# TCS NQT Aptitude Questions with Answers And explanation [60 Questions] 2021

#### What is NQT?

TCS NQT is a multi-level assessment conducted to determine if a graduate has the following competencies and skills that are required by companies while recruiting freshers: Competence on the core cognitive processes required for entry-level jobs, industry-specific knowledge and insights, and specialization on skills required to perform various job roles.

#### TCS NQT Aptitude 60 Questions with Answers And explanation

**Q1. In a potato race, 20 potatoes are placed in a line of intervals of 4 meters with the first potato 24 meters from the starting point. A contestant is required to bring the potatoes back to the starting place one at a time. How far would he run in bringing back all the potatoes?**

A. 2400

B. 1400

C. 2480

D. 1240

**Ans. C. 2480**

**Explanation**

Given, a total number of potatoes = 20.

First potato 24 meters from the

starting point. There are 4 meters

in the intervals.

A contestant is required to bring

the potatoes back to the starting

place one at a time. So for the first potato,

he has to travel 48 meters, for the second 56 meters …

48,56,64………..20 terms.

a = 48, d= 8, n = 20.

Sum of n terms in A.P = Sn=n/2[2a+(n−1)d]

S20=20/2[2×48+(20−1)8]

S20=20/2[96+152]

S20=10×248=2480

∴ 2480 meters he run in bringing back all the potatoes.

**Q2. problemsolvingproblemsolvingprob . . . . . . Find the 2015th term in the series?**

A. n

B. r

C. p

D. q

**Ans. A. n**

**Explanation**

Problemsolving = 14 letter word.

So divide 2015 by 14 and find the remainder.

Here remainder is 13. so 13th letter in problemsolving is ‘n’.

**Q3. 0.397 /(1.2×1.2+0.66+1.1×1.1)= ?**

A. 0.1

B. 1

C. 0.01

D. 0.001

**Ans A. 0.1**

**Q4. What is the sum of all the two-digits numbers which when divided by 7 gives a remainder of 3 ?**

A. 547

B. 252

C. 676

D, 576

**Ans C. 676**

**Explanation**

The First Two-Digit Number, which is divided by 7 and gives the remainder 3, will be 10.

The last Two-Digit Number, which is divided by 7 and gives the remainder 3, will be 94.

The common difference between two consecutive numbers will be 7.

This series will be like →10, 17, 24,………………………. 94.

Here First Number a = 10, Last number tn = 94, Common Difference d = 7 ;

Using the Formula for Last Number tn = a + (n – 1) x d;

⇒ a + (n – 1) x d = tn

Put the value of a , d and tn in above equation.

⇒ 10 + (n – 1) x 7 = 94

⇒ 10 + 7n – 7 = 94

or 7n + 3 = 94

⇒ 7n = 94 – 3 = 91

or n = 91/7

⇒ n = 13

Using the formula for the sum of Arithmetic Progression.

Sn = n/2 [ First Number + Last Number ] ;

Put the value of n, First Number, and Last Number, we will get,

Sn = 13/2 [ 10 + 94 ]

Sn = 104 x 13/2

.or Sn = 52 x 13

Sn = 676

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