Problems on ages for IBPS Clerk

Age problems give you two facts about people's ages, usually at two different times, and ask for one age or a ratio. They look like word puzzles, but underneath each one is a linear equation or a ratio. Clerk papers often include an age question in the arithmetic block, and it is one of the quicker ones to secure once the setup is automatic.

The whole topic rests on three observations. Learn them, and the setup takes ten seconds.

Three rules, and why they hold

1. The difference between two people's ages never changes. Both get one year older every year, so the gap stays fixed. A parent who is 25 years older than a child is 25 years older forever.
2. A shift in time applies to every person. "Five years ago" means subtract 5 from each age. For a group of k people, the sum of ages changes by k × 5.
3. Ratios need a common multiplier. If ages are in the ratio 3 : 4, write them as 3x and 4x, the same parts method used in ratio and proportion.

Rule 1 gives you one of the fastest shortcuts in the topic. If the ratio of two ages changes from a : b to c : d over time, the difference in ratio terms must represent the same number of years in both. Use that to fix the parts.

Setting up with one variable

• Choose the younger person's present age as x. Most expressions then have no negative terms.
• Write every other age in terms of x, now.
• Only then apply the time shift ("after 10 years": add 10 to each expression).
• Form one equation from the remaining fact and solve.

Writing present ages first, and shifting afterwards, prevents the most common error: shifting one person and forgetting the other.

Worked examples

Example 1. The ratio of the ages of A and B is 3 : 4. After 5 years, it will be 4 : 5. Find their present ages.

1. Present ages 3x and 4x. After 5 years: 3x + 5 and 4x + 5.
2. (3x + 5) ÷ (4x + 5) = 4 ÷ 5, so 15x + 25 = 16x + 20 and x = 5.
3. A is 15, B is 20. Check: after 5 years, 20 : 25 = 4 : 5.

Shortcut using rule 1: the difference is 1 part in both ratios (4 − 3 and 5 − 4). So 1 part now equals 1 part later. From 3 parts to 4 parts, A gains 1 part in 5 years, so 1 part = 5 years. A = 15, B = 20.

Example 2. A parent is three times as old as their child. In 10 years, the parent will be twice as old as the child. Find their present ages.

1. Child x, parent 3x. In 10 years: x + 10 and 3x + 10.
2. 3x + 10 = 2(x + 10), so 3x + 10 = 2x + 20 and x = 10.
3. Child 10, parent 30. Check: in 10 years, 40 = 2 × 20.

Example 3. Six years ago, the ages of A and B were in the ratio 5 : 4. The sum of their present ages is 57. What will be the ratio of their ages 4 years from now?

1. Six years ago: 5x and 4x. Present: 5x + 6 and 4x + 6.
2. Sum: 9x + 12 = 57, so x = 5. Present ages 31 and 26.
3. In 4 years: 35 and 30, so the ratio is 7 : 6.

Example 4. Three years ago, the average age of a family of 5 was 17. A baby has been born since, and the family's average age today is still 17. How old is the baby?

1. Three years ago, total = 5 × 17 = 85. Today those 5 people total 85 + 5 × 3 = 100.
2. Today, 6 people average 17, so the total is 102.
3. Baby = 102 − 100 = 2 years.

Example 5. A is as much older than B as B is older than C. A is 40 and C is 24. Find B's age.

1. The two gaps are equal, so B is exactly halfway between A and C.
2. B = (40 + 24) ÷ 2 = 32. Check: gaps of 8 each.

Example 6 (checking options). The ages of two siblings are in the ratio 2 : 3 and differ by 6 years. Options: (a) 10, 15 (b) 12, 18 (c) 14, 21 (d) 8, 12.

1. Every option is in the ratio 2 : 3, so the ratio does not separate them.
2. Differences: 5, 6, 7 and 4. Only (b) differs by 6.
3. Answer: 12 and 18.

Common mistakes

• Shifting one person's age and not the other's.
• In group questions, shifting the sum by the years only, instead of by (number of people × years).
• Setting up the ratio with the time shift already inside x, then shifting again.
• Answering the present age when the question asks for the age "after 5 years", or the reverse.

Practice

Set a 5-minute timer.

1. The sum of the ages of A and B is 50. A is 10 years older than B. Find their ages.
2. Two ages are in the ratio 5 : 7. After 6 years, the ratio will be 3 : 4. Find the present ages.
3. A parent is four times as old as their child. In 20 years, the parent will be twice as old. Find their present ages.
4. The present ages of A and B are in the ratio 3 : 2. Ten years ago, A was twice as old as B. Find their present ages.
5. The average age of 4 people is 30. What will the sum of their ages be after 5 years?
6. The sum of the ages of a parent and child is 50. Five years ago, the parent was 7 times as old as the child. Find their present ages.
7. Two ages are in the ratio 4 : 5 and differ by 7 years. What will the ratio be after 7 years?
8. A is as much older than B as B is older than C. The sum of the ages of A and C is 56. Find B's age.

Answers:

1. A 30, B 20. B = x, A = x + 10; 2x + 10 = 50.
2. 30 and 42. (5x + 6) ÷ (7x + 6) = 3/4 gives 20x + 24 = 21x + 18, so x = 6. Check: 36 : 48 = 3 : 4.
3. Child 10, parent 40. 4x + 20 = 2(x + 20), so x = 10.
4. 30 and 20. 3x − 10 = 2(2x − 10) gives 3x − 10 = 4x − 20, so x = 10. Check: ten years ago, 20 and 10.
5. 140. Present total 120; each of 4 people adds 5 years.
6. Parent 40, child 10. Child x, parent 50 − x. 45 − x = 7(x − 5) gives 8x = 80. Check: five years ago, 35 = 7 × 5.
7. 5 : 6. One part = 7, so the ages are 28 and 35. After 7 years: 35 and 42.
8. 28. B is halfway between A and C, so B = 56 ÷ 2.

What to do next

• Set up ten age problems a day for a week, writing present ages first every time.
• Practise the rule-1 shortcut on every ratio question where the ratio difference is the same before and after.
• Revise ratio and proportion and averages, since most age questions use one or both.