Fractions in ratio problems - Edexcel GCSE Maths (Calculator) Exam Questions

Practise 28 exam-style questions on Fractions in ratio problems for Edexcel GCSE Maths (Calculator). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with an AI tutor on hand if you get stuck.

28 questions
1–5 marks each
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20 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

2 marks

Mortar is made by mixing sand and cement in the ratio 9 : 2 by mass.
Sam needs 46.2 kg of mortar.

Work out the mass of cement Sam needs.

Model answer

Total parts \(= 9 + 2 = 11\)
One part \(= 46.2 \div 11 = 4.2\) kg
Cement \(= 2 \times 4.2 = 8.4\) kg

Explanation

Cement is 2 parts out of a total of 11, so it is \(\frac{2}{11}\) of the mortar. Find one part by dividing by 11, then multiply by 2. A common mistake is to divide 46.2 by 2 or by 9, using one part of the ratio as if it were the total.

Mark scheme

1 mark
M1 for \(46.2 \div 11\) (= 4.2) or \(\frac{2}{11} \times 46.2\)
1 mark
A1 for 8.4 kg
cao (exact); correct answer with no working scores full marks

Question 2

3 marks

Jen and Kim share some money in the ratio 5 : 3
Jen gets £96.50 more than Kim.

How much money does Kim get?

Model answer

Difference \(= 5 - 3 = 2\) parts
2 parts = £96.50, so 1 part \(= 96.50 \div 2 = £48.25\)
Kim gets \(3 \times 48.25 = £144.75\)

Explanation

The £96.50 is the difference between the shares, which is \(5 - 3 = 2\) parts. Once you know one part, multiply by Kim's 3 parts. A common mistake is to treat £96.50 as the total and divide it by 8.

Mark scheme

1 mark
M1 for the difference of 2 parts = £96.50
oe, e.g. \(5 - 3 = 2\)
1 mark
M1 for one part \(= 96.50 \div 2\) (= 48.25)
1 mark
A1 for £144.75
cao; correct answer with no working scores full marks

Question 3

3 marks

A bag contains only red, blue and green counters.
The probability of taking a red counter at random is 0.35
The ratio of blue counters to green counters is 5 : 8

Work out the probability of taking a green counter at random.

Model answer

P(blue or green) \(= 1 - 0.35 = 0.65\)
Green is 8 parts out of \(5 + 8 = 13\)
P(green) \(= 0.65 \times \frac{8}{13} = 0.4\)

Explanation

The probabilities add up to 1, so blue and green together have probability 0.65. That 0.65 is shared in the ratio 5 : 8, so green gets \(\frac{8}{13}\) of it. A common mistake is to give \(\frac{8}{13}\) as the answer, forgetting the red counters.

Mark scheme

1 mark
M1 for \(1 - 0.35 = 0.65\)
correct answer with no working scores full marks
1 mark
M1 for \(0.65 \times \frac{8}{13}\)
oe, e.g. \(0.65 \div 13 \times 8\)
1 mark
A1 for 0.4
oe, e.g. \(\frac{2}{5}\); exact

Question 4

4 marks

Concrete is made by mixing cement, sand and gravel in the ratio 1 : 2 : 4 by mass.
A builder has 375 kg of sand, 640 kg of gravel and plenty of cement.
The builder makes as much concrete as possible.

What fraction of the sand is left over?
Give your answer in its simplest form.

Model answer

Sand : gravel = 2 : 4 = 1 : 2
375 kg of sand would need 750 kg of gravel, but there is only 640 kg, so the gravel runs out first.
One part \(= 640 \div 4 = 160\) kg
Sand used \(= 2 \times 160 = 320\) kg
Sand left \(= 375 - 320 = 55\) kg
Fraction left \(= \dfrac{55}{375} = \dfrac{11}{75}\)

Explanation

First decide which ingredient limits the amount of concrete: the gravel allows 160 kg per part but the sand would allow 187.5 kg per part, so gravel runs out first. Then find how much sand is used and give the leftover as a fraction of the 375 kg, simplified by dividing by 5.

Mark scheme

1 mark
P1 for finding which runs out first, e.g. \(640 \div 4 = 160\) and \(375 \div 2 = 187.5\), or 375 kg of sand would need 750 kg of gravel
correct answer with no working scores full marks
1 mark
P1 for the sand used: \(2 \times 160 = 320\) kg
1 mark
P1 for the sand left: \(375 - 320 = 55\) kg
1 mark
A1 for \(\frac{11}{75}\)
cao

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