Ratio, Proportion, Partnership & Ages

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

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Proportion & proportional division of terms

⏱ 3 min read🧩 5 question types🎯 13 practice Q

The idea in one minute

Four numbers $a, b, c, d$ are in proportion when $\dfrac{a}{b} = \dfrac{c}{d}$ . Cross-multiplying gives $ad = bc$ : the outer pair multiplies to the same as the inner pair.

Three named proportionals: fourth $= \dfrac{bc}{a}$ , third $= \dfrac{b^2}{a}$ , mean $= \sqrt{ab}$ .

Componendo and dividendo turns a sum-over-difference into a plain ratio: $\dfrac{x+y}{x-y} = \dfrac{p}{q}$ gives $\dfrac{x}{y} = \dfrac{p+q}{p-q}$ .

What proportion means

$a : b :: c : d$ means $\dfrac{a}{b} = \dfrac{c}{d}$ , and cross-multiplication gives $ad = bc$ .

$a$ and $d$ are the extremes (outer terms); $b$ and $c$ are the means (middle terms). Product of extremes = product of means. Every "find the missing term" question is one cross-multiplication away.

Rule: $ad = bc$ for $a : b :: c : d$ — outer times outer equals middle times middle.

The three named proportionals

Fourth proportional to $a, b, c$ : solve $\dfrac{a}{b} = \dfrac{c}{d}$ , so $d = \dfrac{bc}{a}$ . Four different numbers.
Third proportional to $a, b$ : solve $\dfrac{a}{b} = \dfrac{b}{x}$ , so $x = \dfrac{b^2}{a}$ . The middle term repeats.
Mean proportional between $a$ and $b$ : $x = \sqrt{ab}$ , from $\dfrac{a}{x} = \dfrac{x}{b}$ .

Worked: fourth to 8, 12, 18 is $\dfrac{12 \times 18}{8} = 27$ . Third to 9, 12 is $\dfrac{144}{9} = 16$ . Mean between 12 and 48 is $\sqrt{576} = 24$ .

Watch: The mean proportional is the geometric mean $\sqrt{ab}$ , never the average $\dfrac{a+b}{2}$ . The average of 3 and 12 is 7.5, but the answer is 6.

Continued proportion

$a, b, c$ are in continued proportion when $\dfrac{a}{b} = \dfrac{b}{c}$ , that is $b^2 = ac$ .

So the middle term is automatically the mean proportional of the outer two, and each member recovers from the others: $a = \dfrac{b^2}{c}$ , $c = \dfrac{b^2}{a}$ , $b = \sqrt{ac}$ .

Three numbers $x, 12, 18$ in continued proportion: $144 = 18x$ , so $x = 8$ . Check $\dfrac{8}{12} = \dfrac{12}{18} = \dfrac{2}{3}$ .

A worked chain

"Three numbers $a, b, c$ are in continued proportion, $b = 12$ and $a + c = 26$ ."

Then $ac = 144$ , and $a, c$ are the factor pair of 144 that sums to 26: 8 and 18. The numbers are 8, 12, 18.

Exam numbers are built so the factor pair is clean. If yours is not, re-read the question, not the method.

Componendo and dividendo

If $\dfrac{a}{b} = \dfrac{c}{d}$ , then $\dfrac{a+b}{a-b} = \dfrac{c+d}{c-d}$ .

The exam use runs it backwards: given $\dfrac{x+y}{x-y} = \dfrac{p}{q}$ , jump to $\dfrac{x}{y} = \dfrac{p+q}{p-q}$ .

$\dfrac{x+y}{x-y} = \dfrac{5}{3}$ gives $\dfrac{x}{y} = \dfrac{8}{2} = 4$ . No solving — one addition and one subtraction.

Tip: A sum-over-difference equal to a ratio is the fingerprint. Componendo-dividendo is the intended route.

Coefficient twists

$\dfrac{5x - 3y}{5x + 3y} = \dfrac{2}{7}$ . Apply the rule to get $\dfrac{5x}{3y} = \dfrac{7+2}{7-2} = \dfrac{9}{5}$ , then rearrange: $25x = 27y$ .

Write the two fractions with plus on top and minus below before adding and subtracting. Sign slips are the only failure here.

Order matters

Fourth proportional to 4, 9, 12 is 27, but fourth proportional to 12, 9, 4 is 3. The terms stay in the order written — copy the proportion exactly as the question lists it.

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

Fourth proportional

How to spot it:

'Find the fourth proportional to a, b, c' — three numbers, the fourth that completes a : b :: c : ?

\frac{a}{b} = \frac{c}{d} \Rightarrow d = \frac{bc}{a}

Method

Write the proportion in the exact order given: a : b :: c : d.
Cross-multiply: ad = bc.
Divide: d = bc/a.

Why it works:

The fourth term must keep the same pairwise ratio as the first three.

Try this

The fourth proportional to 8, 12 and 18 is:

Show solution

$d = \dfrac{12 \times 18}{8} = 27$ .
Check: $\dfrac{8}{12} = \dfrac{18}{27} = \dfrac{2}{3}$ .

Answer

Type 2common2 practice Q

Third proportional

How to spot it:

'Find the third proportional to a and b' — two numbers only; the middle term repeats (a : b :: b : ?).

\frac{a}{b} = \frac{b}{x} \Rightarrow x = \frac{b^2}{a}

Method

Repeat the second term: a : b :: b : x.
Cross-multiply: ax = b².
Divide: x = b²/a.

Why it works:

'Third proportional to a and b' is the next member of a continued proportion starting a, b.

Try this

The third proportional to 9 and 12 is:

Show solution

$x = \dfrac{12^2}{9} = \dfrac{144}{9} = 16$ .
Check: $\dfrac{9}{12} = \dfrac{12}{16} = \dfrac{3}{4}$ .

Answer

Type 3very common2 practice Q

Mean proportional

How to spot it:

'Find the mean proportional between a and b' — the number x with a : x :: x : b.

x = \sqrt{ab}

Method

Multiply the two numbers.
Take the square root (exact in exam numbers).
Sanity check: x² = ab; the average (a+b)/2 is the planted trap.

Why it works:

The mean proportional is the self-repeating middle term, the geometric mean.

Try this

The mean proportional between 12 and 48 is:

Show solution

$\sqrt{12 \times 48} = \sqrt{576} = 24$ .
Check: $\dfrac{12}{24} = \dfrac{24}{48} = \dfrac{1}{2}$ .

Answer

Type 4common2 practice Q

Continued proportion (find a member)

How to spot it:

'a, b, c are in continued proportion; given two, find the third' — sometimes dressed as a geometric progression.

b^2 = ac \ \Rightarrow\ b = \sqrt{ac},\ a = \frac{b^2}{c},\ c = \frac{b^2}{a}

Method

Write the defining equation a/b = b/c, that is b² = ac.
Substitute the two known members.
Solve; check both fractions reduce to the same value.

Why it works:

Continued proportion fixes one exact relationship among the three members.

Try this

The numbers x, 12 and 18 are in continued proportion. Find x.

Show solution

$12^2 = 18x$ .
$x = \dfrac{144}{18} = 8$ .

Answer

Type 5common2 practice Q

Componendo and dividendo

How to spot it:

(x + y)/(x − y) given as a ratio; find x : y — sometimes with coefficients on x and y.

\frac{x + y}{x - y} = \frac{p}{q} \Rightarrow \frac{x}{y} = \frac{p + q}{p - q}

Method

Apply C&D: add and subtract the given ratio's terms.
Read x/y = (p+q)/(p−q) directly.
With coefficients, C&D gives mx : ny first; divide off m : n.

Why it works:

One application converts the sum/difference form into the pure ratio, no solving.

Try this

If $\dfrac{a+b}{a-b} = \dfrac{7}{5}$ , then $a : b$ equals:

Show solution

$\dfrac{a}{b} = \dfrac{7+5}{7-5} = \dfrac{12}{2}$ .
$a : b = 6 : 1$ .

Answer

6 : 1

Formula sheet

Basic proportion

\frac{a}{b} = \frac{c}{d} \iff ad = bc

Fourth proportional

d = \frac{bc}{a}

Third proportional

c = \frac{b^2}{a}

Mean proportional

\text{mean} = \sqrt{ab}

Componendo & dividendo

\frac{a}{b} = \frac{c}{d} \Rightarrow \frac{a+b}{a-b} = \frac{c+d}{c-d}

Invertendo / alternando

\frac{b}{a} = \frac{d}{c},\quad \frac{a}{c} = \frac{b}{d}

Shortcuts that save time

⚡ Extremes times means

Set up the proportion in the exact order given, cross-multiply, done.

Example

Find the fourth proportional to 4, 9 and 12.

Show solution

$\dfrac{4}{9} = \dfrac{12}{d}$ .
$4d = 108$ , so $d = 27$ .

Answer

⚡ Mean proportional = geometric mean

Multiply the two numbers and take the square root. Exam numbers make it a perfect square.

Example

Find the mean proportional between 25 and 81.

Show solution

$\sqrt{25 \times 81} = \sqrt{2025}$ .
$= 45$ .

Answer

⚡ Componendo-dividendo jump

When $(x + y)$ and $(x - y)$ both appear, jump straight to $x/y$ by adding and subtracting the given ratio's terms.

Example

If $\dfrac{x+y}{x-y} = \dfrac{5}{3}$ , find $\dfrac{x}{y}$ .

Show solution

$\dfrac{x}{y} = \dfrac{5+3}{5-3}$ .
$= \dfrac{8}{2} = 4$ .

Answer

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Swapping third and fourth proportional.

Third repeats the middle term (b²/a); fourth uses three different terms (bc/a).

Mistake 02

Taking the arithmetic mean (a+b)/2 as the mean proportional.

Mean proportional is the geometric mean √(ab).

Mistake 03

Reordering the terms before writing the proportion.

Keep the question's order: 4, 9, 12 gives 27 but 12, 9, 4 gives 3.

Mistake 04

Using componendo-dividendo as (p+q)/(p−q) on the wrong side.

Sum-over-difference equals p/q means x/y = (p+q)/(p−q); check which side is the sum.

Mistake 05

Forgetting the middle term repeats in a continued proportion.

b appears in both fractions: a/b = b/c, so b² = ac.

Quick revision

Read this the night before the exam.

$ad = bc$ : extremes times extremes equals means times means.
Fourth $\dfrac{bc}{a}$ , third $\dfrac{b^2}{a}$ , mean $\sqrt{ab}$ .
Continued proportion: $b^2 = ac$ .
Sum-over-difference $\dfrac{p}{q}$ gives $\dfrac{x}{y} = \dfrac{p+q}{p-q}$ .
Order of terms is part of the question.

Practice: 13 questions

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

Topic test · 13 questions

Suggested time 7 min · wrong answers go to your mistake notebook automatically.

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