Direct Proportion

What it means for one quantity to be directly proportional to another, how to test a table of values, and how to find an unknown value by the unitary method and by the ratio method.

Before you start

A rate compares two different quantities: a fine of 15 cents per day, rice at $2.80 per kilogram.
The unitary method finds the value of one unit first. If 4 pens cost $6, then 1 pen costs $6 ÷ 4 = $1.50, and 7 pens cost 7 × $1.50 = $10.50.

What Direct Proportion Means

A public library charges a fine of 15 cents for each day a book is overdue. The table shows the fine, $y$ cents, for a book that is $x$ days overdue. The last row divides each fine by its number of days.

Days, $x$	$1$	$2$	$3$	$4$	$5$	$10$
Fine, $y$ cents	$15$	$30$	$45$	$60$	$75$	$150$
$\dfrac{y}{x}$	$15$	$15$	$15$	$15$	$15$	$15$

For every pair, $\frac{y}{x} = 15$ : $\frac{15}{1} = 15$ , $\frac{30}{2} = 15$ , and so on to $\frac{150}{10} = 15$ . The fine is always 15 times the number of days. This is what makes the fine directly proportional to the number of days.

Direct proportion

$y$ is directly proportional to $x$ if $\frac{y}{x}$ is the same for every pair of values of $x$ and $y$ . We write $y \propto x$ (" $y$ varies directly as $x$ ").

The value of $\frac{y}{x}$ is called the constant of proportionality, $k$ . For the library fines, $k = 15$ cents per day.

Because $y$ is always $k$ times $x$ , multiplying $x$ by a number multiplies $y$ by the same number:

• when $x$ is doubled, $y$ is doubled: 2 days cost 30 cents, twice the 15 cents for 1 day;

• when $x$ is multiplied by 10, so is $y$ : 10 days cost 150 cents, 10 times 15 cents;

• when $x$ is halved, $y$ is halved: 5 days cost 75 cents, half of the 150 cents for 10 days.

These follow from the definition. But "when one goes up, the other goes up" is not the definition: two quantities can both increase without being in proportion, as Example 1(b) shows.

Example 1: Deciding from a table

Is $y$ directly proportional to $x$ in each table?
(a) Rice: mass $x$ kg and cost $ $y$ .
(b) A taxi ride: distance $x$ km and fare $ $y$ . The fare is $4 at the start and $0.60 for each kilometre.

Table (a)

$x$ (kg)	$2$	$5$	$8$
$y$ ($)	$5.60$	$14.00$	$22.40$

Table (b)

$x$ (km)	$1$	$2$	$3$	$5$
$y$ ($)	$4.60$	$5.20$	$5.80$	$7.00$

Solution

(a) Work out $\frac{y}{x}$ for every pair:

$x$ (kg)	$2$	$5$	$8$
$y$ ($)	$5.60$	$14.00$	$22.40$
$\dfrac{y}{x}$	$2.8$	$2.8$	$2.8$

$\frac{y}{x} = 2.8$ for every pair, so the cost is directly proportional to the mass, with $k = 2.8$ : the rice costs $2.80 per kilogram.

(b) Work out $\frac{y}{x}$ for every pair:

$x$ (km)	$1$	$2$	$3$	$5$
$y$ ($)	$4.60$	$5.20$	$5.80$	$7.00$
$\dfrac{y}{x}$	$4.6$	$2.6$	$1.93$	$1.4$

The values of $\frac{y}{x}$ are not the same ( $\frac{5.80}{3} \approx 1.93$ to 2 decimal places), so the fare is not directly proportional to the distance.

The fare does go up as the distance goes up. But doubling the distance from 1 km to 2 km takes the fare from $4.60 to $5.20, not to $9.20. The $4 charged at the start, whatever the distance, is the reason.

(a) Yes: $\frac{y}{x} = 2.8$ for every pair, so $y \propto x$ . (b) No: $\frac{y}{x}$ is not the same for every pair.

The Graph of a Direct Proportion

Plot the library fines with the number of days across and the fine up, and add the point $(0, 0)$ : a book returned on time, 0 days overdue, has no fine.

The graph of a direct proportion

When $y \propto x$ , the points lie on a straight line that passes through the origin, $(0, 0)$ .

Both are needed. The taxi fares of Example 1(b) also lie on a straight line, but it meets the fare axis at $4, not at 0, and the fare is not proportional to the distance.

Lesson 6.2 writes the line as an equation, $y = 15x$ , and uses the graph to find $k$ .

Finding an Unknown Value

When two quantities are directly proportional, three known values give the fourth. There are two methods, and both give the same answer.

Two methods

The unitary method. Find the value for one unit, then multiply by the number of units you want.

The ratio method. Since $\frac{y}{x}$ is the same for every pair, $\frac{y_1}{x_1} = \frac{y_2}{x_2}$ . Put in the three values you know and solve for the fourth.

Example 2: Finding the cost

6 kg of biscuits cost $27. The cost is directly proportional to the mass. Find the cost of 13 kg of biscuits.

Solution

Method 1: the unitary method.

6 kg cost $27, so 1 kg costs $27 ÷ 6 = $4.50.

13 kg cost 13 × $4.50 = $58.50.

Method 2: the ratio method. Let 13 kg cost $ $y$ . Cost divided by mass is the same for both pairs:

$\dfrac{y}{13} = \dfrac{27}{6}$

$y = \dfrac{27 \times 13}{6} = \dfrac{351}{6} = 58.5$

Both methods give $58.50.

13 kg of biscuits cost $58.50.

Example 3: Finding the mass

The biscuits of Example 2 cost $4.50 per kilogram. How many kilograms can be bought for $40.50?

Solution

Unitary method. Each kilogram costs $4.50, so the number of kilograms is 40.50 ÷ 4.50 = 9.

Ratio method. Let the mass be $x$ kg. Cost divided by mass is the same as for 6 kg and $27:

$\dfrac{40.5}{x} = \dfrac{27}{6}$

$27x = 6 \times 40.5 = 243\text{,}$ so $x = \frac{243}{27} = 9$ .

Check: 9 × $4.50 = $40.50.

9 kg of biscuits can be bought for $40.50.

Example 4: Small units, and rounding to the nearest 5 cents

50 g of sweets cost $2.10. Find the cost of 380 g of the same sweets, correct to the nearest 5 cents.

Solution

Unitary method. 1 g costs $2.10 ÷ 50 = $0.042, so 380 g cost 380 × $0.042 = $15.96.

Ratio method. Let 380 g cost $ $y$ : $\frac{y}{380} = \frac{2.10}{50}$ , so $y = \frac{2.10 \times 380}{50} = 15.96\text{.}$

To the nearest 5 cents: $15.96 lies between $15.95 and $16.00. It is 1 cent from $15.95 and 4 cents from $16.00, so it is nearer $15.95.

380 g of sweets cost $15.95, correct to the nearest 5 cents.

Example 5: A fraction of a whole

$\frac{3}{4}$ of a piece of metal has a mass of 15 kg. What is the mass of $\frac{2}{5}$ of the piece?

Solution

The mass of a part is directly proportional to the fraction of the piece it is.

Unitary method, with the whole piece as the unit: $\frac{3}{4}$ of the piece is 15 kg, so the whole piece is $15 \div \frac{3}{4} = 15 \times \frac{4}{3} = 20$ kg.

Then $\frac{2}{5}$ of the piece is $\frac{2}{5} \times 20 = 8$ kg.

Ratio method. Mass ÷ fraction is the same for both parts. Let the mass be $m$ kg: $m \div \frac{2}{5} = 15 \div \frac{3}{4} = 20\text{,}$ so $m = 20 \times \frac{2}{5} = 8$ .

$\frac{2}{5}$ of the piece has a mass of 8 kg.

Example 6: A recipe and an exchange rate

(a) A recipe for 4 people uses 250 g of flour. How much flour is needed for 10 people?
(b) On one day, S$1 = RM3.40 (Malaysian ringgit). How many ringgit does S$45 buy?

Solution

(a) The flour is directly proportional to the number of people. For 1 person: 250 g ÷ 4 = 62.5 g. For 10 people: 10 × 62.5 g = 625 g.

Or by multipliers: 10 people is $\frac{10}{4} = 2.5$ times as many, so the flour is 2.5 × 250 g = 625 g.

(b) The ringgit received are directly proportional to the dollars changed, with $k = 3.40$ ringgit per dollar. S$45 buys 45 × RM3.40 = RM153.

(a) 625 g of flour. (b) RM153.

Saying what you assume

A question in words often says "assuming that the cost is directly proportional to the mass" or "at the same rate". Real prices are not always proportional (a large bag of rice may cost less per kilogram), so the method works only when the question says, or the setting makes clear, that the rate is the same.

Common Mistakes

"Both increase, so it is direct proportion." The taxi fare goes up with the distance but is not directly proportional to it. Work out $\frac{y}{x}$ .
Checking $\frac{y}{x}$ for two pairs only. For $x = 2, 4, 6$ and $y = 6, 12, 20$ , the first two pairs give $3$ but the third gives $\frac{20}{6} \approx 3.33$ . Check every pair.
Calling any straight-line graph a direct proportion. The line must also pass through the origin.
Mixing the order in the ratio method. Write the same quantity on top on both sides: $\frac{\text{cost}}{\text{mass}} = \frac{\text{cost}}{\text{mass}}\text{.}$
Losing the units. In "Fine, $y$ cents", $y = 150$ means 150 cents, which is $1.50, not $150.

Practice Questions

Show the test or the method you use, and give each answer with its unit. The questions get harder as you go.

Question 1

Is $y$ directly proportional to $x$ ? If it is, give the constant of proportionality.

$x$	$3$	$5$	$8$	$12$
$y$	$7.5$	$12.5$	$20$	$30$

ShowHide solution

$x$	$3$	$5$	$8$	$12$
$y$	$7.5$	$12.5$	$20$	$30$
$\dfrac{y}{x}$	$2.5$	$2.5$	$2.5$	$2.5$

$\frac{y}{x} = 2.5$ for every pair.

Yes, $y \propto x$ , with $k = 2.5$ .

Question 2

Is $y$ directly proportional to $x$ ? Give a reason.

$x$	$3$	$5$	$8$	$10$
$y$	$12$	$20$	$30$	$40$

ShowHide solution

$x$	$3$	$5$	$8$	$10$
$y$	$12$	$20$	$30$	$40$
$\dfrac{y}{x}$	$4$	$4$	$3.75$	$4$

Three pairs give $4$ , but $\frac{30}{8} = 3.75$ . One pair that disagrees is enough.

No: $\frac{y}{x}$ is not the same for every pair (the pair $x = 8$ , $y = 30$ gives $3.75$ ).

Question 3

$y \propto x$ , and $y = 18$ when $x = 6$ . Without working out $k$ , find $y$ when (a) $x = 12$ , (b) $x = 3$ , (c) $x = 30$ .

ShowHide solution

(a) $12 = 6 \times 2$ : $x$ is doubled, so $y$ is doubled. $y = 18 \times 2 = 36$ .

(b) $3 = 6 \div 2$ : $x$ is halved, so $y = 18 \div 2 = 9$ .

Check with $k = \frac{18}{6} = 3$ : $\frac{36}{12}$ , $\frac{9}{3}$ and $\frac{90}{30}$ are all $3$ .

(a) $y = 36$ (b) $y = 9$ (c) $y = 90$

Question 4

7 identical notebooks cost $9.80. Find the cost of 12 of these notebooks (a) by the unitary method, (b) by the ratio method.

ShowHide solution

(a) 1 notebook costs $9.80 ÷ 7 = $1.40, so 12 cost 12 × $1.40 = $16.80.

(b) Let 12 cost $ $y$ : $\frac{y}{12} = \frac{9.80}{7}$ , so $y = \frac{9.80 \times 12}{7} = \frac{117.6}{7} = 16.8\text{.}$

12 notebooks cost $16.80.

Question 5

A car uses 6 litres of petrol to travel 75 km. Assuming that the distance it travels is directly proportional to the petrol it uses, how far can it travel on 26 litres?

ShowHide solution

On 1 litre it travels 75 ÷ 6 = 12.5 km.

On 26 litres it travels 26 × 12.5 = 325 km.

Ratio method: $\frac{d}{26} = \frac{75}{6}$ , so $d = \frac{75 \times 26}{6} = \frac{1950}{6} = 325\text{.}$

The car can travel 325 km on 26 litres.

Question 6

300 g of grapes cost $4.85. Find the cost of 700 g of the same grapes, correct to the nearest 5 cents.

ShowHide solution

Let 700 g cost $ $y$ : $\frac{y}{700} = \frac{4.85}{300}\text{,}$ so $y = \frac{4.85 \times 700}{300} = \frac{3395}{300} \approx 11.317\text{.}$

So the cost is about $11.32. It lies between $11.30 and $11.35: it is about 1.7 cents from $11.30 and 3.3 cents from $11.35, so it is nearer $11.30.

700 g of grapes cost $11.30, correct to the nearest 5 cents.

Question 7

When $\frac{2}{3}$ full, a tank holds 48 litres of water. How much water does it hold when it is $\frac{5}{8}$ full?

ShowHide solution

The full tank holds $48 \div \frac{2}{3} = 48 \times \frac{3}{2} = 72$ litres.

$\frac{5}{8}$ of the tank holds $\frac{5}{8} \times 72 = 45$ litres.

45 litres

Question 8

On one day, S$1 = RM3.40. (a) How many ringgit does S$120 buy? (b) How many Singapore dollars does RM85 buy?

ShowHide solution

(a) 120 × RM3.40 = RM408.

(b) Each Singapore dollar buys RM3.40, so RM85 buys S$(85 ÷ 3.40) = S$25.

Check (b): 25 × RM3.40 = RM85.

(a) RM408 (b) S$25

Question 9

Ravi is paid the same amount for each hour he works. For 15 hours he is paid $180. (a) How much is he paid for 22 hours? (b) How many hours must he work to be paid $330?

ShowHide solution

For 1 hour he is paid $180 ÷ 15 = $12.

(a) For 22 hours: 22 × $12 = $264.

(b) $330 ÷ $12 = 27.5, so he must work 27.5 hours.

(a) $264 (b) 27.5 hours

Question 10

A recipe for 6 people uses 450 g of flour and 3 eggs. How much flour and how many eggs are needed for 10 people?

ShowHide solution

Flour: for 1 person, 450 g ÷ 6 = 75 g; for 10 people, 10 × 75 g = 750 g.

Eggs: for 1 person, 3 ÷ 6 = 0.5 egg; for 10 people, 10 × 0.5 = 5 eggs.

750 g of flour and 5 eggs

Question 11

Aisha has the table below. She says: " $\frac{18}{2} = 9$ and $\frac{36}{4} = 9$ , so $y$ is directly proportional to $x$ ." Is her reasoning complete? Is her conclusion right?

Tickets, $x$	$2$	$4$	$5$
Cost, $ $y$	$18$	$36$	$45$

ShowHide solution

Her reasoning is not complete: two pairs that agree do not show that every pair agrees.

Check the third pair: $\frac{45}{5} = 9$ . Now every pair gives $9$ .

Her reasoning is incomplete, because she checked only two pairs; her conclusion is right, because the third pair also gives $\frac{y}{x} = 9$ .

Question 12

The table gives the mass, $y$ g, of $x$ identical coins. Show that $y \propto x$ , and draw the graph of $y$ against $x$ . How does the graph show the direct proportion?

Coins, $x$	$2$	$4$	$6$	$8$
Mass, $y$ g	$13$	$26$	$39$	$52$

ShowHide solution

Coins, $x$	$2$	$4$	$6$	$8$
Mass, $y$ g	$13$	$26$	$39$	$52$
$\dfrac{y}{x}$	$6.5$	$6.5$	$6.5$	$6.5$

$\frac{y}{x} = 6.5$ for every pair, so $y \propto x$ : each coin has a mass of 6.5 g.

The points lie on a straight line, and the line passes through the origin: 0 coins have a mass of 0 g.

$\frac{y}{x} = 6.5$ for every pair. The graph is a straight line through the origin, which shows that $y \propto x$ .

Key Points

$y$ is directly proportional to $x$ ( $y \propto x$ ) if $\frac{y}{x}$ is the same for every pair of values; that value is the constant of proportionality, $k$ .
To test a table, work out $\frac{y}{x}$ for every pair, not just two.
When $y \propto x$ , multiplying $x$ by a number multiplies $y$ by the same number.
"Both increase" is not enough for direct proportion: a fixed charge (as in a taxi fare) spoils it.
The graph of a direct proportion is a straight line through the origin; both are needed.
To find an unknown value, use the unitary method (the value for one unit first) or the ratio method ( $\frac{y_1}{x_1} = \frac{y_2}{x_2}$ ).