%d %d",m,::m);

1. 10 10 2. 10 20 3. 20 10 4. 20 30 5. 30 20 6. None

d. inline function is used to
1. reduce program size
2. reduce memory size

e. void func(float newtotal)
{ newtotal = newtotal-2;
}
main()
{
float tot = 100.34;
float sum = tot;
sum++;
func(sum);
cout< }

1. 100.34 2. 99.34 3. 101.34 4. 102.34 5. None

f. unsigned char snapnib(unsigned char misc)
{
int aloha, hamalo;
hamalo = aloha = misc;
misc = (hamalo << 4)+((aloha & 0x0f)>>4);
return(misc);
}
main()
{
int getinp;
unsig char getmisc;
scanf("%x",&getinp);
getmisc = snapnib(getinp & 0xff);
printf("%x",getmisc);
}

1. Change line 4 to misc = (hamalo >> 4)+((aloha & oxof)<<4);
2. remove the 5
3. Change line 4 to misc = (hamalo >> 4)+((aloha & oxfo)<<4);
4. Change line 11 to getmisc = snapnib(getinp);
5. Nothing

g. main()
{
int getinp = 6;
while(--getinp)
{
printf("%d",getinp);
if(getinp++ != 3);
else
break;
if(getinp-- != 5)
continue;
else
getinp = getinp - 2;
}

1. 5421 2. 5420 3. 541 4. 6543 5. 5410

h. unsigned char inn[2] = "5";
int i,j = 0;
sscanf(inn,"%d",&i);
while(i)
{
i = i - 1;
j++;
printf("%d",++j);
}

1. 14 2. 12 3. 24 4. 35 5. none

i. char *fn(int num)
{
char a[] = "Amen";
return(&a[num]);
}
main()
{
int i;
char *prechar;
scanf("%d",&i);
(input is 12)
prechar = fn(i);
printf("%c",*prechar);
}

1. a bus
2. give string "Amen" as global
3. use return(a[num]) instead of return(&a[num])
4. give main before fn

j. unsigned char u = 32767;
unsigned char y = 32768;
u = u + y;
printf("%d",(signed char)u);

1. 65535 2. -65535 3. -1 4. 1 5. none

k. *name = "ANYTHING";
*foo = "ALRIGHT";
*name1 = "WRONG";
strcat(name,name1);
printf("%s %s \n",name,foo);
return(0);

1. ANYTHING WRONG? WRONG?
2. ANYTHING
3. ANYTHING WRONG? ALRIGHT
4. ERROR
5. NONE

Assembly
--------
i) swap 2 variables without temp variables
(use XOR, OR, AND, NOT)

AMI Placement Paper : AMI Paper Whole Testpaper Kolkatha - 24 February 2004

AMERICAN MEGATRENDS Inc.(AMI) BIOS- 24th FEB 2004 (KOLKATHA)

Logical Reasoning:

1. 3 types of chickens: Baby chickens cost 5 cents, hen chickens cost 3 dollares, rooster chickens cost 5 dollars. Buy 100 chickens for 100 dollars.How many will you have from each?

2. Calculate : (x-a)(x-b)(x-c)....(x-z)?

3. What is the day after 4 days after 2 days before the day before tomorrow?

4. 2 trains each of length 1 mile enter into 2 tunnels of lenth 1 mileand 2 miles apart. speed if trains are 5 and 10 miles/hr. Give the postion as viewed by a helicopter when the trains meet?

5. A says I,m not guilty
B says C is not guilty
C says A is not guilty
If all above are true.(i) Can we say anyone is guilty/innocent?
(ii)Who is most likely to be guilty?

6. 18 story building - Accountant,s office is 5 times the lawyer,s office. Architects office 2 floors up then he is halfway between Dentist,s & Accountant,s office. Architects office halfway down means betweenDentist,s and Lawyer,s office. Give the order

7. How many tennis matches are to be arranged to conduct a tournament with 213 players?

8. In a family 7 children don,t eat spinach, 6 don,t eat carrot, 5 don,t eat beans, 4 don,t eat spinach & carrots, 3 don,t eat carrot & beans, 2 don,t eat beans & spinach. 1 doesn,t eat all 3. Find the no. of children.

C/C++
-----
a. void main()
{
char a[6] = "INDIA";
while(*a)
{
printf("%c",*a);
a++;
}
}

1. india  2. I N D I A  3. no output  4. error  5.Null

b. protected derivation

c. int m = 10;
main()
{
int m = 20;
{int m = 30;}
printf("%d %d",m,::m);

1. 10 10 2. 10 20 3. 20 10 4. 20 30 5. 30 20 6. None

d. inline function is used to
1. reduce program size
2. reduce memory size

e. void func(float newtotal)
{ newtotal = newtotal-2;
}
main()
{
float tot = 100.34;
float sum = tot;
sum++;
func(sum);
cout< }

1. 100.34 2. 99.34 3. 101.34 4. 102.34 5. None

f. unsigned char snapnib(unsigned char misc)
{
int aloha, hamalo;
hamalo = aloha = misc;
misc = (hamalo << 4)+((aloha & 0x0f)>>4);
return(misc);
}
main()
{
int getinp;
unsig char getmisc;
scanf("%x",&getinp);
getmisc = snapnib(getinp & 0xff);
printf("%x",getmisc);
}

1. Change line 4 to misc = (hamalo >> 4)+((aloha & oxof)<<4);
2. remove the 5
3. Change line 4 to misc = (hamalo >> 4)+((aloha & oxfo)<<4);
4. Change line 11 to getmisc = snapnib(getinp);
5. Nothing

g. main()
{
int getinp = 6;
while(--getinp)
{
printf("%d",getinp);
if(getinp++ != 3);
else
break;
if(getinp-- != 5)
continue;
else
getinp = getinp - 2;
}

1. 5421 2. 5420 3. 541 4. 6543 5. 5410

h. unsigned char inn[2] = "5";
int i,j = 0;
sscanf(inn,"%d",&i);
while(i)
{
i = i - 1;
j++;
printf("%d",++j);
}

1. 14 2. 12 3. 24 4. 35 5. none

i. char *fn(int num)
{
char a[] = "Amen";
return(&a[num]);
}
main()
{
int i;
char *prechar;
scanf("%d",&i);
(input is 12)
prechar = fn(i);
printf("%c",*prechar);
}

1. a bus
2. give string "Amen" as global
3. use return(a[num]) instead of return(&a[num])
4. give main before fn

j. unsigned char u = 32767;
unsigned char y = 32768;
u = u + y;
printf("%d",(signed char)u);

1. 65535 2. -65535 3. -1 4. 1 5. none

k. *name = "ANYTHING";
*foo = "ALRIGHT";
*name1 = "WRONG";
strcat(name,name1);
printf("%s %s \n",name,foo);
return(0);

1. ANYTHING WRONG? WRONG?
2. ANYTHING
3. ANYTHING WRONG? ALRIGHT
4. ERROR
5. NONE

Assembly
--------
i) swap 2 variables without temp variables
(use XOR, OR, AND, NOT)
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