5.02a - Direct proportion

5.02a - Direct proportion

Learn to scale quantities together, convert currencies and build an equation for a constant ratio. You will distinguish direct proportion from a relationship that merely increases at a steady rate.

One unit and many units

If you exchange twice as many pounds, must you receive twice as many euros? At a fixed exchange rate with no fee, you do. A fixed fee would change the answer: the conditions matter as much as the calculation.

Suppose three identical notebooks cost £7.20, with no discount for buying more. One notebook costs £2.40, so five cost £12. The unitary method means finding the value for one unit, then multiplying by the number of units wanted. You can also scale directly: six notebooks cost twice as much as three.

Direct proportion

Two quantities are in direct proportion when multiplying one by a factor multiplies the other by the same factor. For nonzero inputs, their ratio is constant. This assumes the same relationship applies throughout the values being considered.

For currency conversion, keep the units beside the amounts. “£1 buys €1.15” means multiply pounds by 1.15 to obtain euros. To travel back through the same conversion, divide euros by 1.15. An exchange rate in a question is a supplied mathematical condition, not a promise about today's rate.

Worked example 1: converting pounds to euros

Question. At a fixed rate, £120 buys €138. There are no fees. How many euros does £200 buy?

Plan. Find euros per pound, then multiply by 200.

138÷120=1.15euros per pound200×1.15=230scale to 200 pounds\begin{aligned} 138\div120&=1.15 &&\text{euros per pound}\\ 200\times1.15&=230 &&\text{scale to 200 pounds} \end{aligned}

Answer. €230.

Check. The pounds increased by a factor of 200/120=5/3200/120=5/3. Multiplying €138 by 5/35/3 also gives €230.

Keep the rate and change only the direction of conversion. Now the euro amount is known, so undo the multiplication.

Worked example 2: converting euros back to pounds

Question. At the same fixed rate, £120 buys €138, with no fees. How many pounds are equivalent to €184?

Plan. Divide by the number of euros bought by one pound.

138÷120=1.15euros per pound184÷1.15=160undo the conversion\begin{aligned} 138\div120&=1.15 &&\text{euros per pound}\\ 184\div1.15&=160 &&\text{undo the conversion} \end{aligned}

Answer. £160.

Check. £160 is 4/34/3 of £120; €184 is 4/34/3 of €138. Both quantities have the same scale factor.

A useful algebraic version is y1/x1=y2/x2y_1/x_1=y_2/x_2, where the two nonzero xx-values have matching yy-values. The subscripts simply label the first and second pairs. Equal ratios let you find a missing value without finding one unit explicitly.

Watch out

The calculation 184×1.15184\times1.15 uses the rate in the wrong direction. Write the units first: pounds ×1.15\times1.15 gives euros; euros ÷1.15\div1.15 gives pounds.

Use the direction of conversion to decide which operation is needed.

Recognising a constant ratio

Quantities can both increase without being directly proportional: you need the same multiplier between each input and its matching output.

For a model with input xx and output yy, divide each yy-value by its matching nonzero xx-value. Equal differences between outputs are not enough. For example, a delivery charge plus a price per item produces a steady increase, but the charge means the cost per item changes.

Worked example 3: testing matching pairs

Question. A table has xx-values 2, 4 and 7, with matching yy-values 6, 12 and 21. Are these pairs consistent with direct proportion?

Plan. Compare y/xy/x for all three pairs.

6÷2=3first pair12÷4=3second pair21÷7=3third pair\begin{aligned} 6\div2&=3 &&\text{first pair}\\ 12\div4&=3 &&\text{second pair}\\ 21\div7&=3 &&\text{third pair} \end{aligned}

Answer. Yes. All three pairs have the constant ratio 3 and fit y=3xy=3x.

Check. Multiplying every input by 3 reconstructs its output: 3×2=63\times2=6, 3×4=123\times4=12, 3×7=213\times7=21.

Change only the last output. One pair can break the common multiplier, even if the values still increase.

Worked example 4: one pair breaks the rule

Question. A table has xx-values 2, 4 and 7, with matching yy-values 6, 12 and 23. Are these pairs consistent with direct proportion?

Plan. Test whether the ratio from the first two pairs also works for the third.

6÷2=12÷4=3candidate multiplier23÷7=237≠3last ratio differs\begin{aligned} 6\div2=12\div4&=3 &&\text{candidate multiplier}\\ 23\div7&=\frac{23}{7}\ne3 &&\text{last ratio differs} \end{aligned}

Answer. No. There is no single ratio for all three pairs.

Check. The multiplier 3 would require y=3×7=21y=3\times7=21, not 23.

On axes, a direct-proportion model forms a straight line through the origin, (0,0)(0,0), whenever zero is included in its domain. The picture compares buying at £3 per item with paying £3 per item plus a £2 charge. You need no graph-drawing method here: notice what happens at zero items.

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