Problems on Ages

JKSSB Assistant Training Officer (Handicrafts & Handlooms) · Numerical Ability · 81 practice questions with answers and explanations: 43 Easy, 37 Medium, 1 Hard.

Sample questions

Sample Easy question 1

A father is four times as old as his son. If the sum of their ages is 50 years, how old is the son?

  1. 12
  2. 8
  3. 40
  4. 10
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Answer: D. Son = x, father = 4x. 5x = 50, so x = 10.

Sample Easy question 2

The ages of A and B are in the ratio 3 : 5 and their sum is 40 years. Find A's age.

  1. 15
  2. 20
  3. 12
  4. 25
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Answer: A. A = 3/8 × 40 = 15 years.

Sample Easy question 3

Zoya's present age is 23 years. What will it be after 13 years?

  1. 36
  2. 10
  3. 299
  4. 37
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Answer: A. 23 + 13 = 36 years.

Sample Medium question 4

The present ages of two brothers are in the ratio 2 : 4. After 7 years, the sum of their ages will be 44. Find the present age of the elder brother.

  1. 13 years
  2. 20 years
  3. 27 years
  4. 10 years
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Answer: B. Let the present ages be 2k and 4k. After 7 years: 2k + 4k + 14 = 44, so k = 5. The elder brother is 4 × 5 = 20 years old.

Sample Medium question 5

The present ages of two brothers are in the ratio 5 : 3. After 4 years, the sum of their ages will be 48. Find the present age of the elder brother.

  1. 15 years
  2. 29 years
  3. 21 years
  4. 25 years
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Answer: D. Let the present ages be 5k and 3k. After 4 years: 5k + 3k + 8 = 48, so k = 5. The elder brother is 5 × 5 = 25 years old.

Sample Medium question 6

The present ages of two brothers are in the ratio 5 : 3. After 6 years, the sum of their ages will be 68. Find the present age of the elder brother.

  1. 35 years
  2. 29 years
  3. 41 years
  4. 21 years
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Answer: A. Let the present ages be 5k and 3k. After 6 years: 5k + 3k + 12 = 68, so k = 7. The elder brother is 5 × 7 = 35 years old.

Sample Hard question 7

Ten years ago a father was 12 times as old as his son; ten years from now he will be twice as old as his son. Find the son's present age.

  1. 14
  2. 15
  3. 10
  4. 12
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Answer: D. f − 10 = 12(s − 10) and f + 10 = 2(s + 10). Eliminating f: 12s − 110 = 2s + 10, so s = 12 (and f = 34).

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