Problems on Ages • Topic 1 of 3

Present & Past Ages

Start every age problem by naming the present age as a single variable, then translate each phrase into algebra. "5 years ago" means subtract 5 from the present age; "was twice as old" links two such shifted expressions. The reliable habit is to write present ages first (x, y, …) and only then subtract t to reach the past — never carry an unknown for the past age directly, or you double your variables. A useful CAT check: the difference between two people’s ages is the same today as it was years ago, so if a relationship like "A was 3 times B" held in the past, the gap (A − B) you compute must stay consistent with the present. Keep equations in present-age terms and solve the linear system; the arithmetic is light once the set-up is clean.

✅ Solved examples

1. A father is 30 years older than his son. 5 years ago the father was 4 times as old as the son. Find their present ages.
Let son = s now, father = s + 30. Five years ago: (s + 30 − 5) = 4(s − 5) ⇒ s + 25 = 4s − 20 ⇒ 3s = 45 ⇒ s = 15. Father = 45.
2. A is 8 years older than B. 4 years ago A was twice B. Find A’s present age.
B = b now, A = b + 8. Four years ago: (b + 8 − 4) = 2(b − 4) ⇒ b + 4 = 2b − 8 ⇒ b = 12. A = 20.
3. Ten years ago a mother was three times as old as her daughter; the mother is now 46. Find the daughter’s present age.
Mother 10 years ago = 36. Daughter then = 36/3 = 12. Daughter now = 12 + 10 = 22.
4. The present ages of P and Q sum to 51. Six years ago P was twice as old as Q. Find P’s present age.
P + Q = 51. Six years ago: (P − 6) = 2(Q − 6) ⇒ P − 6 = 2Q − 12 ⇒ P = 2Q − 6. Substitute: 2Q − 6 + Q = 51 ⇒ 3Q = 57 ⇒ Q = 19, P = 32.

✏️ Practice — try these, take hints as needed

1. A man is 24 years older than his son. In 2 years the man will be three times as old as his son. Find the son’s present age.
Son = s, man = s + 24.
In 2 yrs: s + 26 = 3(s + 2).
s + 26 = 3s + 6.
10 years
2. B is 5 years younger than A. 3 years ago A was twice B. Find A’s present age.
A = b + 5, B = b.
(b + 2) = 2(b − 3).
b + 2 = 2b − 6.
13 years
3. Eight years ago a teacher was 5 times as old as a student; today the gap is 32 years. Find the student’s present age.
Gap is constant = 32.
Then student = e, teacher = 5e, gap 4e = 32.
e = 8, add 8 years.
16 years
4. The sum of present ages of two friends is 44. 4 years ago one was 3 times the other. Find the elder’s present age.
x + y = 44.
(x − 4) = 3(y − 4).
x = 3y − 8, substitute.
31 years
5. A is twice as old as B was 5 years ago. The sum of their present ages is 50. If A is older, find A’s age.
Let B = b now; B 5 yrs ago = b − 5.
A = 2(b − 5) and A + b = 50.
2b − 10 + b = 50.
30 years

📝 Topic test — 8 questions

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