Quantitative Techniques · Micro-test

Problems on Ages

Age problems are equations told as stories. Name one present age x, write every other age in terms of it, and the sentence in the question becomes the equation.

10 questions · 5 minutes · instant scoring

What this topic actually tests

Three facts do most of the work. First, everyone ages at the same rate: in 6 years each person is 6 years older, and 5 years ago each was 5 years younger, so shift every age in the equation, not only one. Second, the difference between two people's ages never changes, while the ratio does. Third, when ages are given as a ratio, write them as multiples of one unknown: a ratio of 4 : 5 means ages of 4x and 5x. Then translate the remaining sentence. 'After 6 years the ratio will be 5 : 6' becomes (4x + 6)/(5x + 6) = 5/6; cross-multiply to get 24x + 36 = 25x + 30, so x = 6 and the ages are 24 and 30. When two conditions are given at different times, anchor everything to the present. 'Five years ago the mother was four times as old as her daughter; now she is three times as old': let the daughter be d now, so the mother is 3d, and five years ago 3d − 5 = 4(d − 5), which gives d = 15. Average questions use totals: if four people average 20, their ages sum to 80. Always finish by reading the question again, because it may ask for an age some years ago or hence, for the other person's age, or for a sum or difference, and by checking the answer against both conditions.

The common trap on this topic

The most common mistake is to shift only one person's age when moving through time: writing 'father's age five years ago' as f − 5 and leaving the son's as s. The second is answering with x when the question asked for 4x, or with the son's age when it asked for the father's; the value of the unknown is among the options for exactly that reason. The third is setting up a ratio after some years as (4x)/(5x) + 6 in place of adding 6 to each age separately. A ten-second check of the final ages against the words of the question catches all three.

Problems on Ages: answers and explanations

A father is three times as old as his son. The sum of their ages is 48 years. How old is the father?
Answer: C. 36 years
Let the son be x; the father is 3x. Then 4x = 48, so x = 12. The father is 36. Twelve is the son's age.
The ages of A and B are in the ratio 3 : 5 and their sum is 64 years. What is the difference between their ages?
Answer: D. 16 years
The ages are 3x and 5x, so 8x = 64 and x = 8. A is 24 and B is 40. The difference is 16 years.
Five years ago a mother was four times as old as her daughter. Today she is three times as old as her daughter. How old is the daughter today?
Answer: A. 15 years
Let the daughter be d today; the mother is 3d. Five years ago: 3d − 5 = 4(d − 5), so 3d − 5 = 4d − 20 and d = 15. Check: the mother is 45; five years ago they were 40 and 10.
The present ages of two brothers are in the ratio 4 : 5. Six years from now the ratio will be 5 : 6. What is the sum of their present ages?
Answer: A. 54 years
(4x + 6)/(5x + 6) = 5/6 gives 24x + 36 = 25x + 30, so x = 6. The brothers are 24 and 30, a sum of 54. In six years they will be 30 and 36, which is 5 : 6.
The average age of four friends is 20 years. A fifth friend joins them and the average age of the five becomes 22 years. How old is the fifth friend?
Answer: A. 30 years
The four friends' ages total 4 × 20 = 80. The five total 5 × 22 = 110. The fifth friend is 110 − 80 = 30 years old.
A is 6 years older than B. Four years ago A was twice as old as B. How old is A now?
Answer: A. 16 years
Let B be b; A is b + 6. Four years ago: b + 2 = 2(b − 4), so b = 10 and A is 16. Check: four years ago they were 12 and 6.
The sum of the ages of a father and his son is 60 years. Six years ago the father was five times as old as the son. How old is the son now?
Answer: D. 14 years
Let the son be s; the father is 60 − s. Six years ago: 54 − s = 5(s − 6), so 54 − s = 5s − 30 and s = 14. The father is 46. Six years ago they were 40 and 8.
Rahul is as much younger than Sameer as he is older than Tina. The sum of the ages of Sameer and Tina is 48 years. How old is Rahul?
Answer: A. 24 years
Sameer − Rahul = Rahul − Tina, so twice Rahul's age equals Sameer's plus Tina's, which is 48. Rahul is 24. The individual ages of Sameer and Tina cannot be found, but Rahul's can.
Ten years from now a man will be twice as old as he was ten years ago. How old is he now?
Answer: C. 30 years
Let his age be x. Then x + 10 = 2(x − 10), so x + 10 = 2x − 20 and x = 30. Check: in ten years he will be 40; ten years ago he was 20.
The present ages of two sisters are in the ratio 2 : 3. Eight years ago their ages were in the ratio 2 : 5. How old is the elder sister now?
Answer: C. 18 years
(2x − 8)/(3x − 8) = 2/5 gives 10x − 40 = 6x − 16, so x = 6. The sisters are 12 and 18, and the elder is 18. Eight years ago they were 4 and 10, which is 2 : 5.

FAQ

What is the best way to set up an age problem?

Let the youngest person's present age be x, express the others in terms of x, and translate the remaining condition into one equation. Use two unknowns only when no simple relation between present ages is given.

Does the ratio of two ages stay the same over time?

No. The difference stays the same; the ratio moves towards 1 as both people grow older. Ages of 10 and 30 are in the ratio 1 : 3, and twenty years later, at 30 and 50, in the ratio 3 : 5.

Are age problems asked in CLAT?

They are basic algebra, which is within the Class 10 mathematics that CLAT's Quantitative Techniques section draws on, where a question of this kind would sit inside a short data set. MH-CET Law's five-year paper has a separate Basic Mathematics section.

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