Multiplicative relationships as ratios and fractions
Practise 19 exam-style questions on Multiplicative relationships as ratios and fractions for Edexcel GCSE Maths (Non-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.
19 questions
1–5 marks each
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19 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
At a concert there are 2.5 times as many adults as children. Everyone at the concert is either an adult or a child.
What fraction of the people at the concert are children?
Model answer
adults : children = 2.5 : 1 = 5 : 2 Total parts = 7 Fraction of children = \(\dfrac{2}{7}\)
Explanation
Turn the multiplicative relationship into a ratio, then double it to remove the decimal. The children are 2 parts out of 7 in total. A common mistake is to give \(\dfrac{2}{5}\), which compares children with adults rather than with everyone.
Mark scheme
1 mark
M1 for adults : children = 2.5 : 1 or 5 : 2
correct answer with no working scores full marks
1 mark
A1 for \(\dfrac{2}{7}\)
oe; \(\dfrac{1}{2.5}\) or \(\dfrac{2}{5}\) scores M1 A0
Question 2
2 marks
The cost of 3 pens is the same as the cost of 5 pencils.
Write down the ratio cost of one pen : cost of one pencil.
Model answer
Let a pen cost \(p\) and a pencil cost \(q\). \(3p = 5q\) \(\dfrac{p}{q} = \dfrac{5}{3}\) Cost of a pen : cost of a pencil = 5 : 3
Explanation
Fewer pens cost the same as more pencils, so each pen must cost more than each pencil: the ratio is 5 : 3, not 3 : 5. For example, if a pen costs 50p and a pencil 30p, then 3 pens and 5 pencils both cost £1.50.
Mark scheme
1 mark
M1 for \(3p = 5q\) where \(p\) is the cost of a pen and \(q\) is the cost of a pencil, or an example such as pens 50p and pencils 30p
correct answer with no working scores full marks
1 mark
A1 for 5 : 3
cao; 3 : 5 scores M0 A0
Question 3
3 marks
There are three containers, A, B and C. The volume of A is \(\dfrac{3}{5}\) of the volume of B. The volume of C is \(1\dfrac{1}{3}\) times the volume of B.
Find the ratio volume of A : volume of B : volume of C. Give your answer in its simplest form.
Model answer
Let B = 15 units (a multiple of 5 and 3). A \(= \dfrac{3}{5} \times 15 = 9\) units C \(= \dfrac{4}{3} \times 15 = 20\) units A : B : C = 9 : 15 : 20
Explanation
Choose a value for B that both fractions divide into exactly, such as 15. Then work out A and C. Since 9, 15 and 20 have no common factor, the ratio is already in its simplest form.
Mark scheme
1 mark
M1 for \(A : B = 3 : 5\) or \(B : C = 3 : 4\)
or choosing a value of B that is a multiple of 15; correct answer with no working scores full marks
1 mark
M1 for both ratios with the same B part, e.g. \(A : B = 9 : 15\) and \(B : C = 15 : 20\)
1 mark
A1 for 9 : 15 : 20
cao
Question 4
4 marks
A biscuit recipe uses flour, butter and sugar only. The mass of flour is 2.5 times the mass of butter. The mass of sugar is \(\dfrac{3}{5}\) of the mass of butter. The total mass of the three ingredients is 820 g.
Work out the mass of flour.
Model answer
flour : butter : sugar = 2.5 : 1 : 0.6 Multiply by 10: 25 : 10 : 6 Total parts = 41 One part = 820 ÷ 41 = 20 g Flour = 25 × 20 = 500 g (Butter 200 g, sugar 120 g; total 820 g)
Explanation
Express every ingredient as a multiple of the butter, then turn this into a ratio with whole numbers. Sharing 820 g in the ratio 25 : 10 : 6 gives the flour. Check that the three masses add up to 820 g.
Mark scheme
1 mark
P1 for expressions in terms of the mass of butter \(b\), flour \(2.5b\) and sugar \(0.6b\)
or flour : butter : sugar = 2.5 : 1 : 0.6; correct answer with no working scores full marks
1 mark
P1 for \(4.1b = 820\), or flour : butter : sugar = 25 : 10 : 6
1 mark
P1 for \(b = 200\), or one part \(= 820 \div 41 = 20\)
1 mark
A1 for 500 (g)
cao
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