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Year 7 Maths Ratios, Rates and Time

Ratio Word Problems

20 practice questions 2 video lessons Theory + worked examples
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Practice questions

Every question with a fully worked solution.

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  • Ratio Word Problems - Video - Ratio Word Problems Watch
  • Ratio Word Problems - Video - Art of Problem Solving: Introducing Ratios Watch
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Theory

To solve a ratio problem, find the value of one part first, then multiply to find each share. If the total is given, divide it by the total number of parts. If one share is given, divide it by its own number of parts. This Year 7 maths page shows both types with bar models.

In the ratio \(3:5\), the first share is \(3\) equal parts and the second share is \(5\) equal parts. Every problem comes down to: what is one part worth?

Type 1: the total is given (divide a quantity in a ratio).
Step 1. Total parts \(=\) add the terms.
Step 2. One part \(=\) total \(\div\) total parts.
Step 3. Each share \(=\) its number of parts \(\times\) one part.
Check: the shares add up to the total.

Type 2: one share is given (the unitary method).
Step 1. Match the numbers to the words.
Step 2. One part \(=\) the share you know \(\div\) its number of parts.
Step 3. Multiply to find the share you want.
Check: the shares simplify back to the ratio.

Units. If the amounts are in different units, change to the same unit first, and give the answer in the units asked for.

Sharing 40 in the ratio 3 to 5 Pia has 3 boxes and Sam has 5 boxes; each box is 5. Together 8 boxes make 40. Pia555 Sam55555 total 40 40 ÷ 8 = 5 in each part
Type 1: Pia and Sam share \(40\) stickers in the ratio \(3:5\). There are \(8\) parts, so one part is \(5\). Pia gets \(3 \times 5 = 15\) and Sam gets \(5 \times 5 = 25\).
Pia has 12 so one part is 4 Pia has 3 boxes making 12, so each box is 4. Sam has 5 boxes of 4, which is 20. Pia44412 known Sam444445 × 4 = 20 12 ÷ 3 = 4 in each part
Type 2: Pia has \(12\) stickers, which is \(3\) parts. One part is \(4\), so Sam has \(5 \times 4 = 20\).

One part, from what you know:

\[ \text{one part} = \text{total} \div \text{total parts} \]
one part = total ÷ total parts
\[ \text{one part} = \text{known share} \div \text{its parts} \]
one part = known share ÷ its parts
\[ \text{share} = \text{its parts} \times \text{one part} \]
share = its parts × one part
Divide by the total number of parts. To share \(40\) in the ratio \(3:5\), one part is \(40 \div 8 = 5\), not \(40 \div 3\) or \(40 \div 5\).
Is it a share or the total? “Boys : girls \(= 3:5\) and there are \(12\) boys” means \(3\) parts \(= 12\). The \(12\) is not the whole class.

How to solve a ratio problem

  1. Read: is the total given, or one share?
  2. Find one part: divide by the total parts, or by the parts of the known share.
  3. Multiply to find each share asked for.
  4. Check: the shares add to the total, or simplify back to the ratio.
Example 1 — Sharing a total
Ruby and Omar share \(42\) marbles in the ratio \(2:5\). How many marbles does each person get?
Solution
\(2 + 5\)\(=\)\(7\)
\(42 \div 7\)\(=\)\(6\)
\(2 \times 6\)\(=\)\(12\)
\(5 \times 6\)\(=\)\(30\)
7 parts , one part 42÷7=6, Ruby 12, Omar 30

Ruby gets \(12\) and Omar gets \(30\). Check: \(12 + 30 = 42\).

Example 2 — One share is known
Bill and Ben’s masses are in the ratio \(2:3\). Bill has a mass of \(56\) kg. What is Ben’s mass?
Solution
\(56 \div 2\)\(=\)\(28\)
\(3 \times 28\)\(=\)\(84\)
2 parts =56, one part =28, Ben =3×28=84

Ben’s mass is \(84\) kg. Check: \(56:84 = 2:3\) (divide by \(28\)).

Example 3 — Concrete mix
Sand and cement are mixed in the ratio \(4:1\) to make concrete. How much cement is needed to make \(120\) kg of concrete?
Solution
\(4 + 1\)\(=\)\(5\)
\(120 \div 5\)\(=\)\(24\)
5 parts =120 kg ,1 part =24 kg

\(24\) kg of cement (and \(4 \times 24 = 96\) kg of sand).

Example 4 — Juice and soda
A punch uses orange juice and soda water in the ratio \(3:5\). Ali uses \(12\) cups of orange juice. How many cups of soda water does he need?
Solution
\(12 \div 3\)\(=\)\(4\)
\(5 \times 4\)\(=\)\(20\)
3 parts =12, one part =4, soda =5×4=20

\(20\) cups of soda water

Example 5 — A three-term ratio
Three friends share a \(\$135\) prize in the ratio \(2:3:4\). How much does each friend get?
Solution
\(2 + 3 + 4\)\(=\)\(9\)
\(135 \div 9\)\(=\)\(15\)
\(2 \times 15\)\(=\)\(30\)
\(3 \times 15\)\(=\)\(45\)
\(4 \times 15\)\(=\)\(60\)
9 parts , one part 1530,45,60

\(\$30\), \(\$45\) and \(\$60\). Check: \(30 + 45 + 60 = 135\).

Example 6 — Units and a difference
A plank \(3.6\) m long is cut into two pieces in the ratio \(4:5\). How much longer is the longer piece, in centimetres?
Solution
\(3.6 \times 100\)\(=\)\(360\)
\(360 \div 9\)\(=\)\(40\)
\(4 \times 40\)\(=\)\(160\)
\(5 \times 40\)\(=\)\(200\)
\(200 - 160\)\(=\)\(40\)
3.6 m =360 cm , one part 40 cm ,160 cm and 200 cm , difference 40 cm

The pieces are \(160\) cm and \(200\) cm, so the longer piece is \(40\) cm longer (one part).

Common pitfalls

Dividing by one term. When the total is given, divide by the total number of parts.
Treating a share as the total. Read whether the number is one share or the whole amount.
Swapping the shares. In flour : sugar \(= 5:2\), flour is the \(5\) parts.
Not checking. The shares should add to the total and simplify back to the ratio.

Frequently asked questions

How do you share an amount in a ratio?

Add the terms, divide the amount by that total to get one part, then multiply. \(42\) in the ratio \(2:5\): one part is \(6\), so \(12\) and \(30\).

What is the unitary method?

Find the value of one part (or one unit) first, then multiply. If \(3\) parts are \(12\), one part is \(4\).

What if one share is given, not the total?

Divide the share by its own number of parts. If Bill’s \(2\) parts are \(56\) kg, one part is \(28\) kg.

How do you check a ratio answer?

Add the shares to get the total, and simplify the shares to get the ratio back.

Do the units matter?

Yes. Change to the same units first and answer in the units asked for.