Proportion problems

A shopkeeper mixes two brands of tea in the ratio $2:3$ and wants a second mixture, in a bigger sack, that tastes exactly the same. How much of each brand goes in? The answer is not "just guess and taste" — it is a precise piece of algebra called proportion: the statement that two ratios are equal. Once four numbers are locked into a proportion, any one of them can be found from the other three by a single cross-multiplication rule. A close cousin, continued proportion, links three numbers so tightly that the middle one — the mean proportional — is pinned down by the other two through a square root. This article builds both ideas from the ratio work you already know, sets out the exact rule for finding unknown terms, and walks through fully solved examples of the kind that show up directly in your ICSE Class 10 paper.

The core insight

Four numbers $a,b,c,d$ are in proportion exactly when $ad=bc$ (product of extremes equals product of means), and this single rule also gives the mean proportional $b=\sqrt{ac}$ for three numbers in continued proportion.

The core idea

A ratio $a:b$ compares two quantities. A proportion goes one step further: it is the equality of two ratios, written

a:b :: c:d, \quad \text{i.e.} \quad \dfrac{a}{b}=\dfrac{c}{d}.

where: $a$ and $d$ , the outer pair, are the extremes; $b$ and $c$ , the inner pair, are the means. Cross-multiplying the fraction form gives the single working rule every question in this topic reduces to: $ad = bc$ , "product of extremes equals product of means." A proportion is a claim that can be true or false for a given set of four numbers, and solving one always means using this rule (or a close variant of it) to pin down whatever term is unknown.

Continued proportion tightens the idea from four numbers to three. Quantities $a,b,c$ are in continued proportion if the ratio of the first pair equals the ratio of the second pair, with $b$ shared between them:

a:b = b:c, \quad \text{i.e.} \quad b^2 = ac.

Here $b$ — playing double duty as the second term of the first ratio and the first term of the second — is called the mean proportional between $a$ and $c$ , and is found as $b=\sqrt{ac}$ (taken positive, since these problems deal with magnitudes). Once $a$ and $b$ are known, the next term $c$ is called the third proportional to $a$ and $b$ , and is found from the very same relation $b^2=ac$ , now solved for $c$ instead of $b$ . It helps to keep the vocabulary straight. A ratio is a comparison of two quantities, $a:b$ , with no claim about any other pair. A proportion is a stronger statement: it says a second ratio, $c:d$ , equals the first. So "solving a proportion" always means you are given three of the four numbers $a,b,c,d$ (or two of the three numbers $a,b,c$ in a continued proportion) and are asked to find the missing one using $ad=bc$ or $b^2=ac$ . These ideas extend beyond three or four terms too — for example, $a,b,c,d$ are in continued proportion if $a:b=b:c=c:d$ — but at Class 10 level, questions are almost always limited to the basic four-term proportion and the three-term continued proportion covered here. The figure below makes the "equal ratios" idea visual: two bars split in the same proportion look identical in shape even though their total lengths differ.

Figure

Two bars split in the same proportion two to three, one bar shorter and one longer, showing that equal ratios keep the same shape at different sizes

Derivation AProduct of extremes equals product of means, and its continued-proportion form

The goal here is to show, starting only from the definition of a proportion as an equality of two fractions, why cross-multiplying always gives "product of extremes = product of means," and then to show that the very same cross-multiplication, applied to a continued proportion, forces the mean proportional to be a square root rather than an ordinary average.

Step 1. Start from the definition. If $a,b,c,d$ are in proportion, then by definition $\dfrac{a}{b}=\dfrac{c}{d}$ . This is simply a statement that two fractions are equal; nothing about proportion has been used yet beyond this equality.

Step 2. Clear denominators. Multiply both sides of $\dfrac{a}{b}=\dfrac{c}{d}$ by $bd$ (the product of the two denominators). The left side becomes $\dfrac{a}{b}\times bd = ad$ , and the right side becomes $\dfrac{c}{d}\times bd = bc$ . So $ad = bc$ — the product of the extremes $a,d$ equals the product of the means $b,c$ . This step is exactly cross-multiplication, and it works in reverse too: if $ad=bc$ for some four numbers, dividing both sides by $bd$ recovers $\dfrac{a}{b}=\dfrac{c}{d}$ , so the two forms are fully equivalent.

Step 3. Specialise to continued proportion. If $a,b,c$ are in continued proportion, then $b$ plays the role of both middle terms at once: $a:b=b:c$ , i.e. $\dfrac{a}{b}=\dfrac{b}{c}$ . Applying Step 2 with the substitution $d\to c$ and $c\to b$ (matching term positions) gives $a\times c = b\times b$ , i.e. $b^2=ac$ . Taking the positive square root, since $b$ represents a magnitude, gives the mean proportional formula $b=\sqrt{ac}$ .

Step 4. Use it to find the third proportional. Once $a$ and $b$ are fixed and we instead want the third term $c$ such that $a,b,c$ continue the proportion, the same equation $b^2=ac$ is simply solved for $c$ : $c=\dfrac{b^2}{a}$ . So one relation, $b^2=ac$ , answers both "find the mean proportional" (solve for $b$ ) and "find the third proportional" (solve for $c$ ) — the only difference is which letter is unknown. Both results — $ad=bc$ for four-term proportion and $b^2=ac$ for continued proportion — are the same cross-multiplication idea applied to a two-term and a three-term chain respectively.

Worked example

Question: Find $x$ if $x,6,8,24$ are in proportion. Then find the mean proportional between $4$ and $25$ , and the third proportional to $4$ and $10$ .

Step 1. Since $x:6::8:24$ , apply product of extremes = product of means: $x\times 24 = 6\times 8$ , so $24x=48$ , giving $x=2$ . Checking: $2:6=1:3$ and $8:24=1:3$ — the two ratios agree, confirming the proportion holds.

Step 2. For the mean proportional between $4$ and $25$ , let $b$ be the middle term of the continued proportion $4,b,25$ . By $b^2=ac$ , $b^2 = 4\times 25 = 100$ , so $b=\sqrt{100}=10$ . Checking: $4:10=2:5$ and $10:25=2:5$ — equal, so $4,10,25$ are genuinely in continued proportion. Note that $10\ne\dfrac{4+25}{2}=14.5$ , confirming a mean proportional is not the same thing as an ordinary average.

Step 3. For the third proportional to $4$ and $10$ , we need $c$ such that $4,10,c$ are in continued proportion, i.e. $10^2=4\times c$ , so $100=4c$ and $c=25$ . Checking directly: $4:10=2:5$ and $10:25=2:5$ — the ratios match, confirming continued proportion.

Answer: $x=2$ ; the mean proportional between $4$ and $25$ is $10$ ; the third proportional to $4$ and $10$ is $25$ .

Competitive stretch

If instead $p,p+3,p+3,p+15$ were said to be in proportion, the same rule applies even with an algebraic term: cross-multiplying, $p(p+15)=(p+3)(p+3)$ , giving $p^2+15p = p^2+6p+9$ , so $9p=9$ and $p=1$ . This shows the extremes-means rule works unchanged when the four terms are algebraic expressions rather than plain numbers — only the bookkeeping gets longer. As a check, substitute back: with $p=1$ the four terms are $1,4,4,16$ , and indeed $1\times16=16=4\times4$ , so the proportion holds. It is worth noticing why these examples were chosen to have neat, whole-number answers. Exam-setters generally build proportion questions around numbers whose product or ratio simplifies cleanly, since the underlying skill being tested is the correct setup of extremes and means, not messy arithmetic. If your own working produces an ugly decimal or a negative value under a square root, that is usually a signal to re-check which terms you assigned as extremes and which as means, rather than a sign that proportion problems are inherently messy.

Common traps

Mistaking the mean proportional ( $b=\sqrt{ac}$ , from continued proportion) for a plain average $\dfrac{a+c}{2}$ — they coincide only when $a=c$ .
Swapping "third proportional to $a$ and $b$ " (find $c$ with $a,b,c$ in continued proportion) with "mean proportional between $a$ and $b$ " (find the middle term of $a,x,b$ ) — these are two different setups solving for different letters in $b^2=ac$ .
Skipping the check that $ad=bc$ actually holds for a claimed proportion before using it to solve for an unknown term.
Taking a negative square root for a mean proportional, when the quantities represent magnitudes and only the positive root is meaningful.
Rushing the cross-multiplication when one term is an algebraic expression like $x+2$ — expand and simplify fully rather than treating it as a single block.

Ask the tutor: want a few more mixed proportion and continued-proportion questions, including ones with algebraic terms, to practise before your test?

Why is the mean proportional a square root and not an average?

Because it comes from a ratio equality, not from adding. In continued proportion $a,b,c$ , the defining condition is $a:b=b:c$ , which cross-multiplies to $b^2=ac$ — a product, not a sum. Solving for $b$ therefore gives a square root, $b=\sqrt{ac}$ . An average $\dfrac{a+c}{2}$ satisfies a completely different condition (equal spacing, not equal ratios), so the two only ever agree in the special case $a=c$ , where both formulas collapse to the same number.

How do I tell mean proportional and third proportional apart quickly on an exam?

Look at what is missing. "Mean proportional between $a$ and $c$ " gives you the two outer numbers and asks for the middle term $b$ in $a,b,c$ — solve $b^2=ac$ for $b$ . "Third proportional to $a$ and $b$ " gives you the first two numbers in order and asks for the next term $c$ — solve the same $b^2=ac$ , but now for $c=\dfrac{b^2}{a}$ . Same equation, different unknown; reading the question carefully to see which letter is missing avoids the mix-up.

Does the order of the four terms in a:b::c:d matter?

Yes. Extremes are the outer terms $a$ and $d$ ; means are the inner terms $b$ and $c$ . Writing the same four numbers in a different order changes which pair is multiplied together, so $a:b::c:d$ is not automatically the same statement as, say, $a:c::b:d$ . Always cross-multiply using the actual positions given in the question — outer times outer, inner times inner — rather than relying on memorised number pairs.

Proportion problems

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads