Ratios of Three Quantities

What a ratio of three quantities such as 5 : 2 : 3 says, how to write it in its simplest form, and how to read two-term ratios, fractions of the whole and missing terms from it.

Before you start

A ratio compares quantities of the same kind, and it has no unit of measurement. Multiplying or dividing every term by the same non-zero number gives an equivalent ratio. A ratio is in its simplest form when its terms are whole numbers with no common factor other than 1: divide the terms by their highest common factor (HCF), as in $12 : 18 = 2 : 3$ (dividing by 6).

Decimals in a ratio: multiply every term by 10, 100 or 1000 to make the terms whole numbers, as in $0.6 : 1.5 = 6 : 15 = 2 : 5$ .

Fractions in a ratio: multiply every term by the lowest common multiple (LCM) of the denominators, as in $\frac{1}{2} : \frac{2}{3} = 3 : 4$ (multiplying by 6). A mixed number is changed to an improper fraction first.

Different units of measurement: change the quantities to the same unit of measurement before writing the ratio. The ratio of 40 cm to 2 m is $40 : 200 = 1 : 5$ .

What a : b : c Means

A ratio can compare three quantities at once. A fruit basket holds 5 pears, 2 plums and 3 peaches. The ratio of the number of pears to the number of plums to the number of peaches is $5 : 2 : 3$ , read "5 to 2 to 3".

The bar model has one bar for each quantity, and every bar is made of units of the same size. The one ratio $5 : 2 : 3$ makes three comparisons at once: pears to plums is $5 : 2$ , plums to peaches is $2 : 3$ , and pears to peaches is $5 : 3$ .

Definition

The ratio of three quantities of the same kind is written $a : b : c$ and read " $a$ to $b$ to $c$ ". The numbers $a$ , $b$ and $c$ are the terms of the ratio.

The terms are written in the order in which the quantities are named: $5 : 2 : 3$ and $3 : 2 : 5$ are different ratios.

Like every ratio, $a : b : c$ has no unit of measurement.

The ratio of two quantities can be described by one fraction (the number of plums is $\frac{2}{5}$ of the number of pears). The ratio of three quantities cannot be described by one fraction.

Example 1: Writing a ratio of three quantities

At the school sports day, Red House won 7 gold medals, 4 silver medals and 9 bronze medals. Write the ratio of
(a) the number of gold medals to the number of silver medals to the number of bronze medals,
(b) the number of bronze medals to the number of gold medals to the number of silver medals,
(c) the number of silver medals to the number of gold medals to the total number of medals.

Solution

(a) Write the numbers in the order of the words: gold, silver, bronze.

(b) The order of the words is now bronze, gold, silver, so the same three numbers are written in a different order.

(a) Gold : silver : bronze $= 7 : 4 : 9$
(b) Bronze : gold : silver $= 9 : 7 : 4$
(c) Silver : gold : total $= 4 : 7 : 20$

Equivalent Ratios and Simplest Form

A larger basket holds 10 pears, 4 plums and 6 peaches, so its ratio is $10 : 4 : 6$ . Put the fruit in twos and let 1 unit stand for 2 fruits. The bars are again 5 units, 2 units and 3 units long.

So $10 : 4 : 6 = 5 : 2 : 3$ . The two baskets have the same mix of fruit, and the two ratios are equivalent ratios. As with two terms, multiplying or dividing every term by the same non-zero number gives an equivalent ratio:
$5 : 2 : 3 = 5 \times 2 : 2 \times 2 : 3 \times 2 = 10 : 4 : 6$
$10 : 4 : 6 = 10 \div 2 : 4 \div 2 : 6 \div 2 = 5 : 2 : 3$

Simplest form

A ratio $a : b : c$ is in its simplest form when its terms are whole numbers and the only factor common to all three terms is 1.

To write a ratio of three whole numbers in its simplest form, divide every term by the HCF of all three terms.

A factor of only two of the terms does not count

Look at $6 : 9 : 10$ . The number 3 is a factor of 6 and of 9, but not of 10. The number 2 is a factor of 6 and of 10, but not of 9. The only number that divides all three terms is 1, so $6 : 9 : 10$ is already in its simplest form.
Dividing only the 6 and the 9 by 3 would give $2 : 3 : 10$ , and that is a different ratio. In $6 : 9 : 10$ the first quantity to the third is $6 : 10 = 3 : 5$ . In $2 : 3 : 10$ it is $2 : 10 = 1 : 5$ .

Example 2: Simplest form by the HCF

Write each ratio in its simplest form.
(a) $18 : 24 : 6$
(b) $36 : 60 : 84$
(c) $10 : 15 : 12$

Solution

(a) The third term, 6, is a factor of 18 and of 24, so the HCF of 18, 24 and 6 is 6. Divide every term by 6: $18 \div 6 : 24 \div 6 : 6 \div 6 = 3 : 4 : 1$

(b) By prime factorisation, $36 = 2^{2} \times 3^{2}$ , $60 = 2^{2} \times 3 \times 5$ and $84 = 2^{2} \times 3 \times 7$ . The HCF is $2^{2} \times 3 = 12$ . Divide every term by 12: $36 \div 12 : 60 \div 12 : 84 \div 12 = 3 : 5 : 7$

Or divide by one common factor at a time until none is left: $36 : 60 : 84 = 18 : 30 : 42$ (dividing by 2) $= 9 : 15 : 21$ (dividing by 2 again) $= 3 : 5 : 7$ (dividing by 3).

(c) 5 is a factor of 10 and of 15, but not of 12. 2 is a factor of 10 and of 12, but not of 15. 3 is a factor of 15 and of 12, but not of 10. No number other than 1 divides all three terms.

(a) $18 : 24 : 6 = 3 : 4 : 1$
(b) $36 : 60 : 84 = 3 : 5 : 7$
(c) $10 : 15 : 12$ is already in its simplest form.

Example 3: A ratio from a class

In a class of 40 students, 12 walk to school, 18 come by bus and the rest come by MRT. Find the ratio of the number of students who walk to the number who come by bus to the number who come by MRT, in its simplest form.

Solution

Number of students who come by MRT $= 40 - 12 - 18 = 10$

Walk : bus : MRT $= 12 : 18 : 10$

The HCF of 12, 18 and 10 is 2. (6 is a common factor of 12 and 18, but 6 is not a factor of 10.)

$12 \div 2 : 18 \div 2 : 10 \div 2 = 6 : 9 : 5$

Check: 6 and 9 have the common factor 3, but 3 is not a factor of 5, so $6 : 9 : 5$ is in its simplest form.

The ratio of the number of students who walk to the number who come by bus to the number who come by MRT is $6 : 9 : 5$ .

Ratios with Decimals and Fractions

The terms of a ratio of three quantities may be decimals, fractions or mixed numbers. Such a ratio is not yet in its simplest form, because the terms of a ratio in its simplest form are whole numbers. The method is the one used for two terms, carried out on all three: first make every term a whole number, then divide by the HCF.

Method

Make every term a whole number. Decimals: multiply every term by 10, 100 or 1000; the term with the most decimal places decides which. Fractions: change every mixed number to an improper fraction, then multiply every term by the LCM of all the denominators.
Divide every term by the HCF of the three whole numbers.

A term that is already a whole number is multiplied too. The ratio does not change, because every term is multiplied by the same number. If a ratio has both a decimal and a fraction, write every term the same way first, as in the lesson on ratios with fractions and decimals; changing the decimal to a fraction always works.

Example 4: Decimals

Write each ratio in its simplest form.
(a) $0.6 : 3 : 1.2$
(b) $0.45 : 1.2 : 0.75$

Solution

(a) The decimals have 1 decimal place, so multiply every term by 10: $0.6 \times 10 : 3 \times 10 : 1.2 \times 10 = 6 : 30 : 12$

The HCF of 6, 30 and 12 is 6: $6 \div 6 : 30 \div 6 : 12 \div 6 = 1 : 5 : 2$

(b) 0.45 and 0.75 have 2 decimal places, so multiply every term by 100: $0.45 \times 100 : 1.2 \times 100 : 0.75 \times 100 = 45 : 120 : 75$

$45 = 3^{2} \times 5$ , $120 = 2^{3} \times 3 \times 5$ and $75 = 3 \times 5^{2}$ , so the HCF is $3 \times 5 = 15$ : $45 \div 15 : 120 \div 15 : 75 \div 15 = 3 : 8 : 5$

(a) $0.6 : 3 : 1.2 = 1 : 5 : 2$
(b) $0.45 : 1.2 : 0.75 = 3 : 8 : 5$

Example 5: Fractions and mixed numbers

Write each ratio in its simplest form.
(a) $\frac{1}{2} : \frac{2}{3} : \frac{3}{4}$
(b) $\frac{2}{3} : \frac{4}{9} : 2$
(c) $2\frac{1}{4} : 1\frac{2}{3} : 3\frac{5}{6}$

Solution

(a) The denominators are 2, 3 and 4, and their LCM is 12. Multiply every term by 12: $\frac{1}{2} \times 12 : \frac{2}{3} \times 12 : \frac{3}{4} \times 12 = 6 : 8 : 9$

No number other than 1 divides 6, 8 and 9, so this is the simplest form. (Multiplying by 6, the LCM of 2 and 3 only, would leave the third term as $\frac{3}{4} \times 6 = 4\frac{1}{2}$ , which is not a whole number. That is why the LCM of all the denominators is used.)

(b) The denominators are 3 and 9, and their LCM is 9. Multiply every term by 9, the whole number 2 as well: $\frac{2}{3} \times 9 : \frac{4}{9} \times 9 : 2 \times 9 = 6 : 4 : 18$

The terms are now whole numbers, but they still have the common factor 2: $6 \div 2 : 4 \div 2 : 18 \div 2 = 3 : 2 : 9$

(c) Change the mixed numbers to improper fractions: $2\frac{1}{4} : 1\frac{2}{3} : 3\frac{5}{6} = \frac{9}{4} : \frac{5}{3} : \frac{23}{6}$

The LCM of 4, 3 and 6 is 12. Multiply every term by 12: $\frac{9}{4} \times 12 : \frac{5}{3} \times 12 : \frac{23}{6} \times 12 = 27 : 20 : 46$

The only prime factor of 27 is 3, and 3 is not a factor of 20, so no number other than 1 divides all three terms.

(a) $\frac{1}{2} : \frac{2}{3} : \frac{3}{4} = 6 : 8 : 9$
(b) $\frac{2}{3} : \frac{4}{9} : 2 = 3 : 2 : 9$
(c) $2\frac{1}{4} : 1\frac{2}{3} : 3\frac{5}{6} = 27 : 20 : 46$

Quantities in Different Units of Measurement

A ratio compares quantities measured in the same unit of measurement. When three quantities are given in different units of measurement, change all of them to one unit of measurement first. Choosing a smaller unit of measurement usually makes every number a whole number. Then leave out the unit of measurement and write the ratio in its simplest form.

Example 6: Three lengths

Three ribbons are 50 cm, 1.2 m and 800 mm long. Find the ratio of their lengths, in that order, in its simplest form.

Solution

Change every length to centimetres: 1.2 m $= 1.2 \times 100$ cm $= 120$ cm, and 800 mm $= 800 \div 10$ cm $= 80$ cm.

The lengths are 50 cm, 120 cm and 80 cm, so the ratio is $50 : 120 : 80$ .

The HCF of 50, 120 and 80 is 10: $50 \div 10 : 120 \div 10 : 80 \div 10 = 5 : 12 : 8$

In millimetres the lengths are 500 mm, 1200 mm and 800 mm, and $500 : 1200 : 800 = 5 : 12 : 8$ as well (dividing by 100). Any one unit of measurement gives the same ratio.

Writing $50 : 1.2 : 800$ from the numbers as they stand would be wrong. It would say that the third ribbon is 16 times as long as the first, when it is less than twice as long.

The ratio of the lengths of the three ribbons is $5 : 12 : 8$ .

The Two-Term Ratios Inside a Three-Term Ratio

Every bar of a bar model is made of units of the same size. So, as with the fruit basket at the start, any two of the three quantities can be compared by their own two terms, with the other term left out.

Rule

If three quantities are in the ratio $a : b : c$ , then the first to the second is $a : b$ , the second to the third is $b : c$ , and the first to the third is $a : c$ . Write each one in its simplest form.

A three-term ratio in its simplest form can hold a two-term ratio that is not. In Example 3, walk : bus : MRT $= 6 : 9 : 5$ is in its simplest form, but walk : bus $= 6 : 9 = 2 : 3$ .

Example 7: Reading the two-term ratios

If $a : b : c = 9 : 12 : 8$ , find each ratio in its simplest form.
(a) $a : b$
(b) $b : c$
(c) $a : c$
(d) $c : a$

Solution

(a) Take the first and second terms: $a : b = 9 : 12$ . The HCF of 9 and 12 is 3, so $a : b = 9 \div 3 : 12 \div 3 = 3 : 4$ .

(b) Take the second and third terms: $b : c = 12 : 8$ . The HCF of 12 and 8 is 4, so $b : c = 12 \div 4 : 8 \div 4 = 3 : 2$ .

(c) Take the first and third terms, leaving out the middle one: $a : c = 9 : 8$ . The only common factor of 9 and 8 is 1.

(d) The same two terms as in (c), in the other order: $c : a = 8 : 9$ .

(a) $a : b = 3 : 4$
(b) $b : c = 3 : 2$
(c) $a : c = 9 : 8$
(d) $c : a = 8 : 9$

Each Quantity as a Fraction of the Whole

The basket of 5 pears, 2 plums and 3 peaches holds no other fruit, so the three bars together are the whole basket: $5 + 2 + 3 = 10$ units.

The pears are 5 of the 10 units, so the pears are $\frac{5}{10} = \frac{1}{2}$ of the fruit in the basket. In the same way the plums are $\frac{2}{10} = \frac{1}{5}$ of the fruit, and the peaches are $\frac{3}{10}$ of the fruit.

Rule

When three quantities in the ratio $a : b : c$ together make up the whole, the whole is $a + b + c$ units.

The first quantity is $\frac{a}{a + b + c}$ of the whole, the second is $\frac{b}{a + b + c}$ of the whole, and the third is $\frac{c}{a + b + c}$ of the whole.

The three fractions add up to 1.

Always say what a fraction is a fraction of. The plums are $\frac{1}{5}$ of all the fruit, but they are $\frac{2}{5}$ of the number of pears, because plums : pears $= 2 : 5$ . And add the terms only when the three quantities, and nothing else, make up the whole. In Example 1(c) the ratio $4 : 7 : 20$ already has the total as its third term, so its terms are not added.

Example 8: Fractions of the whole

Mrs Goh makes fruit punch from orange juice, pineapple juice and soda water only, in the ratio $4 : 3 : 5$ by volume.
(a) What fraction of the punch is orange juice?
(b) What fraction of the punch is pineapple juice?
(c) What fraction of the punch is soda water?
(d) What fraction of the volume of soda water is the volume of pineapple juice?

Solution

The three drinks make up all of the punch, so the punch is $4 + 3 + 5 = 12$ units.

(a) Orange juice is 4 of the 12 units: $\frac{4}{12} = \frac{1}{3}$

(b) Pineapple juice is 3 of the 12 units: $\frac{3}{12} = \frac{1}{4}$

Check: $\frac{4}{12} + \frac{3}{12} + \frac{5}{12} = \frac{12}{12} = 1$

(d) This compares two of the drinks with each other, not with the whole punch. Pineapple juice : soda water $= 3 : 5$ , so there are 3 units of pineapple juice for every 5 units of soda water. 3 units as a fraction of 5 units is $\frac{3}{5}$ , so the pineapple juice is $\frac{3}{5}$ of the soda water.

(a) Orange juice is $\frac{1}{3}$ of the punch.
(b) Pineapple juice is $\frac{1}{4}$ of the punch.
(c) Soda water is $\frac{5}{12}$ of the punch.
(d) The volume of pineapple juice is $\frac{3}{5}$ of the volume of soda water.

Example 9: From fractions of the whole to a ratio

In a school robotics club, $\frac{1}{2}$ of the members are in Secondary 1, $\frac{1}{3}$ are in Secondary 2 and the rest are in Secondary 3. Find the ratio of the number of Secondary 1 members to the number of Secondary 2 members to the number of Secondary 3 members.

Solution

The whole club is 1. Fraction of the members in Secondary 3 $= 1 - \frac{1}{2} - \frac{1}{3} = \frac{6}{6} - \frac{3}{6} - \frac{2}{6} = \frac{1}{6}$

All three are fractions of the same whole, the club, so the numbers of members are in the ratio $\frac{1}{2} : \frac{1}{3} : \frac{1}{6}$ .

The LCM of 2, 3 and 6 is 6. Multiply every term by 6: $\frac{1}{2} \times 6 : \frac{1}{3} \times 6 : \frac{1}{6} \times 6 = 3 : 2 : 1$

Check: $3 + 2 + 1 = 6$ units, and Secondary 1 is 3 of them, which is $\frac{3}{6} = \frac{1}{2}$ of the club, as given.

The ratio of the number of Secondary 1 members to the number of Secondary 2 members to the number of Secondary 3 members is $3 : 2 : 1$ .

Finding Missing Terms

Two equivalent ratios of three terms are linked by one multiplier: every term of the first ratio, multiplied by the same number, gives the matching term of the second. If one pair of matching terms is known, it gives the multiplier, and the multiplier gives the missing terms.

Method

Find the place (first, second or third) where both ratios show a term.
Find the number that one of these two terms is multiplied by to give the other.
Multiply the other terms by the same number. Going the other way, divide by it.

Example 10: Missing terms

Find the missing terms $x$ and $y$ .
(a) $3 : 4 : 7 = 12 : x : y$
(b) $45 : x : 30 = 9 : 4 : y$
(c) $4 : 6 : 10 = 6 : x : y$

Solution

(a) The first terms are both known, and $3 \times 4 = 12$ . So every term of $3 : 4 : 7$ is multiplied by 4: $x = 4 \times 4 = 16$ and $y = 7 \times 4 = 28$ .

Check: $12 : 16 : 28$ , with every term divided by 4, is $3 : 4 : 7$ .

(b) The first terms are both known, and $9 \times 5 = 45$ . So every term on the right, multiplied by 5, gives the matching term on the left, and every term on the left, divided by 5, gives the matching term on the right.

Second terms: $x = 4 \times 5 = 20$ . Third terms: $y = 30 \div 5 = 6$ .

Check: $45 : 20 : 30$ , with every term divided by 5, is $9 : 4 : 6$ .

(c) The first terms are 4 and 6, and 6 is not a multiple of 4. Write $4 : 6 : 10$ in its simplest form first: dividing every term by 2 gives $2 : 3 : 5$ .

Now $2 \times 3 = 6$ , so multiply every term of $2 : 3 : 5$ by 3: $x = 3 \times 3 = 9$ and $y = 5 \times 3 = 15$ .

Check: $6 : 9 : 15$ , with every term divided by 3, is $2 : 3 : 5$ , the simplest form of $4 : 6 : 10$ .

(a) $x = 16$ and $y = 28$
(b) $x = 20$ and $y = 6$
(c) $x = 9$ and $y = 15$

Example 11: Missing terms in a recipe

A drinks stall makes bandung from rose syrup, evaporated milk and water in the ratio $2 : 3 : 10$ by volume. For one large jug the stallholder uses 300 ml of rose syrup. How much evaporated milk and how much water go into the jug?

Solution

The three volumes in millilitres make a ratio equivalent to $2 : 3 : 10$ whose first term is 300.

First terms: $2 \times 150 = 300$ , so the multiplier is 150.

Second term: $3 \times 150 = 450$ . Third term: $10 \times 150 = 1500$ .

$2 : 3 : 10 = 300 : 450 : 1500$

Check: $300 : 450 : 1500$ , with every term divided by 150, is $2 : 3 : 10$ .

The jug takes 450 ml of evaporated milk and 1500 ml of water.

Common Mistakes

Dividing by a factor of only two of the terms. In $4 : 6 : 9$ , 2 divides 4 and 6 but not 9, and 3 divides 6 and 9 but not 4. The ratio is already in its simplest form, and $2 : 3 : 9$ is a different ratio.
Leaving a term out when multiplying. In $0.6 : 3 : 1.2$ the whole number 3 is multiplied by 10 as well: the ratio becomes $6 : 30 : 12$ , not $6 : 3 : 12$ .
Stopping as soon as the terms are whole numbers. Multiplying $\frac{2}{3} : \frac{4}{9} : 2$ by 9 gives $6 : 4 : 18$ , which still has the common factor 2. The simplest form is $3 : 2 : 9$ .
Writing the ratio before the units of measurement are the same. 50 cm, 1.2 m and 800 mm are in the ratio $50 : 120 : 80 = 5 : 12 : 8$ , not $50 : 1.2 : 800$ .
Taking the wrong denominator for a fraction of the whole. In orange juice : pineapple juice : soda water $= 4 : 3 : 5$ , the orange juice is $\frac{4}{12}$ of the punch. The fractions $\frac{4}{3}$ and $\frac{4}{5}$ compare the orange juice with one of the other drinks, not with the whole.
Turning the order round. If $a : b : c = 9 : 12 : 8$ , then $a : c = 9 : 8$ and $c : a = 8 : 9$ . Write the terms in the order the question names the quantities.

Practice

Work each question on paper before you open its solution.

Question 1

A car park has 24 cars, 9 motorcycles and 6 vans. Write each ratio in its simplest form.
(a) The number of cars to the number of motorcycles to the number of vans
(b) The number of vans to the number of cars to the number of motorcycles

ShowHide solution

(a) In the order cars, motorcycles, vans the ratio is $24 : 9 : 6$ .

The HCF of 24, 9 and 6 is 3: $24 \div 3 : 9 \div 3 : 6 \div 3 = 8 : 3 : 2$

8 and 2 have the common factor 2, but 2 is not a factor of 3, so $8 : 3 : 2$ is in its simplest form.

(b) The same three terms in the order vans, cars, motorcycles: $6 : 24 : 9 = 2 : 8 : 3$

(a) Cars : motorcycles : vans $= 8 : 3 : 2$
(b) Vans : cars : motorcycles $= 2 : 8 : 3$

Question 2

Write each ratio in its simplest form.
(a) $14 : 21 : 49$
(b) $54 : 90 : 36$
(c) $15 : 20 : 18$
(d) $125 : 75 : 200$

ShowHide solution

(a) $14 = 2 \times 7$ , $21 = 3 \times 7$ and $49 = 7 \times 7$ , so the HCF is 7: $14 \div 7 : 21 \div 7 : 49 \div 7 = 2 : 3 : 7$

(b) $54 = 2 \times 3^{3}$ , $90 = 2 \times 3^{2} \times 5$ and $36 = 2^{2} \times 3^{2}$ , so the HCF is $2 \times 3^{2} = 18$ : $54 \div 18 : 90 \div 18 : 36 \div 18 = 3 : 5 : 2$

(c) 5 is a factor of 15 and of 20, but not of 18. 2 is a factor of 20 and of 18, but not of 15. 3 is a factor of 15 and of 18, but not of 20. No number other than 1 divides all three terms.

(d) $125 = 5^{3}$ , $75 = 3 \times 5^{2}$ and $200 = 2^{3} \times 5^{2}$ , so the HCF is $5^{2} = 25$ : $125 \div 25 : 75 \div 25 : 200 \div 25 = 5 : 3 : 8$

(a) $14 : 21 : 49 = 2 : 3 : 7$
(b) $54 : 90 : 36 = 3 : 5 : 2$
(c) $15 : 20 : 18$ is already in its simplest form.
(d) $125 : 75 : 200 = 5 : 3 : 8$

Question 3

Ravi says, " $8 : 12 : 15$ is not in its simplest form, because 8 and 12 can both be divided by 4. Its simplest form is $2 : 3 : 15$ ." Is Ravi right? Explain.

ShowHide solution

4 is a factor of 8 and of 12, but it is not a factor of 15. A ratio is made simpler only by dividing every term by the same number.

The factors of 15 are 1, 3, 5 and 15. Of these, only 1 is a factor of 8. So the only factor common to 8, 12 and 15 is 1.

$2 : 3 : 15$ is a different ratio: the first quantity to the third is $8 : 15$ in the one and $2 : 15$ in the other.

Ravi is not right. $8 : 12 : 15$ is already in its simplest form, and $2 : 3 : 15$ is not equivalent to it.

Question 4

Write each ratio in its simplest form.
(a) $0.4 : 2 : 1.6$
(b) $0.25 : 1.5 : 0.75$
(c) $1.2 : 0.36 : 0.6$

ShowHide solution

(a) Multiply every term by 10: $0.4 \times 10 : 2 \times 10 : 1.6 \times 10 = 4 : 20 : 16$

The HCF of 4, 20 and 16 is 4: $4 \div 4 : 20 \div 4 : 16 \div 4 = 1 : 5 : 4$

(b) 0.25 and 0.75 have 2 decimal places, so multiply every term by 100: $0.25 \times 100 : 1.5 \times 100 : 0.75 \times 100 = 25 : 150 : 75$

25 is a factor of 150 and of 75, so the HCF is 25: $25 \div 25 : 150 \div 25 : 75 \div 25 = 1 : 6 : 3$

(c) 0.36 has 2 decimal places, so multiply every term by 100: $1.2 \times 100 : 0.36 \times 100 : 0.6 \times 100 = 120 : 36 : 60$

$120 = 2^{3} \times 3 \times 5$ , $36 = 2^{2} \times 3^{2}$ and $60 = 2^{2} \times 3 \times 5$ , so the HCF is $2^{2} \times 3 = 12$ : $120 \div 12 : 36 \div 12 : 60 \div 12 = 10 : 3 : 5$

(a) $0.4 : 2 : 1.6 = 1 : 5 : 4$
(b) $0.25 : 1.5 : 0.75 = 1 : 6 : 3$
(c) $1.2 : 0.36 : 0.6 = 10 : 3 : 5$

Question 5

Write each ratio in its simplest form.
(a) $\frac{2}{3} : \frac{1}{2} : \frac{5}{6}$
(b) $\frac{3}{4} : \frac{5}{8} : \frac{1}{2}$
(c) $1\frac{1}{2} : 2\frac{1}{4} : 3$
(d) $1.5 : \frac{3}{4} : 2$

ShowHide solution

(a) The LCM of 3, 2 and 6 is 6. Multiply every term by 6: $\frac{2}{3} \times 6 : \frac{1}{2} \times 6 : \frac{5}{6} \times 6 = 4 : 3 : 5$

(b) The LCM of 4, 8 and 2 is 8. Multiply every term by 8: $\frac{3}{4} \times 8 : \frac{5}{8} \times 8 : \frac{1}{2} \times 8 = 6 : 5 : 4$

The LCM of 2 and 4 is 4. Multiply every term by 4: $\frac{3}{2} \times 4 : \frac{9}{4} \times 4 : 3 \times 4 = 6 : 9 : 12$

The HCF of 6, 9 and 12 is 3: $6 \div 3 : 9 \div 3 : 12 \div 3 = 2 : 3 : 4$

(d) Write the decimal as a fraction: $1.5 = 1\frac{1}{2} = \frac{3}{2}$ , so the ratio is $\frac{3}{2} : \frac{3}{4} : 2$ .

The LCM of 2 and 4 is 4. Multiply every term by 4: $\frac{3}{2} \times 4 : \frac{3}{4} \times 4 : 2 \times 4 = 6 : 3 : 8$

6 and 3 have the common factor 3, but 3 is not a factor of 8, so this is the simplest form.

(a) $\frac{2}{3} : \frac{1}{2} : \frac{5}{6} = 4 : 3 : 5$
(b) $\frac{3}{4} : \frac{5}{8} : \frac{1}{2} = 6 : 5 : 4$
(c) $1\frac{1}{2} : 2\frac{1}{4} : 3 = 2 : 3 : 4$
(d) $1.5 : \frac{3}{4} : 2 = 6 : 3 : 8$

Question 6

Find each ratio in its simplest form.
(a) 45 minutes to 1 hour 15 minutes to 2 hours
(b) $1.20 to 90 cents to $3
(c) 750 g to 1.5 kg to 0.6 kg

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(a) In minutes: 1 hour 15 minutes $= 60 + 15 = 75$ minutes, and 2 hours $= 2 \times 60 = 120$ minutes. The ratio is $45 : 75 : 120$ .

The HCF of 45, 75 and 120 is 15: $45 \div 15 : 75 \div 15 : 120 \div 15 = 3 : 5 : 8$

(b) In cents: $1.20 $= 120$ cents and $3 $= 300$ cents. The ratio is $120 : 90 : 300$ .

The HCF of 120, 90 and 300 is 30: $120 \div 30 : 90 \div 30 : 300 \div 30 = 4 : 3 : 10$

(c) In grams: 1.5 kg $= 1.5 \times 1000$ g $= 1500$ g, and 0.6 kg $= 0.6 \times 1000$ g $= 600$ g. The ratio is $750 : 1500 : 600$ .

Divide every term by 10, then by 15: $750 : 1500 : 600 = 75 : 150 : 60 = 5 : 10 : 4$

5 and 10 have the common factor 5, but 5 is not a factor of 4, so this is the simplest form.

(a) $3 : 5 : 8$
(b) $4 : 3 : 10$
(c) $5 : 10 : 4$

Question 7

If $a : b : c = 12 : 20 : 15$ , find each ratio in its simplest form.
(a) $a : b$
(b) $b : c$
(c) $a : c$
(d) $c : b$

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(a) First and second terms: $a : b = 12 : 20$ . The HCF of 12 and 20 is 4, so $a : b = 3 : 5$ .

(b) Second and third terms: $b : c = 20 : 15$ . The HCF of 20 and 15 is 5, so $b : c = 4 : 3$ .

(d) The terms of (b) in the other order: $c : b = 15 : 20 = 3 : 4$ .

(a) $a : b = 3 : 5$
(b) $b : c = 4 : 3$
(c) $a : c = 4 : 5$
(d) $c : b = 3 : 4$

Question 8

One Monday a stall in a school canteen sold 120 plates of chicken rice, 45 plates of nasi lemak and 75 plates of mee goreng, and nothing else.
(a) Find the ratio of the number of plates of chicken rice to the number of plates of nasi lemak to the number of plates of mee goreng, in its simplest form.
(b) Find the ratio of the number of plates of nasi lemak to the number of plates of mee goreng.
(c) What fraction of all the plates sold were chicken rice?

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(a) The ratio is $120 : 45 : 75$ . The HCF of 120, 45 and 75 is 15: $120 \div 15 : 45 \div 15 : 75 \div 15 = 8 : 3 : 5$

(b) Take the second and third terms: nasi lemak : mee goreng $= 3 : 5$

(c) The three dishes make up all the plates sold, so the whole is $8 + 3 + 5 = 16$ units. Chicken rice is 8 of the 16 units: $\frac{8}{16} = \frac{1}{2}$

Check with the numbers of plates: $120 + 45 + 75 = 240$ , and $\frac{120}{240} = \frac{1}{2}$ .

(a) Chicken rice : nasi lemak : mee goreng $= 8 : 3 : 5$
(b) Nasi lemak : mee goreng $= 3 : 5$
(c) $\frac{1}{2}$ of all the plates sold were chicken rice.

Question 9

Of the books in a class library, $\frac{3}{8}$ are storybooks, $\frac{1}{3}$ are science books and the rest are comics. Find the ratio of the number of storybooks to the number of science books to the number of comics.

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The whole library is 1. Fraction of the books that are comics $= 1 - \frac{3}{8} - \frac{1}{3} = \frac{24}{24} - \frac{9}{24} - \frac{8}{24} = \frac{7}{24}$

All three are fractions of the same whole, so storybooks : science books : comics $= \frac{3}{8} : \frac{1}{3} : \frac{7}{24}$

The LCM of 8, 3 and 24 is 24. Multiply every term by 24: $\frac{3}{8} \times 24 : \frac{1}{3} \times 24 : \frac{7}{24} \times 24 = 9 : 8 : 7$

No number other than 1 divides 9, 8 and 7, so this is the simplest form.

Check: $9 + 8 + 7 = 24$ units, and the storybooks are 9 of them, which is $\frac{9}{24} = \frac{3}{8}$ of the library, as given.

The ratio of the number of storybooks to the number of science books to the number of comics is $9 : 8 : 7$ .

Question 10

Find the missing terms $x$ and $y$ .
(a) $2 : 7 : 9 = 10 : x : y$
(b) $x : 24 : y = 5 : 8 : 11$
(c) $12 : 18 : 30 = 10 : x : y$

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(a) First terms: $2 \times 5 = 10$ . Multiply the other terms of $2 : 7 : 9$ by 5: $x = 7 \times 5 = 35$ and $y = 9 \times 5 = 45$ .

(b) The second terms are both known: $8 \times 3 = 24$ . Multiply the other terms of $5 : 8 : 11$ by 3: $x = 5 \times 3 = 15$ and $y = 11 \times 3 = 33$ .

(c) 10 is not a multiple of 12, so write $12 : 18 : 30$ in its simplest form first. The HCF of 12, 18 and 30 is 6, and dividing every term by 6 gives $2 : 3 : 5$ .

Now $2 \times 5 = 10$ . Multiply the other terms of $2 : 3 : 5$ by 5: $x = 3 \times 5 = 15$ and $y = 5 \times 5 = 25$ .

Check: $10 : 15 : 25$ , with every term divided by 5, is $2 : 3 : 5$ , the simplest form of $12 : 18 : 30$ .

(a) $x = 35$ and $y = 45$
(b) $x = 15$ and $y = 33$
(c) $x = 15$ and $y = 25$

Question 11

Mr Tan cooks fried rice with rice, chicken and vegetables in the ratio $5 : 2 : 3$ by mass. He uses 360 g of vegetables. Find the mass of rice and the mass of chicken that he uses.

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The three masses in grams make a ratio equivalent to $5 : 2 : 3$ whose third term is 360.

Third terms: $3 \times 120 = 360$ , so the multiplier is 120.

First term: $5 \times 120 = 600$ . Second term: $2 \times 120 = 240$ .

$5 : 2 : 3 = 600 : 240 : 360$

Check: $600 : 240 : 360$ , with every term divided by 120, is $5 : 2 : 3$ .

Mr Tan uses 600 g of rice and 240 g of chicken.

Question 12

A shopkeeper mixes 1.2 kg of peanuts, 900 g of cashew nuts and 0.6 kg of almonds, and nothing else.
(a) Find the ratio of the mass of peanuts to the mass of cashew nuts to the mass of almonds, in its simplest form.
(b) Find the ratio of the mass of cashew nuts to the mass of almonds.
(c) What fraction of the mixture is peanuts?
(d) What fraction of the mass of peanuts is the mass of almonds?

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(a) Change every mass to grams: 1.2 kg $= 1200$ g and 0.6 kg $= 600$ g. The ratio is $1200 : 900 : 600$ .

The HCF of 1200, 900 and 600 is 300: $1200 \div 300 : 900 \div 300 : 600 \div 300 = 4 : 3 : 2$

(b) Take the second and third terms: cashew nuts : almonds $= 3 : 2$

(c) The three kinds of nut make up the whole mixture, so the whole is $4 + 3 + 2 = 9$ units. Peanuts are 4 of the 9 units: $\frac{4}{9}$

(d) Almonds : peanuts $= 2 : 4 = 1 : 2$ , so there is 1 unit of almonds for every 2 units of peanuts. 1 unit as a fraction of 2 units is $\frac{1}{2}$ , so the almonds are $\frac{1}{2}$ of the peanuts.

(a) Peanuts : cashew nuts : almonds $= 4 : 3 : 2$
(b) Cashew nuts : almonds $= 3 : 2$
(c) Peanuts are $\frac{4}{9}$ of the mixture.
(d) The mass of almonds is $\frac{1}{2}$ of the mass of peanuts.

Key Points

A ratio $a : b : c$ compares three quantities of the same kind at once. Its terms follow the order of the words, and it has no unit of measurement.
Multiplying or dividing every term by the same non-zero number gives an equivalent ratio.
Simplest form: whole-number terms, with 1 the only factor common to all three. Divide every term by the HCF of all three terms. A factor of only two of the terms does not count.
Decimals: multiply every term by 10, 100 or 1000. Fractions: multiply every term by the LCM of all the denominators. Then divide by the HCF.
Different units of measurement: change every quantity to the same unit of measurement before writing the ratio.
From $a : b : c$ read the two-term ratios $a : b$ , $b : c$ and $a : c$ , and write each in its simplest form.
When the three quantities make up the whole, the whole is $a + b + c$ units, and the first quantity is $\frac{a}{a + b + c}$ of the whole.
Missing terms: find the multiplier from the pair of matching terms that are both known, and use it on the other terms.

Two ratios such as $a : b$ and $b : c$ can be joined into one ratio $a : b : c$ , and a quantity can be shared among three in a given ratio. These are the subjects of the lessons on combining two ratios and on dividing a quantity in a given ratio.