R10, R13, R14 - Direct and inverse proportion

Some quantities change by the same factor, while others change by reciprocal factors. You will solve numerical proportion problems, interpret given equations and graphs, and use a straight line's gradient as a rate of change.

Direct proportion and constant unit rates

If four notebooks cost £6, must eight cost £12? Yes, if every notebook has the same price and there is no fixed fee or discount. But “both quantities increase” alone is not enough to guarantee that doubling rule.

You already know

Equivalent ratios use equal scale factors. A unit rate finds the amount for one item, then scales to the number needed.

The equal-scaling rule has a precise name.

Direct proportion

Two quantities are directly proportional if multiplying one by a factor multiplies the other by the same factor. Their output-to-input ratio is constant for non-zero inputs, and the graph is a straight line through the origin.

Zero input gives zero output in this model. The given equation $C=1.5n$ , where $C$ is cost in pounds and $n$ is notebook count, has constant unit rate £1.50. For example, $C=18$ gives $18=1.5n$ , so $n=12$ . This uses a supplied relationship. Direct-proportion unit-rate methods are not on the formula sheet: learn them.

Worked example 1: scaling a constant price

Question. Four identical notebooks cost £6 at a constant unit price. Find the cost of ten.

Plan. Find the price of one notebook, then multiply by ten.

\begin{aligned} \text{unit price}&=6\div4=1.50 && \text{divide by the item count}\\ \text{cost of ten}&=1.50\times10=15 && \text{scale the unit price} \end{aligned}

Why this step? The constant price means each item contributes equally to the total.

Answer. £15.

Check. $10/4=2.5$ and $15/6=2.5$ : both quantities scale equally.

Change only the requested notebook count.

Worked example 2: using a given proportional equation

Question. The same notebooks follow $C=1.5n$ . Find the cost of eight.

Plan. Substitute the new count into the supplied relationship.

\begin{aligned} n&=8 && \text{identify the input}\\ C&=1.5\times8=12 && \text{apply the unit rate} \end{aligned}

Why this step? Multiplication by the coefficient charges £1.50 for each notebook.

Answer. £12, so the opening doubling works under the stated conditions.

Check. Eight is twice four and £12 is twice £6.

Recipes scale in the same way, but every ingredient has to last. Checking only one stock can produce portions you cannot actually make.

Worked example 3: a limiting ingredient

Question. Five portions need $300\,\mathrm{ml}$ milk and $200\,\mathrm g$ flour. There are $1100\,\mathrm{ml}$ milk and $900\,\mathrm g$ flour. Find the greatest whole number of portions.

Plan. Find each ingredient's separate portion limit and satisfy both.

\begin{aligned} \text{milk per portion}&=300/5=60\,\mathrm{ml} && \text{find a unit amount}\\ \text{flour per portion}&=200/5=40\,\mathrm g && \text{find a unit amount}\\ \text{milk limit}&=1100/60=18\frac13 && \text{test milk stock}\\ \text{flour limit}&=900/40=22.5 && \text{test flour stock} \end{aligned}

Why this step? Both ingredients are required, so the smaller limit controls the feasible count.

Answer. 18 whole portions.

Check. Eighteen need $1080\,\mathrm{ml}$ milk and $720\,\mathrm g$ flour. Nineteen need $1140\,\mathrm{ml}$ milk, more than available.

Change only the available milk, leaving the recipe and flour stock unchanged.

Worked example 4: changing a resource limit

Question. The same recipe has $1200\,\mathrm{ml}$ milk and $900\,\mathrm g$ flour available. Find the greatest whole number of portions.

Plan. Use the same unit amounts to recalculate the limits.

\begin{aligned} \text{milk limit}&=1200/60=20 && \text{use the new milk stock}\\ \text{flour limit}&=900/40=22.5 && \text{unchanged flour stock}\\ \text{whole portions}&=20 && \text{choose the smaller limit} \end{aligned}

Why this step? Extra milk increases production only until another ingredient becomes the limiting resource.

Answer. 20 portions.

Check. They need $1200\,\mathrm{ml}$ milk and $800\,\mathrm g$ flour, both available.

On a cost graph, $(0,0)$ , $(4,6)$ , $(8,12)$ and $(10,15)$ lie on one line. A continuous line can show the model, but actual notebook purchases use whole-number inputs. A fixed delivery fee would keep the graph straight while breaking direct proportion.

Watch out

Wrong: 2 items cost £8 and 4 cost £14, so cost is directly proportional because both increase. Doubling 2 to 4 should double £8 to £16. Correct: reject direct proportion; $8/2=4$ differs from $14/4=3.5$ .

Change only the milk stock once more and test both ingredients.

Inverse proportion and constant products

For a fixed job shared by equally effective workers, doubling the workforce halves the time in an ideal model.

Inverse proportion

Two positive quantities are inversely proportional if multiplying one by a factor divides the other by that factor. Their product stays constant. Equivalently, the output is directly proportional to the reciprocal of the input, with input non-zero.

This model assumes a fixed job that can be shared, equal worker rates and no extra delay from adding workers. It will not describe every real workplace.

Worked example 5: a constant work total

Question. Three workers complete a fixed job in eight hours. Time is inversely proportional to worker number. Find the time for six workers.

Plan. Keep total worker-hours fixed, then divide by the new workforce.

\begin{aligned} \text{work total}&=3\times8=24\text{ worker-hours} && \text{find the constant product}\\ \text{time}&=24\div6=4\text{ hours} && \text{share the fixed work} \end{aligned}

Why this step? Doubling workers divides the time by two; it does not double the amount of work.

Answer. 4 hours.

Check. $6\times4=24=3\times8$ .

Change only the number of workers to five. A whole-number workforce need not give a whole-number time.

Worked example 6: a given reciprocal equation

Question. The same job follows the given equation $t=24/w$ , with time in hours. Find the time for five workers in hours and minutes.

Plan. Substitute $w=5$ , then convert the fractional hour into minutes.

\begin{aligned} t&=24/5=4.8\text{ hours} && \text{use the given relationship}\\ \text{extra minutes}&=0.8\times60=48 && \text{convert the hour fraction} \end{aligned}

Why this step? $0.8$ of an hour is $0.8$ of 60 minutes, rather than 80 minutes.

Answer. 4 hours 48 minutes.

Check. $5\times4.8=24$ ; fewer than six workers need more than four hours.

For the given rule $y=12/x$ , inputs $1,2,4$ have reciprocals $1,1/2,1/4$ , giving outputs $12,6,3$ . Each output is twelve times the reciprocal. At $x=3$ , $y=4$ and $xy=12$ . Plotting $y$ against $1/x$ would give a straight line through the origin. Zero workers or $x=0$ is excluded: division by zero is undefined. The constant-product and reciprocal interpretations are not on the formula sheet: learn them.

Watch out

Wrong: a decreasing relationship must be inverse proportion. For $y=20-x$ , $x=2$ gives $xy=36$ , while $x=4$ gives $xy=64$ . Correct: check a constant product, or use the inverse relationship stated in the question.

Keep the job and change just the workforce again.

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R10, R13, R14 - Direct and inverse proportion

Direct proportion and constant unit rates

Inverse proportion and constant products

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