Ratio, Proportion, Partnership & Ages

high importance~2 Q in Tier 123 formulas⚡ 15 shortcuts5 subtopics

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Problems on ages

⏱ 4 min read🧩 4 question types🎯 12 practice Q

The idea in one minute

Three facts never change: each age grows by 1 every year, the sum of two ages grows by 2 per year, and the difference between two ages is constant forever.

Standard template: present ages in ratio $a : b$ , a second ratio $n$ years later (or earlier). Write $ax$ , $bx$ and solve $\dfrac{ax \pm n}{bx \pm n} = \dfrac{p}{q}$ .

One multiplier is always enough.

The three facts that never change

Each person's age grows by 1 every year. Shift n years: add n to both ages.
A sum of two ages grows by 2n over n years.
The difference between two ages never changes. A 32-year gap now is a 32-year gap forever.

Rule: Shift the ages, never the ratio. $5 : 7$ today is not $5+n : 7+n$ tomorrow.

The multiplier template

Present ages in ratio $a : b$ — write $ax$ and $bx$ . A shifted ratio becomes one equation:

\frac{ax + n}{bx + n} = \frac{p}{q} \quad (\text{after } n \text{ years}), \qquad \frac{ax - n}{bx - n} = \frac{p}{q} \quad (n \text{ years ago})

One cross-multiplication gives x. $5 : 7$ now and $3 : 4$ after 8 years: $\dfrac{5x+8}{7x+8} = \dfrac{3}{4}$ gives $20x + 32 = 21x + 24$ , so $x = 8$ . Ages 40 and 56, sum 96.

If the question gives a sum or difference of present ages instead, $(a+b)x$ or $(b-a)x$ is that value — no shift needed.

Sum and difference shortcuts

Ratio $3 : 4$ with sum 35: seven parts of 5, so 15 and 20. The younger is 15.

The sum grows by 2 a year: "the sum is 40 now; what was it 5 years ago?" — 30, no ratio needed. Read which year the question means before dividing by parts.

k-times questions

"Father is k times the son" — write father $= k \times$ son, with the son as the single unknown.

Mother 4 times the daughter; after 5 years, 3 times: $4d + 5 = 3(d + 5)$ , so $d = 10$ and the mother is 40. The gap of 30 never moved: 45 and 15 give exactly 3 times.

Watch: For an older-younger pair the multiple falls over time — 4 times becomes 3 times. If your answer grows the multiple, flip the equation.

Two shifted ratios

A ratio n years ago AND n years hence give two equations in the same x.

Ten years ago $2 : 3$ ; ten years hence $4 : 5$ . Present ages are $2x + 10$ and $3x + 10$ . The second snapshot: $\dfrac{2x + 20}{3x + 20} = \dfrac{4}{5}$ gives $10x + 100 = 12x + 80$ , so $x = 10$ and A is 30.

The 2n years between the two snapshots is the easiest slip: each age changes by 2n, not n.

The constant-difference check

After solving, verify the gap is the same at both times. Father 48, son 16: gap 32. Eight years ago, 40 and 8 — gap 32, ratio 5, matching "5 times as old".

The check also runs forwards. A man is 25 years older than his son and will be exactly twice his age in ten years. Then $s + 35 = 2(s + 10)$ , so the son is 15 now. In ten years the pair is 25 and 50 — gap still 25.

Tip: Compute the age gap once and reuse it at every time point. It kills wrong options without solving.

Three people

Ratios among three people work the same: $4x, 5x, 6x$ . Exams add one pairwise fact ("A is 6 years older than C" gives $2x = 6$ ) because more would over-determine the problem.

Question types you will see

Each type: how to recognise it, the method step by step, and one question to try.

Type 1very common2 practice Q

Present ratio with sum or difference

How to spot it:

'Ages are in ratio a : b and their sum (or difference) is S' — no time shift at all.

(a + b)x = S \ \text{or} \ (b - a)x = D

Method

Total parts = a + b (or the gap b − a for a difference).
One part = S ÷ total parts.
Multiply by the asked person's parts.

Why it works:

With no shift, this is a plain ratio division of the given total.

Try this

The ages of two brothers are in the ratio 3 : 4 and the sum of their ages is 35 years. The younger brother's age is:

Show solution

7 parts $= 35$ , so 1 part $= 5$ .
Younger $= 3 \times 5 = 15$ years.

Answer

15 years

Type 2very common2 practice Q

Ratio now and ratio after or before n years

How to spot it:

'The present ages are in ratio a : b; after (or before) n years the ratio will be p : q.'

\frac{ax \pm n}{bx \pm n} = \frac{p}{q}

Method

Write present ages as ax, bx.
Add or subtract n on BOTH, equate to p : q.
Cross-multiply once and solve for x.

Why it works:

Both ages move by the same n, so the ratio shift pins x exactly.

Try this

The present ages of P and Q are in the ratio 5 : 6. Four years from now, the ratio will become 6 : 7. P's present age is:

Show solution

$\dfrac{5x + 4}{6x + 4} = \dfrac{6}{7}$ .
$35x + 28 = 36x + 24$ , so $x = 4$ .
P $= 20$ years.

Answer

20 years

Type 3very common2 practice Q

'k times as old' questions

How to spot it:

'A mother is 4 times as old as her daughter; after 5 years she will be 3 times as old' — one multiple now, another later.

F = kS, \quad F \pm n = k'(S \pm n)

Method

Take the younger age as the single unknown; the elder is k times it.
Apply the second condition at the shifted time.
Solve; check the constant difference.

Why it works:

Both conditions describe the same two people at two dates.

Try this

A mother is 4 times as old as her daughter. After 5 years, the mother will be 3 times as old as the daughter. The mother's present age is:

Show solution

$4d + 5 = 3(d + 5)$ .
$d = 10$ ; mother $= 40$ years.
Check: gap 30; 45 and 15 give 3 times.

Answer

40 years

Type 4common2 practice Q

Two shifted ratios (ago and hence)

How to spot it:

'n years ago the ratio was a : b, and n years hence it will be p : q' — two snapshots bracket the present.

\frac{ax - n}{bx - n} = \frac{a'}{b'}, \qquad \frac{ax + n}{bx + n} = \frac{p}{q}

Method

Write present ages as ax and bx.
Build both snapshots: subtract n for 'ago', add n for 'hence'.
Equate the second snapshot's ratio and solve for x.

Why it works:

Both snapshots describe the same present ages, so one x serves both.

Try this

Ten years ago the ages of A and B were in the ratio 2 : 3. Ten years from now the ratio will be 4 : 5. A's present age is:

Show solution

Present ages $2x + 10$ , $3x + 10$ ; hence $2x + 20$ , $3x + 20$ .
$\dfrac{2x + 20}{3x + 20} = \dfrac{4}{5}$ gives $x = 10$ .
A $= 30$ years.

Answer

30 years

Formula sheet

Present = ax, bx

\frac{ax \pm n}{bx \pm n} = \frac{p}{q}

Ratio after (or before) n years.

Constant difference

A - B \text{ is the same at every time}

Sum grows by 2 per year

(A + B)_{t+n} = (A + B)_t + 2n

Multiple of age

F = kS \Rightarrow F \pm n = k'(S \pm n)

k falls over time for elder-younger pairs.

Shortcuts that save time

⚡ Multiplier plus shift

Ages $ax$ , $bx$ now; shift both by n; equate the new ratio; solve for x.

Example

The present ages of A and B are in the ratio 5 : 7. After 8 years the ratio becomes 3 : 4. Find the sum of their present ages.

Show solution

$\dfrac{5x+8}{7x+8} = \dfrac{3}{4}$ gives $x = 8$ .
Ages 40 and 56; sum 96.

Answer

96 years

⚡ Use the constant difference

The gap between two ages never changes — compute it once and reuse it at any time point.

Example

A father is 3 times as old as his son. Eight years ago he was 5 times as old. Find their present ages.

Show solution

Gap $= 2s$ . Eight years ago: $\dfrac{3s-8}{s-8} = 5$ .
$3s - 8 = 5s - 40$ , so $s = 16$ ; father 48 (gap 32).

Answer

Father 48, son 16

⚡ Bracket the present with two ratios

With 'n years ago' and 'n years hence' ratios, both snapshots share one x — the 2n shift separates them.

Example

Six years ago A : B was 5 : 6, and six years hence it will be 6 : 7. Find B's present age.

Show solution

$7(5x + 12) = 6(6x + 12)$ .
$x = 12$ ; B $= 6 \times 12 + 6 = 78$ .

Answer

78 years

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Shifting the ratio terms: 5 : 7 after 8 years read as 13 : 15.

Shift the ages: (5x + 8) and (7x + 8).

Mistake 02

Adding n to only one person's age.

Every living person's age moves by n; the sum of two ages moves by 2n.

Mistake 03

Treating the ratio gap (b − a)x as the age gap at a shifted time.

The age gap (b − a)x is fixed for all time — that is exactly why it is useful.

Mistake 04

Letting the multiple rise over time for elder-younger pairs.

The multiple falls: 4 times now, 3 times later, closer to 1 eventually.

Mistake 05

Solving the two-snapshot system with n instead of 2n.

From 'n years ago' to 'n years hence' each age moves 2n years.

Quick revision

Read this the night before the exam.

Ages $ax$ , $bx$ ; shift both by $\pm n$ .
Age difference constant; sum grows by 2 per year.
k times: write the larger as k $\times$ the smaller.
Two shifted ratios: same x, 2n apart.
Check every answer against the constant gap.

Practice: 12 questions

Sets of 10, mixed across the question types above. Every answer has a step-by-step explanation.

Topic test · 12 questions

Suggested time 8 min · wrong answers go to your mistake notebook automatically.

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