✖️ Multiplicative Relationships

GCSE Maths · Ratio, Proportion & Rates of Change (R6)

Ages 15–16 · Foundation & Higher

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1 What is a Multiplicative Relationship?

Two quantities are linked multiplicatively when you get from one to the other by multiplying (or dividing) — not by adding.

The core statement
$A = k \times B$   where $k$ is the multiplier
Three ways of saying exactly the same thing. If $A = 15$ and $B = 6$:
Multiplier: $A = 2.5 \times B$  (because $15 \div 6 = 2.5$)
Fraction: $A$ is $\dfrac{5}{2}$ of $B$
Ratio: $A : B = 15 : 6 = 5 : 2$
Learning to slide between these three forms is the whole of this subtopic.
B = 6 (the "of" one) × 2.5 A = 15 ÷ 2.5  (or × 0.4)
Additive thinking is the enemy. "$15$ is $9$ more than $6$" is true but almost never useful in this topic — the number $9$ tells you nothing that scales. The multiplier $2.5$ does.
2 Finding the Multiplier
Worked Example 1 — Both directions

A shop's takings rose from £$800$ on Monday to £$1000$ on Tuesday.

Monday $\to$ Tuesday: $1000 \div 800 = 1.25$
Tuesday $\to$ Monday: $800 \div 1000 = 0.8$
Note that $1.25 \times 0.8 = 1$ — the two multipliers are reciprocals of each other.

So Tuesday's takings are $1.25$ times Monday's, and Monday's are $0.8$ of Tuesday's.

Useful check: if the multiplier is bigger than $1$ the quantity has grown; if it is between $0$ and $1$ the quantity has shrunk.
Worked Example 2 — Multiplier as a fraction

A pencil is $18$ cm long and a pen is $12$ cm long. Express the pen's length as a multiple of the pencil's.

Multiplier $= \dfrac{12}{18} = \dfrac{2}{3}$
So pen $= \dfrac{2}{3} \times$ pencil.
As a decimal: $0.666\ldots$, which is awkward — here the fraction is the better answer.
3 Switching Between Ratio, Fraction and Multiplier
GivenTo a ratioTo a fractionTo a multiplier
$A : B = 3 : 4$$A = \dfrac{3}{4}B$$A = 0.75B$
$A = \dfrac{5}{2}B$$A : B = 5 : 2$$A = 2.5B$
$A = 1.6B$$A : B = 8 : 5$$A = \dfrac{8}{5}B$
$A$ is $40\%$ of $B$$A : B = 2 : 5$$A = \dfrac{2}{5}B$$A = 0.4B$
Worked Example 3 — Decimal multiplier to ratio

$A = 1.6B$. Write $A : B$ in its simplest form.

Take $B = 1$, so $A = 1.6$. That gives $A : B = 1.6 : 1$.
Multiply both by $10$ to clear the decimal: $16 : 10$.
Divide by the HCF $2$: $\;8 : 5$.

Answer: $A : B = 8 : 5$

Worked Example 4 — Ratio to multiplier

The ratio of Ella's savings to Fred's is $7 : 4$. Fred has £$260$. Find Ella's savings using a multiplier.

Ella $= \dfrac{7}{4} \times$ Fred, so the multiplier is $1.75$.
Ella $= 260 \times 1.75 = £455$

Check with the parts method: one part $= 260 \div 4 = 65$, so Ella $= 7 \times 65 = £455$ ✓

4 Chains of Multipliers

If you apply one multiplier and then another, the overall multiplier is found by multiplying them together.

Combining multipliers
$\text{overall multiplier} = m_1 \times m_2 \times \ldots$
Worked Example 5 — Two changes in a row

A price is increased by $20\%$ and then reduced by $15\%$. Find the overall multiplier and describe the overall change.

Increase by $20\%$: multiplier $= 1.2$
Decrease by $15\%$: multiplier $= 0.85$
Overall $= 1.2 \times 0.85 = 1.02$

An overall increase of $2\%$.

Note that this is not the same as $20 - 15 = 5\%$. Percentages must be combined by multiplying, never by adding.

Worked Example 6 — A chain of three quantities

$P$ is $\dfrac{3}{5}$ of $Q$, and $Q$ is $\dfrac{10}{9}$ of $R$. Express $P$ as a fraction of $R$.

$P = \dfrac{3}{5}Q$  and  $Q = \dfrac{10}{9}R$
Substitute: $P = \dfrac{3}{5} \times \dfrac{10}{9} \times R$
$\dfrac{3 \times 10}{5 \times 9} = \dfrac{30}{45} = \dfrac{2}{3}$

$P = \dfrac{2}{3}R$, so $P : R = 2 : 3$.

5 Reversing a Multiplier

To undo a multiplication you divide — equivalently, you multiply by the reciprocal.

Forward multiplierReverse multiplier
$\times 4$$\div 4$  (that is, $\times 0.25$)
$\times 1.15$$\div 1.15$  ($\approx \times 0.8696$)
$\times \dfrac{3}{7}$$\times \dfrac{7}{3}$
$\times 0.8$$\times 1.25$
Worked Example 7 — Working backwards

After a $35\%$ discount, a bike costs £$266.50$. What was the original price?

Forward multiplier $= 1 - 0.35 = 0.65$
So original $\times 0.65 = 266.50$
Original $= 266.50 \div 0.65 = £410$

Check: $410 \times 0.65 = 266.50$ ✓

6 Writing the Relationship Algebraically

Exam questions increasingly ask you to write the link as an equation, because that is what you need for proportion questions later.

Worked Example 8 — From words to an equation

"There are $5$ times as many sheep as cows." Let $s$ be the number of sheep and $c$ the number of cows. Write an equation.

Sheep are the bigger group, so sheep $=$ multiplier $\times$ cows.
$s = 5c$
Very common error: writing $5s = c$. Test with numbers — if there are $2$ cows there are $10$ sheep, and $s = 5c$ gives $10 = 5 \times 2$ ✓ while $5s = c$ gives $50 = 2$ ✗.
Worked Example 9 — Two relationships combined

$a : b = 4 : 3$ and $b : c = 9 : 2$. Write $a$ in terms of $c$.

$a = \dfrac{4}{3}b$  and  $b = \dfrac{9}{2}c$
$a = \dfrac{4}{3} \times \dfrac{9}{2} \times c = \dfrac{36}{6}c = 6c$

$a = 6c$, so $a : c = 6 : 1$.

7 Quick Reference

Multiplier

$k = \dfrac{\text{quantity you want}}{\text{quantity you start from}}$.

Bigger or smaller?

$k > 1$ means growth; $0 < k < 1$ means shrinkage.

Ratio → fraction

$A : B = a : b$ means $A = \dfrac{a}{b}B$.

Decimal → ratio

Write $k : 1$, then clear the decimal and simplify.

Chains

Multiply the multipliers together; never add them.

Reversing

Divide by the multiplier — the reciprocal undoes it.

Algebra

"$A$ is $n$ times $B$" $\Rightarrow A = nB$. Always test with real numbers.

Percentages

$p\%$ of $B$ means $\dfrac{p}{100} \times B$ — just another multiplier.

8 Practice Questions
Question 1

$A = 24$ and $B = 15$. Write (a) the multiplier from $B$ to $A$, (b) $A$ as a fraction of $B$, (c) $A : B$ in its simplest form.

▶ Show solution

(a) $24 \div 15 = 1.6$

(b) $\dfrac{24}{15} = \dfrac{8}{5}$

(c) $24 : 15 = 8 : 5$ (divide both by $3$)

Question 2

The number of visitors to a museum fell from $4500$ to $3600$. Find the multiplier, and state the percentage change.

▶ Show solution

Multiplier $= 3600 \div 4500 = 0.8$

$0.8 = 80\%$, so visitors are now $80\%$ of what they were.

That is a decrease of $100 - 80 = 20\%$.

Question 3

$P = 0.35Q$. Write $P : Q$ in its simplest form.

▶ Show solution

$P : Q = 0.35 : 1$

Multiply both by $100$: $35 : 100$

Divide by the HCF $5$: $\;7 : 20$

Question 4

The ratio of Jaz's height to Kim's height is $6 : 5$. Kim is $145$ cm tall. Find Jaz's height.

▶ Show solution

Jaz $= \dfrac{6}{5} \times$ Kim $= 1.2 \times 145$

$= 174$ cm

Question 5

A quantity is multiplied by $1.5$ and then by $0.6$. Find the single multiplier that has the same effect, and say whether the quantity has grown or shrunk.

▶ Show solution

$1.5 \times 0.6 = 0.9$

Since $0.9 < 1$, the quantity has shrunk — it is now $90\%$ of the original, a $10\%$ decrease.

Question 6

$X$ is $\dfrac{4}{7}$ of $Y$, and $Y$ is $\dfrac{7}{2}$ of $Z$. Express $X$ in terms of $Z$.

▶ Show solution

$X = \dfrac{4}{7} \times \dfrac{7}{2} \times Z = \dfrac{28}{14}Z = 2Z$

$X = 2Z$, so $X : Z = 2 : 1$.

Question 7

After a $12\%$ increase, a salary is £$33\,600$. Find the salary before the increase.

▶ Show solution

Forward multiplier $= 1.12$.

Original $= 33\,600 \div 1.12 = £30\,000$

Check: $30\,000 \times 1.12 = 33\,600$ ✓

Question 8

There are $\dfrac{3}{8}$ as many buses as cars in a depot. Write an equation linking the number of buses $b$ and the number of cars $c$, and find $c$ when $b = 21$.

▶ Show solution

$b = \dfrac{3}{8}c$

When $b = 21$: $\;21 = \dfrac{3}{8}c$

Multiply both sides by $8$: $168 = 3c$, so $c = 56$ cars.

Question 9

A shop reduces all prices by $30\%$, then adds $20\%$ VAT to the reduced price. A jacket was originally £$95$ before VAT. Find the final price and the overall multiplier.

▶ Show solution

Overall multiplier $= 0.7 \times 1.2 = 0.84$

Final price $= 95 \times 0.84 = £79.80$

The overall effect is a $16\%$ decrease on the original.

Question 10

$m : n = 5 : 2$ and $n : p = 6 : 7$.

(a) Write $m$ in terms of $p$.   (b) If $p = 42$, find $m$ and $n$.

▶ Show solution

(a) $m = \dfrac{5}{2}n$ and $n = \dfrac{6}{7}p$.

$m = \dfrac{5}{2} \times \dfrac{6}{7} \times p = \dfrac{30}{14}p = \dfrac{15}{7}p$

(b) $n = \dfrac{6}{7} \times 42 = 36$

$m = \dfrac{15}{7} \times 42 = 90$

Check: $m : n = 90 : 36 = 5 : 2$ ✓  and  $n : p = 36 : 42 = 6 : 7$ ✓

Multiplicative Relationships (R6) · GCSE Maths Revision · Created with MathJax