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TCS Placement Questions & Answers :: TCS

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Tot. Mock Test: 40


Total Qs: 652+

NA
SHSTTON
2
Solv. Corr.
4
Solv. In. Corr.
6
Attempted
0 M:0 S
Avg. Time

481 / 652

Choose the correct option.

A calculator has a key for squaring and one for inverting. So if x is the displayed number, then pressing the square key will replace x by x2 and pressing the invert key will replace x by 1/x. If initially the number displayed is 6 and one alternately presses the invert and square key 16 times each, then the final number displayed (assuming no roundoff or overflow errors) will be


A(6^(-2))^16

B(6^(2))^16

C6^32

D632

Answer: Option B

Explanation:

Even number of inverse key has no effect on the number.
For example, Initially the given number is 6. Square key makes it 6^2 and invert key makes it 1/6^2. Now again square key makes it (1/6^2)^2=1/6^4 and invert key makes it 6^4.
Now observe clearly, after pressing square key 2 times, the power of 6 became 4.
By pressing the square key, the value got increased like 2, 4, 8, .... Which are in the format of 2^n.
So after the 16 pressings the power becomes 2^16
So the final number will be (6^2)^16=6^65536

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NA
SHSTTON
72
Solv. Corr.
244
Solv. In. Corr.
316
Attempted
0 M:2 S
Avg. Time

482 / 652

What will be the result of the following program?
#include <stdio.h>
int main()
{
static int i;
int j;
for(j=0;j<10;j++)
{
i= i+2;
i = i-j;
}
printf("%d",i);
return 0;
}

A25

B-25

C20

D-20

ENone of these

Answer: Option B

Explanation:

Here is no explanation for this answer

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NA
SHSTTON
14
Solv. Corr.
120
Solv. In. Corr.
134
Attempted
1 M:41 S
Avg. Time

483 / 652

What will be the result of the following program?
#include <stdio.h>
int main()
{
       int ones, twos, threes, others;
       int c;
       ones = twos = threes = others = 0;
       while ((c = getchar ()) != EOF)
       {
                  switch (c)
                  {
                     case '1': ++ones;
                     case '2': ++twos;
                     case '3': ++threes;
                                       break;
                     default: ++others;
                                       break;
                  }
       }
       printf ("%d  %d", ones, others);
       return 0;
    }

A1 1

B3 4

C3 2

D3 3

EError

Answer: Option D

Explanation:

Here is no explanation for this answer

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NA
SHSTTON
5
Solv. Corr.
22
Solv. In. Corr.
27
Attempted
0 M:21 S
Avg. Time

484 / 652

What will be the result of the following program?
#include <stdio.h>
void print(char *p);
int main()
{
char s[] = "T.C.S", *A;
print(s);
return 0;
}
void print(char *p)
{
while (*p != '\0')
{
if (*p != '.')
printf ("%c", *p);
p++;
}
}

ATCS

BT.C.S

CSegmentation fault

DNone of the above

Answer: Option A

Explanation:

Here is no explanation for this answer

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NA
SHSTTON
75
Solv. Corr.
139
Solv. In. Corr.
214
Attempted
0 M:27 S
Avg. Time

485 / 652

Choose the correct option.

Which of the choices is true for the mentioned declaration?

const char *p;
and
char * const p;

Choose one of them:


AYou can't change the character in both

BIn first case, you can't change the character and second case you can’t change the pointer

CYou can't change the pointer in both

DIn first case you can't change the pointer and in second case you can't change the character

ENone of these

Answer: Option B

Explanation:

Here is no explanation for this answer

ShortCut By :: sudhan

In C programming language, *p represents the value stored in a pointer and p represents the address of the value, is referred as a pointer. const char* and char const* says that the pointer can point to a constant char and value of char pointed by this pointer cannot be changed.

 const char *ptr : This is a pointer to a constant character. You cannot change the value pointed by ptr, but you can change the pointer itself. “const char *” is a (non-const) pointer to a const char.

char *const ptr : This is a constant pointer to non-constant character. You cannot change the pointer p, but can change the value pointed by ptr. 


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NA
SHSTTON
15
Solv. Corr.
64
Solv. In. Corr.
79
Attempted
0 M:40 S
Avg. Time

486 / 652

What will be the result of the following program?
#include<stdio.h>
int main(){
char p[]="String";
int x;
if(p=="String"){
printf("Pass 1");
if(p[sizeof(p)-2]=='g')
printf("Pass 2");
else
printf("Fail 2");
}
else{
printf("Fail 1");
if(p[sizeof(p)-2]=='g')
printf("Pass 2");
else
printf("Fail 2");
}
return 0;
}

APass 1 Pass 2

BFail 1 Fail 2

CPass 1 Fail 2

DFail 1 Pass 2

ESyntax error during compilation

Answer: Option D

Explanation:

Here is no explanation for this answer

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NA
SHSTTON
12
Solv. Corr.
37
Solv. In. Corr.
49
Attempted
0 M:58 S
Avg. Time

487 / 652

What will be result of the following program?
#include<stdio.h>
#include<malloc.h>
int myalloc(char *x, int n){
x= (char *)malloc(n*sizeof(char));
memset(x,\0,n*sizeof(char));
}
int main(){
char *g="String";
myalloc(g,20);
printf("The string is %s",g);
return 0;
}

ARun time error/Core dump

BThe string is: String

CThe string is: Oldstring

DSyntax error during compilation

ENone of these

Answer: Option B

Explanation:

Here is no explanation for this answer

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NA
SHSTTON
42
Solv. Corr.
59
Solv. In. Corr.
101
Attempted
0 M:45 S
Avg. Time

488 / 652

What will be the result of the following program?
#include<stdio.h>
#include<string.h>
char *gxxx(){
static char xxx[1024];
return xxx;
}
int main(){
char *g="string";
strcpy(gxxx(),g);
g = gxxx();
strcpy(g,"oldstring");
printf("The string is : %s",gxxx());
return 0;
}

AThe string is: string

BThe string is: Oldstring

CRun time error/Core dump

DSyntax error during compilation

ENone of these

Answer: Option B

Explanation:

Here is no explanation for this answer

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NA
SHSTTON
6
Solv. Corr.
12
Solv. In. Corr.
18
Attempted
1 M:20 S
Avg. Time

489 / 652

In the following code segment what will be the result of the function
#include<stdio.h>
int main(){
unsigned int x=-1;
int y;
y = ~0;
if(x == y)
printf("same");
else
printf("not same");
printf("%u %d",x,y);
return 0;
}

Asame, and x=MAXINT, y=-1

Bnot same, and x= MAXINT, y= -MAXINT

Csame , and x=MAXUNIT,y -1

Dsame, iand x=y=MAXUNIT

Enot same, and x=MAXINT, y=MAXUNIT

Answer: Option A

Explanation:

Here is no explanation for this answer

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NA
SHSTTON
4
Solv. Corr.
13
Solv. In. Corr.
17
Attempted
1 M:18 S
Avg. Time

490 / 652

What will the following program do?
#include<stdio.h>
#include<string.h>
#include<malloc.h>
int main(){
int i;
char a[]="String";
char *p="New Sring";
char *Temp;
Temp=a;
a=malloc(strlen(p) + 1);
strcpy(a,p); //Line no:9//
p = malloc(strlen(Temp) + 1);
strcpy(p,Temp);
printf("(%s, %s)",a,p);
free(p);
free(a);
return 0;
} 

Chose correct option

ASwap contents of p and a and print

BGenerate compilation error in line number 8

CGenerate compilation error in line number 5

DGenerate compilation error in line number 7

EGenerate compilation error in line number 1

Answer: Option B

Explanation:

Here is no explanation for this answer

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Tags: TCS


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