Problems on Ages: Shortcuts and 10 Solved MCQs

Age problems look wordy, but each one reduces to a single linear equation. They appear in JKSSB and SSC maths papers and are worth practising because the method never changes.

Five patterns

• Ratio and sum: ages in ratio a : b with sum S. One part = S ÷ (a + b)\n- Ratio and difference: one part = difference ÷ (b − a)\n- Multiples: 'father is 4 times as old as his son, sum 50' means 4x + x = 50\n- After or before t years: set up the equation with (age + t) or (age − t)\n- Averages: total age = average × number of people; add or remove a person from the total

A worked example

A father is 40 and his son is 10. After how many years will the father be twice the son's age? Let the number of years be t: 40 + t = 2(10 + t), so 40 + t = 20 + 2t and t = 20.

More worked examples

• Ratio and sum: the ages of two brothers are in the ratio 3 : 5 and their sum is 40. One part is 40 ÷ 8 = 5, so the ages are 15 and 25
• Ratio and difference: two ages are in the ratio 4 : 7 and differ by 18. One part is 18 ÷ 3 = 6, so the ages are 24 and 42
• Past ages: the sum of the ages of a mother and her daughter is 50 years. Five years ago the mother was 7 times as old as the daughter. Let the daughter's age be x. The mother's age is 50 − x. Five years ago: 50 − x − 5 = 7(x − 5), so 45 − x = 7x − 35 and x = 10. The daughter is 10 and the mother is 40
• Future ages: A is 6 years older than B. In 4 years, A will be 1.5 times as old as B. Let B be x and A be x + 6. Then x + 10 = 1.5(x + 4), so x + 10 = 1.5x + 6 and x = 8. B is 8 and A is 14
• Averages: the average age of four people is 30. A fifth person joins and the average becomes 32. The fifth person's age is 32 × 5 − 30 × 4 = 160 − 120 = 40

A step-by-step method

• Pick the younger person's present age as x, or the smallest unit in a ratio
• Write every other age in terms of x
• Adjust for the time shift by adding or subtracting the same number of years from every person
• Form one equation from the condition in the question
• Solve and check by putting the answers back into the statement

Using the unchanging difference

The gap between two people's ages never changes. If a father is 30 years older than his son now, he was 30 years older when the son was born and will be 30 years older in 20 years. This fact lets you solve many questions without algebra. If in 5 years the father is three times the son, then the difference of 30 is two times the son's age at that time, so the son will be 15 and is now 10.

Common mistakes

• Adding the time shift to one person and forgetting the other
• Writing 'twice as old as' in the wrong direction, so check which person has the larger age
• Forgetting that 'five years ago' means subtracting
• Using the ratio numbers as ages. A ratio of 3 : 5 means 3x and 5x, not 3 and 5

Practice tips

Age problems can be solved in under a minute if you set them up quickly. Write the equation in the first 15 seconds. If you cannot form it, reread the sentence for the clue about time. Practise one of each type: ratio, past, future, average and multiple.

Practise

Try the 10 MCQs below, then continue with age problem sets for Junior Assistant and Class-IV. See also the Class-IV maths guide.