Problems for Ratios (harder)
This subtopic takes ratio problems a step further: the amount you know might be a total, one share or the difference between two shares. NSW Year 7 maths students still ask the same question each time, namely what one part is worth (NESA outcome MA4‑RAT‑C‑01).
You will scale recipes with the unitary method, use map scales such as \(1:25\,000\) to find real distances in kilometres, share among three or four people, find the new ratio after amounts change, and work out which ingredient runs out first. Planning a bushwalk from a map is one real use. This subtopic is part of Year 7 Ratios, Rates and Time.
Every question with a fully worked solution.
- Problems for Ratios (harder) - Video - Using ratios to solve three-term word problems with changing quantities Watch
Theory
Harder ratio problems still come down to one question: what is one part worth? Decide whether you know one share, the total or a difference, count the parts it stands for, and divide. This Year 7 maths page covers recipes, map scales, ratios that change, and ratios from percentages.
What do you know? One share, the total, or a difference. Count the parts that amount stands for. One part \(=\) amount \(\div\) its number of parts. Then multiply to find what you want.
Recipes (the unitary method). Find the amount for \(1\) first. If \(250\) mL of milk serves \(5\) people, \(1\) person needs \(50\) mL, so \(8\) people need \(400\) mL.
Map scales. A scale of \(1:50\,000\) means \(1\) cm on the map stands for \(50\,000\) cm in real life. Multiply, then change units: cm \(\div 100\) gives m, and m \(\div 1000\) gives km.
Ratios that change. Work out the actual amounts first, change them, then write the new ratio and simplify. If one quantity does not change, make its number of parts the same in both ratios (equivalent ratios) so one part stays the same size.
More part counting. In \(1:2:6\), the third share is \(6 - (1 + 2) = 3\) parts more than the other two together. To see which ingredient runs out first, divide what you have of each by its number of parts: the smaller answer runs out first.
From percentages. Numbers that are \(10\%\) and \(40\%\) more than the same number are \(110\%\) and \(140\%\) of it, so they are in the ratio \(110:140 = 11:14\).
The amount you know tells you how many parts to divide by (ratio \(3:5\)):
| You know | Parts | One part |
|---|---|---|
| first share \(= 12\) | \(3\) | \(12 \div 3 = 4\) |
| total \(= 40\) | \(3 + 5 = 8\) | \(40 \div 8 = 5\) |
| difference \(= 10\) | \(5 - 3 = 2\) | \(10 \div 2 = 5\) |
How to solve a harder ratio problem
- Decide what you know: a share, the total or a difference.
- Count the parts it stands for, and divide to find one part.
- Multiply to find what is asked.
- Check with the original ratio.
How to use a map scale \(1:n\)
- Map to real: multiply the map length (cm) by \(n\), then change units.
- Real to map: change the real length to cm, then divide by \(n\).
(i) How many exotic plants are there?
(ii) How many plants are there altogether?
| \(35 \div 5\) | \(=\) | \(7\) |
| \(2 \times 7\) | \(=\) | \(14\) |
| \(7 \times 7\) | \(=\) | \(49\) |
(i) \(14\)
(ii) \(49\) plants
| \(7 - 4\) | \(=\) | \(3\) |
| \(18 \div 3\) | \(=\) | \(6\) |
| \(4 \times 6\) | \(=\) | \(24\) |
| \(7 \times 6\) | \(=\) | \(42\) |
Noah has \(24\) and Isla has \(42\). Check: \(42 - 24 = 18\).
| \(175 \div 5\) | \(=\) | \(35\) |
| \(4 + 5 + 6\) | \(=\) | \(15\) |
| \(15 \times 35\) | \(=\) | \(525\) |
\(\$525\)
| \(4.5 \times 1000\) | \(=\) | \(4500\) |
| \(4500 \times 100\) | \(=\) | \(450\,000\) |
| \(450\,000 \div 50\,000\) | \(=\) | \(9\) |
\(9\) cm on the map.
(i) How many juniors are in the club?
(ii) How many seniors were there at first?
The juniors do not change, so make their parts the same: \(3:2 = 6:4\) and \(2:1 = 6:3\). The seniors fell by \(4 - 3 = 1\) part, which is \(8\).
| \(6 \times 8\) | \(=\) | \(48\) |
| \(4 \times 8\) | \(=\) | \(32\) |
(i) \(48\) juniors
(ii) \(32\) seniors. Check: \(48:32 = 3:2\) and \(48:24 = 2:1\).
(ii) A biscuit mix uses flour and sugar in the ratio \(3:2\). Lena has \(900\) g of flour and \(500\) g of sugar. Which runs out first, and how much mix can she make?
| \(250 \div 5\) | \(=\) | \(50\) |
| \(8 \times 50\) | \(=\) | \(400\) |
| \(900 \div 3\) | \(=\) | \(300\) |
| \(500 \div 2\) | \(=\) | \(250\) |
| \(5 \times 250\) | \(=\) | \(1250\) |
(i) \(400\) mL
(ii) The sugar gives the smaller part (\(250\) g), so it runs out first. She uses \(750\) g of flour and \(500\) g of sugar: \(1250\) g of mix.
| \(100 + 25\) | \(=\) | \(125\) |
| \(100 + 50\) | \(=\) | \(150\) |
\(125:150 = 5:6\) (divide by \(25\)).
Common pitfalls
Frequently asked questions
How do you solve a ratio problem with a difference?
Divide the difference by the difference in parts. In \(4:7\), \(18\) more is \(3\) parts, so one part is \(6\).
How do you use a map scale?
Multiply the map length by the scale number, then change units. At \(1:50\,000\), \(9\) cm is \(450\,000\) cm \(= 4.5\) km.
How do you scale a recipe?
Find the amount for one person first, then multiply. \(250\) mL for \(5\) people is \(50\) mL each, so \(8\) people need \(400\) mL.
What if one quantity in a ratio stays the same?
Make its number of parts the same in both ratios using equivalent ratios. Then one part is the same size before and after.
How do you find which ingredient runs out first?
Divide what you have of each ingredient by its number of parts. The smaller answer runs out first.