Direct and Inverse Proportion

Proportion is about how two quantities change together. If buying more apples costs more money at a steady rate, that is direct proportion; if more workers means a job finishes in less time, that is inverse proportion. This topic runs across the whole of OCR GCSE Mathematics (J560), from simple recipe scaling on Foundation to the algebraic $y = kx$y=kx and $y = \dfrac{k}{x}$y=xk​ relationships on Higher. This lesson covers the unitary method, recognising direct versus inverse proportion, and — for Higher tier — setting up and using proportion equations with a constant of proportionality.

This lesson is rich in AO1 technique (the unitary method and substituting into a proportion equation), AO2 reasoning when you decide whether a situation is direct or inverse, and AO3 problem-solving in multi-step contexts.

Key Vocabulary

Term	Meaning
Direct proportion	As one quantity increases, the other increases at the same rate
Inverse proportion	As one quantity increases, the other decreases proportionally
Unitary method	Find the value of one unit first, then scale up or down
Constant of proportionality ($k$k)	The fixed multiplier linking two proportional quantities
$y = kx$y=kx	The equation of direct proportion ($y$y is proportional to $x$x)
$y = \dfrac{k}{x}$y=xk​	The equation of inverse proportion ($y$y is inversely proportional to $x$x)

Direct Proportion and the Unitary Method

Two quantities are in direct proportion when doubling one doubles the other, and halving one halves the other. More generally, whatever you do to one quantity (multiply by $3$3, divide by $5$5, and so on) you do to the other. A defining feature is that when both quantities are zero together: no apples bought means no money spent. The most reliable way to solve direct-proportion problems — especially on the non-calculator Paper 1 — is the unitary method: find the value of one unit, then multiply by however many you need. The word "unitary" comes from "unit", a reminder that the first step is always to get down to a single one of something.

Worked Example 1

$6$6 identical pens cost £4.50. Work out the cost of $10$10 pens.

Cost of $1$1 pen $= 4.50 \div 6 = 0.75$=4.50÷6=0.75, i.e. £0.75.

Cost of $10$10 pens $= 10 \times 0.75 = 7.50$=10×0.75=7.50, i.e. £7.50.

Answer: £7.50.

Common error: dividing by $10$10 and multiplying by $6$6 (the wrong way round). Find the cost of one pen first, then scale to the number you want.

Worked Example 2

A recipe for $8$8 scones needs $320$320 g of flour. Work out how much flour is needed for $20$20 scones.

Flour for $1$1 scone $= 320 \div 8 = 40$=320÷8=40 g.

Flour for $20$20 scones $= 20 \times 40 = 800$=20×40=800 g.

Answer: $800$800 g.

Worked Example 3

A car travels $135$135 miles on $15$15 litres of fuel. Assuming fuel use is in direct proportion to distance, work out how far it travels on $24$24 litres.

Distance per litre $= 135 \div 15 = 9$=135÷15=9 miles.

Distance on $24$24 litres $= 24 \times 9 = 216$=24×9=216 miles.

Answer: $216$216 miles.

Worked Example 3b

$9$9 identical tiles cover an area of $1.44$1.44 m². Work out the area covered by $15$15 of these tiles.

Area covered by $1$1 tile $= 1.44 \div 9 = 0.16$=1.44÷9=0.16 m².

Area covered by $15$15 tiles $= 15 \times 0.16 = 2.4$=15×0.16=2.4 m².

Answer: $2.4$2.4 m². As a sanity check, $15$15 tiles is more than $9$9, so the area should be larger than $1.44$1.44 m² — and $2.4 > 1.44$2.4>1.44, as expected.

Inverse Proportion

Two quantities are in inverse proportion when increasing one decreases the other in the same ratio: double one, the other halves; treble one, the other is divided by three. This is the opposite behaviour to direct proportion, and the contrast is worth holding firmly in mind, because mixing the two up is the single most common error in this topic. The key idea behind inverse proportion is that the product of the two quantities stays constant. A typical context is "more workers, less time": if a job is a fixed amount of work, then doubling the workforce halves the time, but the total work — workers multiplied by time — does not change. Defining that fixed total (in "worker-hours", "pump-minutes" or similar) and then dividing by the new quantity is the most reliable route through any inverse-proportion problem.

Worked Example 4

It takes $4$4 painters $9$9 hours to paint a hall. Working at the same rate, how long would $6$6 painters take?

Find the total work in "painter-hours": $4 \times 9 = 36$4×9=36 painter-hours.

With $6$6 painters: time $= 36 \div 6 = 6$=36÷6=6 hours.

Answer: $6$6 hours.

Common error: treating this as direct proportion and scaling up. More painters means less time, so the answer must be smaller than $9$9 hours.

Worked Example 5

A tank is filled by $5$5 identical pumps in $12$12 minutes. How long would $4$4 pumps take?

The total amount of work needed to fill the tank is fixed. Measuring that work in "pump-minutes" gives a constant we can rely on.

Total work $= 5 \times 12 = 60$=5×12=60 pump-minutes.

With $4$4 pumps: $60 \div 4 = 15$60÷4=15 minutes.

Answer: $15$15 minutes — and as a sanity check, fewer pumps gives a longer time, which is correct. If you had mistakenly treated this as direct proportion and scaled $12$12 down, you would have got a shorter time, which makes no physical sense.

Deciding Direct or Inverse

The hardest part of a proportion question is often not the arithmetic but deciding which type of proportion you are dealing with. Spend a moment on this before you calculate, because choosing wrongly leads to dividing when you should multiply (or vice versa). Before calculating, ask: "If I increase one quantity, does the other go up or down?" If it goes up, the relationship is direct and you scale both quantities the same way; if it goes down, it is inverse and you should think in terms of a fixed total (the product). The table below shows some common situations.

Situation	Increase one...	The other...	Type
Number of items and total cost	more items	more cost	Direct
Speed and time for a fixed journey	faster speed	less time	Inverse
Number of workers and time for a job	more workers	less time	Inverse
Hours worked and pay	more hours	more pay	Direct

Algebraic Proportion [Higher]

On Higher tier you express proportion as an equation using a constant of proportionality, $k$k. This is a more powerful approach than the unitary method, because once you have the equation you can answer any question about the relationship — find $y$y from $x$x, find $x$x from $y$y, or describe how the quantities behave in general. The symbol $\propto$∝ means "is proportional to", and turning a proportionality statement into an equation always involves introducing the constant $k$k.

• Direct proportion: "$y$y is proportional to $x$x", written $y \propto x$y∝x, means $y = kx$y=kx.
• Inverse proportion: "$y$y is inversely proportional to $x$x", written $y \propto \dfrac{1}{x}$y∝x1​, means $y = \dfrac{k}{x}$y=xk​.

The method is always: use a given pair of values to find $k$k, write the full equation, then use it to answer the question.

Worked Example 6 [Higher]

$y$y is directly proportional to $x$x. When $x = 4$x=4, $y = 22$y=22. Work out $y$y when $x = 10$x=10.

Since $y = kx$y=kx, substitute the known pair: $22 = k \times 4$22=k×4, so $k = \dfrac{22}{4} = 5.5$k=422​=5.5.

The equation is $y = 5.5x$y=5.5x. When $x = 10$x=10:

$y = 5.5 \times 10 = 55.$y=5.5×10=55.

Answer: $y = 55$y=55.

Worked Example 7 [Higher]

$y$y is inversely proportional to $x$x. When $x = 3$x=3, $y = 8$y=8. Work out $y$y when $x = 12$x=12.

Since $y = \dfrac{k}{x}$y=xk​, substitute: $8 = \dfrac{k}{3}$8=3k​, so $k = 8 \times 3 = 24$k=8×3=24.

The equation is $y = \dfrac{24}{x}$y=x24​. When $x = 12$x=12:

$y = \dfrac{24}{12} = 2.$y=1224​=2.

Answer: $y = 2$y=2.

Common error: using $y = kx$y=kx for an inverse relationship. For inverse proportion, $k$k is the product $xy$xy, not the quotient.

Extended Worked Examples

Worked Example 8: Choosing the cheaper option

A shop sells rice in two packs: a $2$2 kg pack for £3.20 and a $5$5 kg pack for £7.50. Using proportion, work out which is better value.

Cost per kg for the small pack $= 3.20 \div 2 = 1.60$=3.20÷2=1.60, i.e. £1.60 per kg.

Cost per kg for the large pack $= 7.50 \div 5 = 1.50$=7.50÷5=1.50, i.e. £1.50 per kg.

Since £1.50 < £1.60, the $5$5 kg pack is better value.

Answer: the $5$5 kg pack, at £1.50 per kg.

Worked Example 9: A two-step unitary problem

$12$12 workers can build a wall in $10$10 days. Work out how long $8$8 workers would take, then how many workers are needed to finish in $5$5 days.

This is inverse proportion. Total work $= 12 \times 10 = 120$=12×10=120 worker-days.