Problems on Ages • Topic 3 of 3

Future Ages

Future-age problems add the same constant t to every present age, exactly mirroring the past case but with a plus sign. Translate "t years from now" as (x + t) and form the equation the statement describes. Two ideas make these fast under CAT pressure. First, the age gap is invariant, so a clause like "in t years A will be twice B" combined with the fixed difference (A − B) lets you skip heavy algebra: if A − B = d and A + t = 2(B + t), then B + t = d, giving B directly. Second, when both a "years ago" and a "years hence" clause appear, keep everything in present-age variables and write two clean equations rather than juggling shifted unknowns. The arithmetic stays simple as long as you never introduce a separate symbol for a future age.

✅ Solved examples

1. A is 4 years older than B. In 6 years A will be twice as old as B was 4 years ago. Find B’s present age.
A = b + 4. In 6 yrs A = b + 10. B four years ago = b − 4. So b + 10 = 2(b − 4) ⇒ b + 10 = 2b − 8 ⇒ b = 18.
2. A man is now 6 times as old as his son. In 4 years he will be 5 times as old as his son. Find the son’s present age.
Man = 6s now. In 4 yrs: 6s + 4 = 5(s + 4) ⇒ 6s + 4 = 5s + 20 ⇒ s = 16. Son = 16, man = 96.
3. In 10 years a girl will be twice as old as she was 5 years ago. Find her present age.
Present = x. (x + 10) = 2(x − 5) ⇒ x + 10 = 2x − 10 ⇒ x = 20.
4. The sum of the present ages of a father and son is 60. After 6 years the father will be three times as old as the son. Find the son’s present age.
f + s = 60. (f + 6) = 3(s + 6) ⇒ f = 3s + 12. Then 3s + 12 + s = 60 ⇒ 4s = 48 ⇒ s = 12.

✏️ Practice — try these, take hints as needed

1. A is 6 years older than B. In 4 years A will be twice as old as B. Find B’s present age.
A = b + 6.
b + 10 = 2(b + 4).
b + 10 = 2b + 8.
2 years
2. In 8 years a man will be three times as old as he was 8 years ago. Find his present age.
Present = x.
x + 8 = 3(x − 8).
x + 8 = 3x − 24.
16 years
3. A father is now 5 times his son’s age. In 12 years he will be twice as old. Find the son’s present age.
Father = 5s.
5s + 12 = 2(s + 12).
5s + 12 = 2s + 24.
4 years
4. The sum of present ages of two brothers is 40. After 5 years the elder will be 1.5 times the younger. Find the elder’s present age.
e + y = 40.
e + 5 = 1.5(y + 5).
e = 1.5y + 2.5.
23 years
5. A is twice as old as B will be in 5 years; the sum of their present ages is 40. Find A’s present age.
B in 5 yrs = b + 5.
A = 2(b + 5) and A + b = 40.
2b + 10 + b = 40.
30 years

📝 Topic test — 8 questions

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