# 100+ TNPSC Aptitude Questions and Answers Pdf - 1

Question: 1

If the arithmetic mean of seventy five numbers is calculated, it is 35. If each number is increased by 5, then mean of new numbers is

(A) 30

(B) 40

(C) 50

(D) 50

Ans: B

A.M. of 75 numbers = 35.

Sum of 75 numbers = (75 × 35) = 2625.

Total increase = (75 × 5) = 375.

Increased sum = (2625 + 375) = 3000.

Increased average = \$\${3000}/{75}\$\$ = 40.

Question: 2

Four years ago, the average age of a family of four persons was 18 years. During this period, a baby was born. Today if the average age of the family is still 18 years, the age of the baby is

(A) 1 years

(B) 1.2 years

(C) 2 years

(D) 2.5 years

Ans: C

Sum of the ages of 4 members, 4 years ago = (18 × 4) years = 72 years.

Sum of the ages of 4 members now = (72 + 4 × 4) years = 88 years.

Sum of the ages of 5 members now = (18 × 5) years = 90 years.

∴ Age of the baby = (90 - 88) years = 2 years.

Question: 3

The average of 12 numbers is 15 and the average of the first two is 14. What is the average of the rest?

(A) 14

(B) \$\$14{1}/{5}\$\$

(C) 15

(D) \$\$15{1}/{5}\$\$

Ans: D

Average of 12 number = 15

Total of 12 number = 15 × 12 = 180

Average of first two number = 14

Total of first two number = 14 × 2 = 28

Total of remaining ten numbers = 180 - 28 = 152.

Required average of remaining ten number = \$\${152}/{10}\$\$ = \$\${76}/{5}\$\$ = \$\$15{1}/{2}\$\$

Question: 4

The average of 6 numbers is 7. The average of three numbers of them is 5. What will be the average of remaining numbers?

(A) 9

(B) 10

(C) 15

(D) 30

Ans: A

Average of 6 numbers = 7

Sum of 6 numbers = 6 × 7 = 42

Average of three numbers = 5

Sum of three numbers = 5 × 3 = 15

∴ Sum of the remaining three numbers = 42 – 15 = 27

∴ Required average = \$\${27}/{3}\$\$ = 9.

Question: 5

Find the average of first 40 natural numbers.

(A) 19.2

(B) 20.1

(C) 20.4

(D) 20.5

Ans: D

Sum of first n natural numbers = \$\${n(n+1)}/{2}\$\$

So, sum of first 40 natural numbers = \$\${40 × 41} / {2}\$\$= 820.

∴ Required average = \$\${820}/{40}\$\$ = 20.5.

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