Ratio Word Problems
Ratio word problems put ratios into real situations, and the key step is always working out what one part is worth. In NSW Year 8 maths this subtopic brings together equivalent ratios, simplifying and sharing (NESA outcome MA4‑RAT‑C‑01).
You will match the numbers in a story to the terms of a ratio, use one share or the difference between two shares to find the rest, scale recipes up or down and find which ingredient runs out first, and use map scales such as \(1:40\,000\) in both directions. You will also solve problems where a ratio changes because people join, leave, swap money or grow older. Tiling, sports results and Anzac biscuits all feature. It is part of Year 8 Rates and Ratios.
Theory
In a ratio word problem, find the value of one part first; every share, total and difference is then a number of parts. Sand and cement mixed \(4:1\) to make \(120\) kg of concrete gives \(5\) parts of \(24\) kg, so \(24\) kg of cement. This Year 8 maths page shows how.
One share given. In \(4:9\) the first share is \(4\) parts. If it is \(36\), one part is \(36 \div 4 = 9\); the other share is \(9 \times 9 = 81\) and the total is \(13 \times 9 = 117\).
The difference given. In \(4:9\) the second share has \(9 - 4 = 5\) more parts, so a difference of \(45\) gives one part \(= 45 \div 5 = 9\).
Three or four terms work the same way. Change different units to the same unit first.
Recipes. Find the amount for one serve or item, then multiply by the new number. To see which ingredient runs out first, work out how many serves each could make alone; the smallest is the answer.
Map scales. A scale \(1:n\) means \(1\) cm on the map stands for \(n\) cm in real life. Real \(=\) map \(\times n\); map \(=\) real \(\div n\), working in centimetres. For the scale itself, write map : real in centimetres and divide both by the map length.
Ratios that change. Work out the actual amounts, add or take away, then write the new ratio. If the total is unknown but nothing is gained or lost, write both ratios with the same total number of parts (the LCM of the two totals). Ages all grow together, so the difference between two ages never changes.
Rewriting \(2:1\) and \(3:1\) with the same total number of parts (the LCM of \(3\) and \(4\) is \(12\)):
| Before | After | |
|---|---|---|
| Ratio | \(2:1\) | \(3:1\) |
| Parts | 3 | 4 |
| Scale by | \(\times 4\) | \(\times 3\) |
| New ratio | \(8:4\) | \(9:3\) |
| Parts | 12 | 12 |
| Length | In centimetres |
|---|---|
| 1 m | 100 cm |
| 1 km | \(100\,000\) cm |
Solving a ratio word problem
- Write the ratio in the order of the words.
- Match the given number to a share, a total or a difference.
- Find one part by dividing by the matching number of parts.
- Multiply to get what is asked, then check with the ratio.
| \(\text{Parts}\) | \(=\) | \(4 + 1 = 5\) |
| \(\text{One part}\) | \(=\) | \(120 \div 5 = 24\) |
Cement is \(1\) part: \(24\) kg.
Let the third number be \(x\). The first is \(1.25x\) and the second is \(1.5x\).
| \(1.25x:1.5x:x\) | \(=\) | \(1.25:1.5:1\) |
| \(\) | \(=\) | \(125:150:100\) |
| \(\) | \(=\) | \(5:6:4\) |
\(5:6:4\) (multiply by \(100\), then divide by \(25\)).
(i) How many does 8A sell?
(ii) How many altogether?
| \(\text{One part}\) | \(=\) | \(68 \div 4 = 17\) |
| \(5 \times 17\) | \(=\) | \(85\) |
| \(9 \times 17\) | \(=\) | \(153\) |
(i) \(85\)
(ii) \(153\)
(i) the oats in grams
(ii) the honey in litres.
For one piece: \(300 \div 12 = 25\) g of oats and \(240 \div 12 = 20\) mL of honey.
| \(28 \times 25\) | \(=\) | \(700\) |
| \(28 \times 20\) | \(=\) | \(560\) |
| \(560 \div 1000\) | \(=\) | \(0.56\) |
(i) \(700\) g
(ii) \(560\) mL \(= 0.56\) L
(i) A river is \(8\) cm long on the map. How long is it in kilometres?
(ii) A ridge is \(54\) km long. How long is it on the map?
| \(8 \times 400\,000\) | \(=\) | \(3\,200\,000 \text{ cm}\) |
| \(3\,200\,000 \div 100\,000\) | \(=\) | \(32 \text{ km}\) |
| \(54 \times 100\,000\) | \(=\) | \(5\,400\,000 \text{ cm}\) |
| \(5\,400\,000 \div 400\,000\) | \(=\) | \(13.5\) |
(i) \(32\) km
(ii) \(13.5\) cm
| \(3 \text{ km}\) | \(=\) | \(300\,000 \text{ cm}\) |
| \(300\,000 \div 12\) | \(=\) | \(25\,000\) |
map : real \(= 12:300\,000 = 1:25\,000\).
(i) How many muffins can he make?
(ii) How much flour is left?
Muffins each ingredient could make alone: flour \(700 \div 200 \times 12 = 42\); eggs \(5 \div 2 \times 12 = 30\); milk \(600 \div 150 \times 12 = 48\).
| \(\text{Flour used}\) | \(=\) | \(30 \div 12 \times 200 = 500\) |
| \(\text{Left}\) | \(=\) | \(700 - 500 = 200\) |
(i) \(30\) muffins: the eggs run out first.
(ii) \(200\) g of flour
(ii) Ana is \(15\) and her brother is \(4\). In how many years will their ages be in the ratio \(2:1\)?
(i) The total is unchanged. With \(20\) parts (LCM of \(4\) and \(5\)): \(1:3 = 5:15\) and \(2:3 = 8:12\). Adults gain \(3\) parts.
(ii) The age difference stays \(11\), and in \(2:1\) the difference is \(1\) part.
| \(\text{One part}\) | \(=\) | \(15 \div 3 = 5\) |
| \(\text{Members}\) | \(=\) | \(20 \times 5 = 100\) |
| \(\text{Brother then}\) | \(=\) | \(11\) |
| \(\text{Years}\) | \(=\) | \(11 - 4 = 7\) |
(i) \(100\) members (adults \(25 \to 40\), juniors \(75 \to 60\))
(ii) In \(7\) years (\(22\) and \(11\)).
Common pitfalls
Frequently asked questions
How do I start a ratio word problem?
Write the ratio in the order of the words, decide whether the given number is a share, the total or a difference, then find one part.
How do I use a map scale of 1:50 000?
Multiply the map distance in centimetres by 50 000 for the real distance in centimetres, then change to metres or kilometres.
How do I find the scale of a map?
Write map distance : real distance in the same unit, then divide both by the map distance to get 1:n.
How do I scale a recipe?
Find the amount for one serve, then multiply by the number of serves you want.
What if the ratio changes after something moves?
If nothing is added or removed, the total stays the same. Rewrite both ratios with the same total number of parts.