Problems on Ages Notes

Last reviewed 22 Sept 2026 · 6 min read

Two facts that solve most of them

The difference between two people's ages never changes. If a father is 24 years older than his son, he is 24 years older at every point in the past and future.
Everybody ages equally. In $t$ years, every age in the question increases by the same $t$ ; $t$ years ago, each was smaller by $t$ .

So a ratio of ages changes with time, but a difference does not — which is why a question that gives one ratio and one difference is solvable in one line.

Setting the problem up

FormulaThe standard set-up

Let the present ages be the unknowns (use one variable where a ratio is given: $4 x$ and $3 x$ ).
" $t$ years ago" → subtract $t$ from every age.
" $t$ years hence" → add $t$ to every age.
"A is $k$ times B" → $A = k B$ at that same moment.

For ages in the ratio $a : b$ now that become $p : q$ after $t$ years:

\frac{a x + t}{b x + t} = \frac{p}{q} \Rightarrow x = \frac{t ( p - q )}{a q - b p}

Useful relations

Statement	Equation
A is $n$ years older than B	$A = B + n$
A is $k$ times as old as B	$A = k B$
$t$ years ago A was $k$ times B	$A - t = k (B - t)$
$t$ years hence A will be $k$ times B	$A + t = k (B + t)$
Sum of present ages is $S$	$A + B = S$
Average age of $n$ people is $A$	Sum of ages $= n A$

For a group, remember that in $t$ years the sum of $n$ ages rises by $n t$ , so the average rises by exactly $t$ .

Worked examples

Worked ExampleExample 1 — father and son

A father is three times as old as his son. Twelve years hence he will be twice as old. Find their present ages.

Solution. $F = 3 S$ and $3 S + 12 = 2 (S + 12) \Rightarrow 3 S + 12 = 2 S + 24 \Rightarrow S = 12$ . Son 12 years, father 36 years.

Worked ExampleExample 2 — a ratio that changes

Two ages are in the ratio 4 : 3. Six years hence the ratio will be 6 : 5. Find the ages.

Solution. $\frac{4 x + 6}{3 x + 6} = \frac{6}{5} \Rightarrow 20 x + 30 = 18 x + 36 \Rightarrow x = 3$ . Ages: 12 and 9 years (check: $18 : 15 = 6 : 5$ ).

Worked ExampleExample 3 — looking back

Ten years ago A was half as old as B. Their present ages are in the ratio 3 : 4. Find the sum of their present ages.

Solution. Let the ages be $3 k$ and $4 k$ . Then $3 k - 10 = \frac{1}{2} (4 k - 10) \Rightarrow 6 k - 20 = 4 k - 10 \Rightarrow k = 5$ . Ages 15 and 20, sum 35 years.

Worked ExampleExample 4 — a constant difference

A mother is 24 years older than her daughter. In two years she will be twice as old. Find their present ages.

Solution. $M = D + 24$ and $M + 2 = 2 (D + 2) \Rightarrow D + 26 = 2 D + 4 \Rightarrow D = 22$ . Daughter 22, mother 46 years.

Worked ExampleExample 5 — sum given

The sum of the present ages of two people is 64 years. Five years ago their ages were in the ratio 2 : 1. Find their present ages.

Solution. $A + B = 64$ and $A - 5 = 2 (B - 5) \Rightarrow A = 2 B - 5$ . $3 B - 5 = 64 \Rightarrow B = 23$ , $A = 41$ (check: 36 and 18, ratio 2 : 1).

Worked ExampleExample 6 — an average that hides a birth

Five years ago the average age of a family of four was 28 years. A baby has been born since, and the average age of the five is now 27 years. Find the baby's age.

Solution. Five years ago the four totalled $4 \times 28 = 112$ ; today those four total $112 + 4 \times 5 = 132$ . The family of five totals $5 \times 27 = 135$ , so the baby is 3 years old.

Worked ExampleExample 7 — present from future

A man's age will be twice his son's in 8 years, and eight years ago it was four times. Find their present ages.

Solution. $M + 8 = 2 (S + 8)$ gives $M = 2 S + 8$ , and $M - 8 = 4 (S - 8)$ gives $M = 4 S - 24$ . Equating: $2 S + 8 = 4 S - 24 \Rightarrow S = 16$ , so $M = 40$ . Son 16, man 40 years (check: in 8 years, 48 and 24; eight years ago, 32 and 8).

Worked ExampleExample 8 — three people

A is 2 years older than B, who is twice as old as C. The sum of the three ages is 27. Find B's age.

Solution. $B = 2 C$ , $A = B + 2 = 2 C + 2$ . Sum: $2 C + 2 + 2 C + C = 27 \Rightarrow 5 C = 25 \Rightarrow C = 5$ . B is 10 years.

Worked ExampleExample 9 — ratio then and now

The present ages of two brothers are in the ratio 5 : 3, and four years ago they were in the ratio 9 : 5. Find the present ages.

Solution. $\frac{5 x - 4}{3 x - 4} = \frac{9}{5} \Rightarrow 25 x - 20 = 27 x - 36 \Rightarrow x = 8$ . Ages 40 and 24 years (check: $36 : 20 = 9 : 5$ ).

Had the second ratio been 3 : 2, the same working would give $x = - 4$ — a negative answer, which means no pair of ages satisfies both statements. An impossible answer is a reading check, not a calculation slip.

Worked ExampleExample 10 — average of a group

The average age of a class of 30 students is 16 years. If the teacher's age is included, the average rises by 1 year. Find the teacher's age.

Solution. $30 \times 16 + T = 31 \times 17 \Rightarrow 480 + T = 527 \Rightarrow T =$ 47 years.

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Problems on Ages