R10 - Direct and inverse proportion

R10 - Direct and inverse proportion

Solve direct and inverse proportion problems by deciding what stays fixed. You will scale quantities, use tables and given equations, and read the same relationships from graphs.

Scaling quantities in direct proportion

Two identical pumps empty a tank in 1818 minutes. Would four pumps take 3636 minutes or 99 minutes? More pumps give less time. Before choosing a calculation, decide whether the quantities should move together or in opposite directions.

You already know

Equivalent ratios multiply both entries by the same number. A fixed ratio gives a straight line through the origin. You can use a unitary method: find the amount for one unit, then multiply.

Begin with a different situation: identical notebooks cost £3\pounds 3 each. Two cost £6\pounds 6, four cost £12\pounds 12, and six cost £18\pounds 18. Doubling the number doubles the cost. The cost per notebook stays fixed.

Direct proportion

Two quantities are in direct proportion when multiplying one by a factor multiplies the other by the same factor. For positive values, their ratio stays constant. For example, cost and number bought are directly proportional at a fixed unit price with no extra charge. A delivery fee would break that relationship.

The phrase “both increase” is not enough. A person's age and height can both increase without a constant ratio. Check a multiplier or unit rate, not just a direction.

Worked example 1: a whole-number scale factor

Question. 44 identical notebooks cost £12\pounds 12. There is no extra charge. Work out the cost of 1212 notebooks. Non-calculator.

Plan. Scale the number of notebooks by 33, then scale the cost by 33.

12÷4=3quantity multipliercost=12×3same multiplier=36evaluate\begin{aligned} 12\div4&=3 &&\text{quantity multiplier}\\ \text{cost}&=12\times3 &&\text{same multiplier}\\ &=36 &&\text{evaluate} \end{aligned}

Why use the same multiplier? Every notebook has the same price, so three times as many notebooks cost three times as much.

Answer. £36\pounds 36.

Check. Both purchases cost £3\pounds 3 per notebook: 12÷4=36÷12=312\div4=36\div12=3.

Change only the number required to 77. The scale factor is no longer a whole number, so going through one notebook is convenient.

Worked example 2: going through one unit

Question. 44 identical notebooks cost £12\pounds 12. There is no extra charge. Work out the cost of 77 notebooks. Non-calculator.

Plan. Divide by 44 to find one notebook's cost, then multiply by 77.

12÷4=3pounds per notebook3×7=21cost of seven\begin{aligned} 12\div4&=3 &&\text{pounds per notebook}\\ 3\times7&=21 &&\text{cost of seven} \end{aligned}

Why not add 33 to the cost because there are three more notebooks? The extra cost is three lots of the unit price, not three pounds simply because the count rose by three.

Answer. £21\pounds 21.

Check. Seven is between four and eight, so its cost lies between £12\pounds 12 and £24\pounds 24.

A rate stays fixed only under the conditions stated. Discounts, waste or a one-off fee may change the model.

Watch out

Wrong: 44 items cost £12\pounds 12, so 77 items cost £15\pounds 15. Adding the same amount preserves a difference, not a ratio. Right: 12÷4×7=2112\div4\times7=21.

Use the unitary method for a different pack size.

Scaling quantities in inverse proportion

Sharing a fixed job between more equal workers reduces the time, provided they can work independently at the same steady rate.

Suppose 33 machines finish a job in 2020 hours. The job needs 3×20=603\times20=60 machine-hours: one machine would take 6060 hours, while 66 machines would take 1010. The number of machines times the time stays fixed.

Inverse proportion

Two quantities are in inverse proportion when multiplying one by a positive factor divides the other by that factor. Their product stays constant. For example, the time for a fixed journey is inversely proportional to a constant speed. “One rises while the other falls” alone is not enough.

This is an ideal model. More people do not always finish a real job faster: they may get in each other's way. Use inverse proportion when the question states it or gives suitable equal-rate assumptions.

A fixed amount of work

For identical workers working independently at a constant rate on the same fixed job,

number of workers×time=constant amount of work\text{number of workers}\times\text{time}=\text{constant amount of work}

The same units of time must be used throughout.

Not on the formulae sheet: learn it.

First change the number of workers. The total work stays unchanged.

Worked example 3: finding a new time

Question. 55 identical printers print an order in 1212 minutes. They work independently at the same constant rate. Work out the time for 33 printers to print the same order. Non-calculator.

Plan. Find the printer-minutes needed, then divide between 33 printers.

5×12=60printer-minutes for the order60÷3=20minutes with three printers\begin{aligned} 5\times12&=60 &&\text{printer-minutes for the order}\\ 60\div3&=20 &&\text{minutes with three printers} \end{aligned}

Why multiply first? 5×125\times12 represents the whole fixed job. Dividing that work between fewer printers must increase the time.

Answer. 2020 minutes.

Check. 3×20=60=5×123\times20=60=5\times12.

Change only what is unknown. This time a deadline is given and you must find the number of printers.

Worked example 4: finding how many machines

Question. 55 identical printers print an order in 1212 minutes. They work independently at the same constant rate. How many printers are needed to print the same order in 1515 minutes? Non-calculator.

Plan. Divide the fixed 6060 printer-minutes by the allowed time.

5×12=60same order60÷15=4number of printers\begin{aligned} 5\times12&=60 &&\text{same order}\\ 60\div15&=4 &&\text{number of printers} \end{aligned}

Why divide by the time? Four printers each work for fifteen minutes, giving the required sixty printer-minutes.

Answer. 44 printers.

Check. 4×15=604\times15=60. The longer deadline allows fewer printers.

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