Skip to content

Proportions

Ratio, rates and money · level 5, lesson 1 of 7

A question to start

A car uses 7 litres of fuel to drive 100 km. How much fuel does it need for a trip of 260 km?

Three friends solve it.

  • Amir says: "100 km is 7 litres, so 200 km is 14 litres. 60 km is a bit more than half of 100, so... wait, I can do it exactly: 260 is 2.6 times 100, so the fuel is 2.6 times 7, which is 18.2 litres."
  • Bea says: "One kilometre uses \(7 \div 100 = 0.07\) litres. So 260 km uses \(260 \times 0.07 =\) 18.2 litres."
  • Chen writes \(\dfrac{x}{260} = \dfrac{7}{100}\), then \(100x = 7 \times 260 = 1820\), so \(x =\) 18.2 litres.

Three different methods, one answer. Is that luck? No. All three are the same idea seen from three sides. This lesson is about that idea, called a proportion, and about why each method works, especially Chen's, which looks like a magic trick until you see what is underneath.

The idea

What a proportion is

You already know that a ratio compares two amounts, like 7 litres for every 100 km, and that equivalent ratios describe the same comparison: 7 : 100 and 14 : 200 and 18.2 : 260 all mean "7 litres for every 100 km" (level 3).

A proportion is a statement that two ratios are equal. Written with fractions:

\[\frac{a}{b} = \frac{c}{d}\]

Read it as "\(a\) is to \(b\) as \(c\) is to \(d\)". For example, \(\dfrac{7}{100} = \dfrac{14}{200}\) is a true proportion, and \(\dfrac{3}{4} = \dfrac{5}{6}\) is a false one.

Most proportion problems give you three of the four numbers and ask for the fourth. The missing number is usually called \(x\). Before any method, there is one thing to check.

First check: is it really a proportion?

A proportion only fits when the two amounts grow together at a fixed rate: double one and the other doubles, and zero of one means zero of the other. Fuel and distance (at a steady speed) behave like that. Many things do not.

  • A taxi charges 4 dollars to start plus 2 dollars per km. A 3 km ride costs 10 dollars. A 6 km ride does not cost 20 dollars: it costs \(4 + 12 = 16\) dollars, because the 4 dollar start is paid only once.
  • A 12-year-old is 150 cm tall. A 24-year-old is not 300 cm tall.

So always ask first: "If I doubled this, would that double too?" If yes, carry on. You will look at this question much more closely in the next lesson.

Method 1: scaling

Put the numbers in a ratio table or on a double number line, with the matching amounts lined up.

L km 0 7 14 18.2 ? 0 100 200 260

Then find the number that takes you from the known row to the new row. Here 100 km becomes 260 km by multiplying by 2.6, so the litres must also be multiplied by 2.6: \(7 \times 2.6 = 18.2\). This number, 2.6, is called the scale factor (the "between" multiplier, because it goes between the two situations).

Scaling is fastest when the scale factor is a nice number: 3 notebooks to 12 notebooks is \(\times 4\), so the price is \(\times 4\) too. It gets clumsy when the scale factor is ugly, like going from 7 to 11.

Method 2: the unit rate

Find how much goes with one unit, then multiply. This is also called the unitary method. Bea found 0.07 litres for 1 km. That number does not change for the whole trip; it is the unit rate (level 3), the "within" multiplier, because it turns km into litres inside one situation:

\[\text{litres} = 0.07 \times \text{km}.\]

The unit rate is the most powerful of the three methods because, once you have it, every question about the same situation is one multiplication. 50 km? \(0.07 \times 50 = 3.5\) litres. 1000 km? 70 litres.

Method 3: cross multiplying, and why it works

Chen's method uses a rule:

\[\frac{a}{b} = \frac{c}{d} \quad \text{exactly when} \quad a \times d = b \times c.\]

The products \(a \times d\) and \(b \times c\) are called the cross products, because if you draw lines joining the numbers you multiply, they make a cross. The numbers \(b\) and \(d\) must not be zero (you can never divide by zero).

Why it works (reason 1: same denominator). To compare two fractions, give them the same denominator. Multiply the top and bottom of \(\dfrac{a}{b}\) by \(d\), and the top and bottom of \(\dfrac{c}{d}\) by \(b\):

\[\frac{a}{b} = \frac{a \times d}{b \times d}, \qquad \frac{c}{d} = \frac{b \times c}{b \times d}.\]

Now both fractions have the same bottom, \(b \times d\). Two fractions with the same bottom are equal exactly when their tops are equal. So the fractions are equal exactly when \(a \times d = b \times c\). That is the whole trick: cross multiplying is just "put both fractions over the same denominator and compare the tops", with the bottoms left out.

Why it works (reason 2: balancing). Start from \(\dfrac{a}{b} = \dfrac{c}{d}\). An equation stays true if you do the same thing to both sides. Multiply both sides by \(b\), then by \(d\):

\[\frac{a}{b} \times b \times d = \frac{c}{d} \times b \times d \quad\Longrightarrow\quad a \times d = c \times b.\]

On the left the \(b\)'s cancel; on the right the \(d\)'s cancel. Same answer, from algebra instead of fractions.

A quick check with numbers: \(\dfrac{7}{100} = \dfrac{14}{200}\), and \(7 \times 200 = 1400 = 100 \times 14\). ✓ And \(\dfrac{3}{4} \ne \dfrac{5}{6}\), because \(3 \times 6 = 18\) but \(4 \times 5 = 20\).

Setting up the proportion

The biggest danger is putting the numbers in the wrong places. A safe habit: write the units next to the numbers and keep the same unit in the same place on both sides.

\[\frac{\text{litres}}{\text{km}}: \qquad \frac{x}{260} = \frac{7}{100}\]

Litres on top on both sides, km underneath on both sides. You could also write \(\dfrac{x}{7} = \dfrac{260}{100}\) (trip compared with trip), which is just as correct, because it says the scale factor is the same for litres and km. What you must not write is \(\dfrac{x}{260} = \dfrac{100}{7}\), which mixes litres-per-km with km-per-litre.

Which method when?

Situation Good method
The scale factor is a whole number or a simple fraction Scaling
You will answer several questions about the same rate Unit rate
The numbers are awkward, or the unknown is in a strange place Cross multiplying

All three give the same answer, so one method can check another.

Worked examples

1. Notebooks, three ways. 5 notebooks cost 8 dollars. How much do 12 cost?

  • Unit rate: one notebook costs \(8 \div 5 = 1.60\) dollars, so 12 cost \(12 \times 1.60 =\) 19.20 dollars.
  • Scaling: 5 to 12 is \(\times 2.4\), and \(8 \times 2.4 = 19.2\).
  • Cross multiplying: \(\dfrac{x}{12} = \dfrac{8}{5}\) (dollars over notebooks), so \(5x = 96\) and \(x = 19.2\).

2. A bare proportion. Solve \(\dfrac{x}{12} = \dfrac{5}{8}\).

  • Cross products: \(8 \times x = 12 \times 5\), so \(8x = 60\).
  • Divide both sides by 8: \(x = 7.5\).
  • Check: \(7.5 \div 12 = 0.625\) and \(5 \div 8 = 0.625\). ✓

3. The unknown on the bottom. A printer prints 45 pages in 3 minutes. How long does it take to print 120 pages?

  • Minutes over pages: \(\dfrac{x}{120} = \dfrac{3}{45}\).
  • Cross multiply: \(45x = 360\), so \(x = 8\) minutes.
  • Unit rate check: 15 pages per minute, and \(120 \div 15 = 8\). ✓

4. Is it a proportion? Two orange drinks: one is 4 cups of juice to 10 cups of water, the other is 6 cups of juice to 15 cups of water. Do they taste the same?

  • They taste the same if \(\dfrac{4}{10} = \dfrac{6}{15}\).
  • Cross products: \(4 \times 15 = 60\) and \(10 \times 6 = 60\). Equal, so yes. (Both simplify to \(\dfrac{2}{5}\).)

5. An unknown on both sides. Solve \(\dfrac{x}{x + 4} = \dfrac{3}{5}\).

  • Cross multiply: \(5x = 3(x + 4) = 3x + 12\).
  • Subtract \(3x\): \(2x = 12\), so \(x = 6\).
  • Check: \(\dfrac{6}{10} = \dfrac{3}{5}\). ✓ (Brackets matter here: the whole of \(x + 4\) is multiplied by 3.)

Try it

  1. Solve \(\dfrac{x}{9} = \dfrac{4}{6}\).
  2. Solve \(\dfrac{3}{x} = \dfrac{12}{20}\).
  3. A recipe for 6 muffins needs 150 g of sugar. How much sugar for 15 muffins? Solve it two different ways.
  4. Is \(\dfrac{4}{10} = \dfrac{6}{15}\) a true proportion? Is \(\dfrac{5}{8} = \dfrac{7}{11}\)?
  5. A 4-litre can of paint covers 36 square metres. Paint is sold in 1-litre cans. How many cans do you need for a wall of 50 square metres?
  6. A taxi charges 4 dollars plus 2 dollars per km. Sam says: "3 km costs 10 dollars, so 6 km costs 20 dollars." Is he right? Explain.
  7. Solve \(\dfrac{x + 2}{5} = \dfrac{6}{10}\).
  8. Explain, without cross multiplying, why \(\dfrac{a}{b} = \dfrac{c}{d}\) also means \(\dfrac{a}{c} = \dfrac{b}{d}\). (Hint: what do the cross products of the second one say?)
Answers
  1. Cross products: \(6x = 36\), so \(x = 6\). Check: \(\dfrac{6}{9} = \dfrac{2}{3} = \dfrac{4}{6}\). ✓
  2. Cross products: \(12x = 60\), so \(x = 5\). Check: \(\dfrac{3}{5} = 0.6\) and \(\dfrac{12}{20} = 0.6\). ✓
  3. Unit rate: \(150 \div 6 = 25\) g per muffin, so \(15 \times 25 = 375\) g. Scaling: 6 to 15 is \(\times 2.5\), and \(150 \times 2.5 = 375\) g. Same answer.
  4. First: \(4 \times 15 = 60\) and \(10 \times 6 = 60\), so yes. Second: \(5 \times 11 = 55\) and \(8 \times 7 = 56\), so no (very close, but not equal).
  5. Litres over square metres: \(\dfrac{x}{50} = \dfrac{4}{36}\), so \(36x = 200\) and \(x = 5.55...\) litres. You cannot buy part of a can, so you need 6 cans (5 would leave part of the wall bare).
  6. No. 6 km costs \(4 + 2 \times 6 = 16\) dollars. The 4 dollar start is paid once, so the cost is not proportional to the distance: doubling the distance does not double the cost.
  7. Cross products: \(10(x + 2) = 30\), so \(x + 2 = 3\) and \(x = 1\). Check: \(\dfrac{3}{5} = \dfrac{6}{10}\). ✓
  8. The cross products of \(\dfrac{a}{c} = \dfrac{b}{d}\) are \(a \times d\) and \(c \times b\). Those are the same two products as for \(\dfrac{a}{b} = \dfrac{c}{d}\). So one proportion is true exactly when the other is. (In words: if the litres-per-km match, then the trip-to-trip scale factors match too.)

Common mistake

Scaling something that is not proportional. Before you set up a proportion, check that doubling one amount doubles the other and that zero goes with zero. Fixed fees (the taxi), starting amounts and things like age and height break the rule.

Mixing the order. \(\dfrac{x}{260} = \dfrac{100}{7}\) puts litres-per-km on one side and km-per-litre on the other. Write the units beside the numbers and make them match on both sides. A quick sense check also helps: 260 km is more than 100 km, so the answer must be more than 7 litres, not less.

Where it leads