Ratio, Proportion, Partnership & Ages

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Problems on ages

Golden rule: the difference of ages is constant over time; the sum grows by 2 per year; each person's age grows by 1 per year.

Standard template: present ratio given (say a : b), and a ratio after (or before) n years given. Write present ages ax, bx and form the equation from the second ratio.

If one age is a multiple of the other ('father is 3 times the son'), write the larger as k × smaller and use the time-shifted condition. Two unknowns are never needed — one multiplier x or k suffices.

Double-ratio questions (a ratio n years ago AND a ratio n years hence) give two equations in x; eliminate x.

📖 Notes 🧩 4 question types ⚡ Formulas 🎯 12 practice Q

Detailed notes

The three invariants

Each person's age grows by 1 every year — add n to BOTH ages for a shift of n years, and 2n to a sum of two ages.
The difference of two ages never changes (32-year gap now is 32 forever).
The ratio keeps changing — that is what the questions exploit.

The multiplier template

Present ages in ratio $a : b$ → write them as $ax$ and $bx$ . The shifted condition becomes one equation: $\frac{ax + n}{bx + n} = \frac{p}{q} \quad \text{(n years hence)}, \qquad \frac{ax - n}{bx - n} = \frac{p}{q} \quad \text{(n years ago)}.$ One cross-multiplication gives x. If the question gives a sum or difference instead of a ratio, $(a+b)x$ or $(b-a)x$ is that value — no shift needed.

The constant-difference checker

After solving, verify: the age difference now must equal the difference at any other time. For a father 3 times the son with son 16: gap = 32; eight years ago it was still 32 (24 vs 8 — father 5 times ✓, matching the given 5 : 1). This check kills half the wrong options without solving anything.

'k times' questions

"Father is k times the son" → father = k·son. Use the son as the variable; the second condition (after/before n years) gives one linear equation. Remember the ratio number DROPS as time moves forward (k decreases: 5 times → 3 times), never grows, for an older–younger pair.

Double-ratio questions

A ratio 'n years ago' AND a ratio 'n years hence' → ages $(ax - n, bx - n)$ and $(ax + n, bx + n)$ : two equations, one unknown x — eliminate by subtracting or cross-multiplying both. The 2n gap between the two time points is the easiest slip.

The sum trick for two people

Sum of ages grows by exactly 2 per year (two people) or 2n per n years. So "the sum of present ages is 40; what was it 5 years ago?" — answer 30, no ratios needed. Combined with a ratio, the sum pins the pair instantly: ratio 3 : 5 with sum 40 → parts of 5 → 15 and 25.

Three-person ages

Ratios among three people (A : B : C = 4 : 5 : 6) work exactly like two — write 4x, 5x, 6x. Only pairwise conditions appear in exams ("A is 6 years older than C" → 2x = 6), because three independent shifts would over-determine the problem.

Typical wrong options, decoded

Shifting the ratio instead of the ages: "ratio 4 : 5 now, 5 : 6 after n years" answered with ages $4 + n, 5 + n$ — the 4 : 5 ratio never applies again.
Adding n to only one person's age.
For 'k times', setting the ELDER as the unknown k times over — the younger must carry the single x.

Quick revision

Ages = ax, bx; shift adds/subtracts n on both.
Difference constant; sum changes by 2 per year.
k times → write the larger as k × smaller.
Two shifted ratios → two equations in the same x.

Types of questions asked

Every way this subtopic shows up in exams — how to recognise it, the formula or logic to use, and a solved example.

Type 1: Present ratio with sum or differencevery common2 practice Q

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

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

Example: 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:

7 parts = 35 → 1 part = 5 → younger $= 3 \times 5 = 15$ years.

Type 2: Ratio now and ratio after/before n yearsvery common2 practice Q

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}

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

Why: both ages move by the same n, so the ratio shifts in a way that pins x exactly.

Example: 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:

$\frac{5x + 4}{6x + 4} = \frac{6}{7}$ → $35x + 28 = 36x + 24$ → x = 4 → P = 20 years.

Type 3: 'k times as old' questionsvery common2 practice Q

How to spot it:

'A father is 4 times as old as his daughter; after 5 years he will be 3 times as old' — one multiple now, another later.

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

Take the younger age as S; the elder is kS.
Apply the second condition at the shifted time.
Solve the small linear equation; check the constant difference.

Why: both conditions are statements about the same two numbers at two dates.

Example: 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:

$4d + 5 = 3(d + 5)$ → $d = 10$ → mother $= 40$ years. (Gap 30 is constant: 40/10 now, 45/15 later ✓.)

Type 4: Two shifted ratios (ago and hence)common2 practice Q

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}

Write both equations in the same x.
Cross-multiply each (or subtract the two statements — the 2n shift cancels x cleanly).
Solve for x, then build the present ages.

Why: both snapshots describe the same present ages, x links them.

Example: 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:

$\frac{2x + 20}{3x + 20} = \frac{4}{5}$ → $10x + 100 = 12x + 80$ → x = 10 → A $= 2(10) + 10 = 30$ years.

Formulas

Present = a x, b x

\frac{ax + n}{bx + n} = \frac{p}{q} \quad (\text{ratio after } n \text{ years})

Ratio n years ago

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

Constant difference

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

Multiple of age

F = kS \Rightarrow \frac{F - n}{S - n} = \frac{p}{q}

Shortcut tricks

⚡ Multiplier + shift equation

Ages = ax and bx now; add/subtract the shift; equate the new ratio; solve for x.

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

$\frac{5x+8}{7x+8} = \frac{3}{4}$ ⇒ $20x + 32 = 21x + 24$ ⇒ x = 8. Ages 40 and 56 ⇒ sum 96.

⚡ Use the constant difference

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

Example: Father is 3 times as old as his son. Eight years ago he was 5 times as old. Find their present ages.

Gap = 3s − s = 2s. Eight years ago gap = 2s still: $\frac{3s-8}{s-8} = 5$ ⇒ s = 16, father 48 (gap 32, checks out).

⚡ Two ratios bracket the answer

With 'n years ago' and 'n years hence' ratios, form both equations and eliminate x by subtraction.

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

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

Where students lose marks

Shifting the ratio by n (ratio 5:7 now does NOT become 5:7 + n).
Adding n to the sum of ages once instead of twice.
For 'n years ago', subtracting from the ratio terms but forgetting both persons.
Assuming the ratio difference (b − a)x equals the age difference at a shifted time.

Practice sets — 12 questions

Sets of 10, mixed across the question types above. Each answer comes with a step-by-step explanation.

Topic test · 12 questions

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