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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