TCS NextStep Interview Questions | TCS NextStep Syllabus | TCS Nextstep Aptitude Questions

a) 55
b) 66
c) 77
d) 44
e) None of these

Answer: Option b
Given that a student scores 55% marks in English in 8 papers of 100 marks each.
Hence, his total marks = (55 x 800)/100 = 440
15% of his 440 marks = 440*(15/100) = 66 = Marks scored in English.

a) 1/70
b) 1/840
c) 1/8
d) 1/40320


Answer: Option a
Total no of possible ways are 8!
The total number of possible ways to find even numbers at the last 4 digits are 4!
The total number of possible ways to find odd numbers at the last 4 digits are 4!
So, the probability that the first 4 digits of the code are even number =4!*4!/8! = 1/70

a)37 secs
b)48 secs
c)72 secs
d)68 secs

Answer: Option c
Relative speed = 60-35 = 25 Km/hr.
For the car to be ahead of the bus by 250 meters, it needs to cover = 250+250 =500m = 0.5kms.
Time taken to cover 0.5 km = 0.5/25 =1/50 hrs = 72 secs

a) 12 days
b) 24 days
c) 36 days
d) 48 days


Answer: Option b
A’s 1 day work = 1/30.
A’s 3 day work = 3/30
(A+B+C) 1 day work = (1/30+1/45+1/90) = 1/15
4 days work = 3/30+1/15 = 1/6.
Total work will be done in= 4*6 = 24 days

a) 5000
b) 7500
c) 9000
d)9500


Answer: Option c
Let x be the capital invested by B.
Profit will be divided in the ratio of 2:3 = (12*3500) : 7x = 6000 : x.
2/3 = 6000/x
x = 9000

a) x^2
b) x
c) x^(3/2)
d) x^3

Answer: Option c
The term is x ^ (1/2 + 1/4 + 1/8 + …). Sum of infinite series with first term 1/2 and common ratio of 1/2 is 1. So the answer is ‘x’.

a) 234
b) 235
c) 236
d) 237

Answer:Option c
I = 5! =120
R= 5!/2 =60
SI = 4! =24
SR = 4!/2 = 12
ST = 4!/2 = 12
SUI = 3! = 6
= 234 Words

So, words with SUR begins from 235. SURIIT is the 235th word and SURITI is the 236rd word.

a) 50
b) 100
c) 120
d) 200

Answer: Option d

Possible ways in which he can attempt 6 questions are
5C4*5C2 = 50
5C3*5C3 = 100
5C2*5C4 = 50
50+100+50 = 200

a) 25 min
b) 30 min
c) 28 min
d) 20 min

Answer: Option a

M1H1 = M2H2
5*60 = 12*H2
H2 = 25 min

(a) 75
(b) 95
(c) 85
(d) 65

She gave most of her time to music.
a) Spent
b) lent
c) devoted
d) No improvement

a) Neutrality
b) Indifference
c) All-knowing
d) Ignorance
e) timeliness

a) Armageddon
b) Flagrant
c) Tenacious
d) Inconsistent

a) rain
b) stream
c) pool
d) canal

a) Huge
b) Invisible
c) Zero
d) Tiny

a) Fresh
b) Modern
c) Ancient
d) Present

a) Trape
b) Discard
c) Complement
d) Berate

a) Credible
b) Discipline
c) Gullible
d) Incredible

a) Merge
b) Amalgamate
c) Split
d) Federate

a) flexibility
b) animosity
c) incompatibility
d) Concord

Program:

include
include
int main()
{
int a,b,gcd;
printf(“\nEnter two numbers : “);
scanf(“%d %d”,&a,&b);
int i;
for(i = 1; i <= a && i <= b; i++)
{
if((a % i == 0) && (b % i == 0))
{
gcd = i;
}
}
printf(“\nGCD of %d and %d is %d “,a,b,gcd);
printf(“\n”);
return 0;
}

Output:

Input- Enter two numbers:20 28 Output- GCD of 4

Program:

include
include
int main()
{
int number, temp, remainder, result = 0, n = 0 ;
printf(“Enter an integer: “);
scanf(“%d”, &number);
temp = number;
while (temp != 0)
{
temp /= 10;
++n;
}
temp = number;
while (temp != 0)
{
remainder = temp%10;
result += pow(remainder, n);
temp /= 10;
}
if(result == number)
printf(“%d is an Armstrong number\n”, number);
else
printf(“%d is not an Armstrong number\n”, number);
return 0;
}

Program:

include
long int decimal_to_binary(int n)
{
long int binary = 0;
int remainder, i, flag = 1;
for(i = 1; n != 0; i = i * 10)
{
remainder = n % 2;
n /= 2;
binary += remainder * i;
}
return binary;
}
int main()
{
int n;
printf(“Enter a decimal number: “);
scanf(“%d”, &n);
printf(“Equivalent binary number: %d\n”, decimal_to_binary(n));
return 0;
}

Output:

Input- Enter a decimal number:255
Output- Equivalent binary number: 11111111

Program:

include
int main()
{
int n, rev = 0, rem;
printf(“\nEnter a number : “);
scanf(“%d”, &n);
printf(“\nReversed Number : “);
while(n != 0)
{
rem = n%10;
rev = rev*10 + rem;
n /= 10;
}
printf(“%d\n”, rev);
return 0;
}

Output:

Enter a number : 12345
Reversed Number : 54321

Program:

include
int check_vowel(char);
int main()
{
char s[100], t[100];
int c, d = 0;
gets(s);
for(c = 0; s[c] != ‘\0’; c++)
{
if(check_vowel(s[c]) == 0)
{
t[d] = s[c];
d++;
}
}
t[d] = ‘\0’;
strcpy(s, t);
printf(“%s\n”, s);
return 0;
}
int check_vowel(char ch)
{
if (ch == ‘a’ || ch == ‘A’ || ch == ‘e’ || ch == ‘E’ || ch == ‘i’ || ch == ‘I’ || ch ==’o’ || ch==’O’ || ch == ‘u’ || ch == ‘U’)
return 1;
else
return 0;
}

Output:

Input- qwertyuiop Output- qwrtyp

Program:

include
include
int binary_to_octal(long int binary)
{
int octal = 0, decimal = 0, i = 0;
while(binary != 0)
{
decimal += (binary%10) * pow(2,i);
++i;
binary/=10;
}
i = 1;
while (decimal != 0)
{
octal += (decimal % 8) * i;
decimal /= 8;
i *= 10;
}
return octal;
}
int main()
{
long int binary;
printf(“\nEnter a binary number: “);
scanf(“%lld”, &binary);
printf(“\nOctal Equivalent : %d\n”, binary_to_octal(binary));
return 0;
}

Output:

Input- Enter a binary number:1111 Output- Octal Equivalent :17

A linked list consists of two parts: information and the link. In the single connected listening, the beginning of the list is marked by a unique pointer named start. This pointer does point to the first element of the list and the link part of each node consists of an arrow looking to the next node, but the last node of the list has null pointer identifying the previous node. With the help of start pointer, the linked list can be traversed easily.

In, Object-Oriented Programming, inheritance is a mechanism based on classes.
Inheritance refers to inhering the data members and properties of a parent class to a child class. A class which is derived from another level is often called as a sub-class or a child class, and the type from which the child class is obtained is known as super-class or parent class.

Polymorphism is a concept in OOPS which means having many forms. In simple words, it means that different actions will be performed in different instances. Polymorphism is of two types:
Method overloading
Operator overloading

Types of Inheritance:
Single inheritance
Multiple Inheritance
Multi-level Inheritance
Multi-path Inheritance
Hierarchical Inheritance
Hybrid Inheritance

Software development life-cycle is steps involved in the life cycle of software development phase. Generally, it is followed by the development team which develops the software in the organization. It consists of a clear explanation of developing and maintaining the software.