E2.8 - Direct and inverse proportion | Cambridge IGCSE Mathematics

The same change in an input can double, quadruple or halve an output, depending on the relationship. You will translate proportion statements into equations, find their constants and calculate unknown quantities. This Extended lesson includes proportion involving squares, cubes and roots.

Direct proportion and its constant

If you double an input, must its output double? A price charged per kilogram does double when the mass doubles, but a square's area becomes four times as large when its side doubles. “They both increase” does not tell you the whole relationship.

Start with a fixed price of $3$ dollars per kilogram. Masses of $2,4,6$ kilograms cost $6,12,18$ dollars. Cost divided by mass is always $3$ : this fixed ratio is what makes the relationship direct proportion.

Direct proportion

$y$ is directly proportional to $x$ , written $y\propto x$ , when

y=kx

for a fixed constant of proportionality $k$ . For $x\ne0$ , the ratio $y/x$ equals $k$ . If zero is included in the model's domain, $x=0$ gives $y=0$ .

The symbol $\propto$ states the type of relationship; it does not state that $k=1$ . Find $k$ from one known pair, then keep that same constant for the new pair. Units must remain consistent. In positive quantity models $k>0$ ; algebraic proportionality can also have a negative constant.

Worked example 1: finding a new output

Question. $y$ is directly proportional to $x$ . When $x=6$ , $y=18$ . Find $y$ when $x=10$ .

Plan. Write the equation, determine its constant, then substitute the new input.

\begin{aligned} y&=kx && \text{direct proportion}\\ 18&=6k\quad\Rightarrow\quad k=3 && \text{use the known pair}\\ y&=3(10)=30 && \text{keep the same constant} \end{aligned}

Answer. $y=30$ .

Check. The input changes by the factor $10/6=5/3$ ; the output changes by $30/18=5/3$ too.

Keep the model and known pair, but change which quantity is missing. You now solve the equation instead of evaluating its right-hand side.

Worked example 2: finding the input

Question. $y$ is directly proportional to $x$ . When $x=6$ , $y=18$ . Find $x$ when $y=42$ .

Plan. Find the same constant, then undo multiplication by it.

\begin{aligned} y&=kx,\qquad k=18\div6=3 && \text{known pair fixes the model}\\ 42&=3x && \text{use the new output}\\ x&=14 && \text{divide both sides by three} \end{aligned}

Answer. $x=14$ .

Check. $42/14=3$ , matching the original ratio $18/6=3$ .

Watch out

An increasing relationship need not be direct proportion. For $y=3x+2$ , the pairs $(1,5)$ and $(2,8)$ have different ratios, so $y\not\propto x$ . The fixed extra $2$ prevents direct proportion.

Inverse proportion

A fixed task takes fewer hours when more equally productive workers share it. In an ideal model, doubling the number of workers halves the time. The product “workers times hours” stays fixed. Real tasks need not obey that model: it is the stated relationship that licenses the calculation.

Inverse proportion

$y$ is inversely proportional to $x$ , written $y\propto 1/x$ , when

y=\frac{k}{x}\qquad(x\ne0)

for a fixed constant $k$ . Equivalently $xy=k$ . For a non-zero constant, multiplying $x$ by a non-zero factor divides $y$ by that factor, within the model's domain.

To find $k$ , multiply the known pair. To find an unknown input, rearrange $xy=k$ . For positive quantities use positive inputs; zero cannot be used in the denominator.

Worked example 3: a fixed product

Question. $y$ is inversely proportional to $x$ . When $x=6$ , $y=18$ . Find $y$ when $x=9$ .

Plan. Find the constant product, then divide it by the new input.

\begin{aligned} y&=\frac{k}{x} && \text{inverse proportion}\\ k&=6(18)=108 && \text{multiply the known pair}\\ y&=\frac{108}{9}=12 && \text{use the new input} \end{aligned}

Answer. $y=12$ .

Check. $9(12)=108=6(18)$ . Increasing $x$ by $3/2$ divides $y$ by $3/2$ .

Change only the missing quantity. The product method works whichever variable you need.

Worked example 4: finding an inverse input

Question. $y$ is inversely proportional to $x$ . When $x=6$ , $y=18$ . Find $x$ when $y=27$ .

Plan. Use the known product and solve for the new input.

\begin{aligned} xy&=108 && \text{constant from }6(18)\\ 27x&=108 && \text{substitute the new }y\\ x&=4 && \text{divide by twenty-seven} \end{aligned}

Answer. $x=4$ .

Check. $4(27)=108$ . The output is $3/2$ times its original value, while the input is $2/3$ times its original value.

Merely decreasing is not enough. The rule $y=10-x$ decreases, but gives products $1(9)=9$ and $2(8)=16$ , so it is not inverse proportion.

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E2.8 - Direct and inverse proportion

Direct proportion and its constant

Inverse proportion

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