Sec 2 Direct and Inverse Proportion: Concepts & Worked Examples

Secondary 2 G3 Mathematics · Study guide

Choose a proportional model, find its constant and handle powers, roots, work rates and percentage changes through four worked examples and a graph explorer.

Concept → worked examples → practice16 min read · practice linked below

First identify what stays constant

Two quantities changing together does not automatically make them proportional. Direct proportion keeps a ratio constant. Inverse proportion keeps a product constant. Start with the exact relationship in the question, then use a known pair to find its constant.

Two basic models
Relationship	Equation	What stays constant?
y directly proportional to x	$y=kx$	$y/x=k$ , for $x\ne0$
y inversely proportional to x	$y=k/x$ , $x\ne0$	$xy=k$

In the positive-quantity models here, $k>0$ . Multiplying x by a positive factor multiplies y by that factor in direct proportion, and divides y by it in inverse proportion.

Direct proportion and the origin

The graph of $y=kx$ is a straight line through the origin. A fixed starting charge gives a different model: $C=2+0.15n$ is linear, but C is not directly proportional to n because $C/n$ is not constant and the graph does not pass through the origin.

Worked example 1 — find the unit cost

A printer charges $1.80 for 12 pages, $4.50 for 30 pages and $7.50 for 50 pages. Check a direct-proportion model and find the cost of 84 pages.

Divide each cost by its corresponding page count:

$\frac{1.80}{12}=\frac{4.50}{30}=\frac{7.50}{50}=0.15.$

The constant cost per page is $0.15, so the data are consistent with $C=0.15n$ , where C is in dollars and n is the number of pages. Assuming the same pricing continues, $C=0.15(84)=\$12.60$ .

Check: 84 is seven times 12, so the cost must be seven times $1.80: $12.60. A table supports the model over the stated conditions; a new bulk discount would require a new rule.

Compare a constant ratio with a constant product

Both graphs below pass through $(2,6)$ . When x doubles from 2 to 4, direct proportion doubles y from 6 to 12. Inverse proportion halves y from 6 to 3. The different changes preserve different quantities.

The axes use the same scales in both panels. Direct proportion gives a line through the origin. The positive inverse curve continues towards both axes without touching them; the arrow shows it continuing above the displayed window.

For the first graph, $y/x=3$ whenever $x\ne0$ . For the second, $xy=12$ , with $x>0$ here. Test the stated ratio or product instead of deciding from “increases” or “decreases” alone.

Match the whole expression, including powers and roots

“Directly proportional to the square of x” means $y=kx^2$ , not $y=kx$ . Similarly, $y\propto x^3$ means $y=kx^3$ , and $y\propto\sqrt{x}$ means $y=k\sqrt{x}$ with $x\ge0$ . The graph of y against x² is a line through the origin for the square model; the graph of y against x is a curve.

Worked example 2 — a square law and a percentage increase

A model gives braking distance d metres directly proportional to speed v km/h squared. At 40 km/h, the braking distance is 8 m. Find the distance at 60 km/h and its percentage increase.

Translate the relationship: $d=kv^2$ .
Find the constant using the known pair: $8=k(40)^2$ , so $k=8/1600=1/200$ . Thus $d=v^2/200$ , with these stated units.
At $v=60$ , $d=3600/200=18$ m.
The increase is $18-8=10$ m, so the percentage increase is $10/8\times100\%=125\%$ .

Check by factors: the speed becomes $60/40=1.5$ times as large, so distance becomes $1.5^2=2.25$ times as large. This is 225% of the original, which is a 125% increase.

Root connection: rearranging gives $v=\sqrt{200d}$ for non-negative speed. At d = 12 m, $v=\sqrt{2400}\approx49.0$ km/h to 3 significant figures. Use the positive root because speed cannot be negative.

Inverse proportion in a fixed job

For positive x and k, the graph of $y=k/x$ is a decreasing curve in the first quadrant. It approaches the axes without meeting them: x cannot be zero, and k/x cannot equal zero. If instead you plot y against $1/x$ , the graph is a straight line through the origin with gradient k, because $y=k(1/x)$ .

More workers take less time only under suitable assumptions: the amount of work stays fixed, each worker has the same rate, and they do not obstruct one another. Then workers × time is constant. A decreasing relationship alone is insufficient to prove inverse proportion.

Worked example 3 — workers and the remaining work

Six workers complete a job in 15 days. After they work for 5 days, four more equally productive workers join them. How many more days are needed?

The whole job requires $6\times15=90$ worker-days. One worker-day means the work done by one worker in one day at this rate.
The first five days complete $6\times5=30$ worker-days, leaving $90-30=60$ worker-days.
There are now $6+4=10$ workers. The remaining time is $60/10=6$ days.

Answer: 6 more days, or 11 days from the start. Check the total work: $6(5)+10(6)=90$ worker-days. Applying inverse proportion to the whole job after five days would count finished work twice.

Inverse powers and transformed quantities

If y is inversely proportional to x², write $y=k/x^2$ . If it is inversely proportional to $\sqrt{x}$ , write $y=k/\sqrt{x}$ , requiring $x>0$ . For a complete expression such as $x+2$ , keep it together: $y=k/(x+2)$ , with $x\ne-2$ . The denominator must never be zero.

Worked example 4 — an inverse square with a changed input

A quantity F is inversely proportional to the square of a positive distance r. F = 18 when r = 2. Find F when r = 3, then the percentage change in F when r increases by 20%.

Write $F=k/r^2$ . The known pair gives $18=k/4$ , so $k=72$ and $F=72/r^2$ . When $r=3$ , $F=72/9=8$ .

For the percentage part, the new distance is $1.2r$ . Compare new and old values before substituting a specific r:

$\frac{F_{\text{new}}}{F_{\text{old}}}=\frac{72/(1.2r)^2}{72/r^2}=\frac{1}{1.44}=\frac{25}{36}.$

The decrease is $(1-25/36)\times100\%=30.555\ldots\%$ , or 30.6% to 3 significant figures. It is not a 40% decrease: square the multiplicative factor, then take its reciprocal.

Check: increasing a positive distance reduces F in this model. Substitution at r = 2.4 gives $F=12.5$ , and $(18-12.5)/18=11/36$ , the same fractional decrease.

Explore the effect of changing x

Same starting pair, different models

Every model passes through $(2,12)$ . Change the relationship, then move x. The dark point is the fixed known pair; the red point is the new pair. The graph shows only positive x-values for inverse models.

The graph window shows 0 ≤ x ≤ 6 and 0 ≤ y ≤ 36. The explanation still reports a point that falls above the window.

Direct proportion: y = 6x. At x = 4, y = 24. x is multiplied by 2 and y is multiplied by 2, a 100% increase.

Common mistakes and checks

Missing the constant: replace $\propto$ with an equals sign and k, then find k.
Choosing a model from direction alone: test constant ratio for direct proportion and constant product for inverse proportion.
Ignoring a power: doubling x quadruples $kx^2$ , divides $k/x^2$ by four, and multiplies $k\sqrt{x}$ by $\sqrt2$ .
Mixing units: convert before finding or using k. Its numerical value depends on the units chosen.
Confusing “increases by” and “becomes”: an increase of 200% makes the input three times its original value.
Rounding too soon: keep the exact factor through the calculation; round the requested final answer.
Missing context restrictions: time and distance are non-negative, workers are whole numbers, and inverse denominators are non-zero.

Practise with our topic questions

Use the Sec 2 direct and inverse proportion question bank to apply these methods. Attempt each question before opening its worked solution. Compare your reasoning as well as the final answer.