Practise 28 exam-style questions on Direct and inverse proportion 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
7 pens cost £6.09
Work out the cost of 12 of these pens.
Model answer
Cost of 1 pen = 6.09 ÷ 7 = £0.87 Cost of 12 pens = 12 × 0.87 = £10.44
Explanation
The cost is directly proportional to the number of pens, so find the cost of one pen (the unitary method) and multiply by 12. Check the answer is sensible: 12 pens should cost a bit less than twice the cost of 7 pens.
Mark scheme
1 mark
M1 for \(6.09 \div 7\) (= 0.87)
oe, e.g. \(\frac{12}{7} \times 6.09\)
1 mark
A1 for £10.44
cao; correct answer with no working scores full marks
Question 2
3 marks
\(y\) is directly proportional to \(x^2\). When \(x = 4\), \(y = 48\)
Find the value of \(y\) when \(x = 6.5\)
Model answer
\(y = kx^2\) \(48 = k \times 16\), so \(k = 3\) \(y = 3x^2\) When \(x = 6.5\): \(y = 3 \times 6.5^2 = 3 \times 42.25 = 126.75\)
Explanation
Write the proportion as an equation with a constant \(k\), use the given pair of values to find \(k\), then substitute. Remember to square \(x\) both when finding \(k\) and when finding the new value of \(y\). Using \(y = kx\) instead gives 78.
Mark scheme
1 mark
M1 for \(y = kx^2\)
oe; correct answer with no working scores full marks
1 mark
M1 for \(48 = k \times 4^2\), giving \(k = 3\)
1 mark
A1 for 126.75
cao; accept 127 (3 s.f.)
Question 3
3 marks
\(y\) is directly proportional to \(x^2\). \(x\) is increased by 20%.
Work out the percentage increase in \(y\).
Model answer
\(y = kx^2\) New \(x = 1.2x\) New \(y = k(1.2x)^2 = 1.44kx^2 = 1.44y\) The percentage increase in \(y\) is 44%.
Explanation
Because \(y\) depends on \(x^2\), the multiplier for \(y\) is the square of the multiplier for \(x\): \(1.2^2 = 1.44\). A common mistake is to say 40% (doubling 20%) - squaring a multiplier is not the same as doubling the percentage.
Mark scheme
1 mark
M1 for \(y = kx^2\) and the multiplier 1.2 for \(x\)
oe, e.g. a numerical example with x = 10 becoming 12; correct answer with no working scores full marks
1 mark
M1 for \(1.2^2\) (= 1.44)
oe, e.g. y changes from 100k to 144k
1 mark
A1 for 44%
cao
Question 4
4 marks
The table shows the time, \(t\) minutes, it takes to fill a tank using \(n\) identical pipes.
Show that \(t\) is inversely proportional to \(n\). Work out the least number of pipes needed to fill the tank in less than 10 minutes.
Model answer
2 × 45 = 90, 3 × 30 = 90, 5 × 18 = 90, 6 × 15 = 90 \(nt = 90\) for every pair, so \(t\) is inversely proportional to \(n\): \(t = \frac{90}{n}\) For \(t < 10\): \(\frac{90}{n} < 10\), so \(n > 9\) 9 pipes take exactly 10 minutes, so the least number is 10 pipes (9 minutes).
Explanation
For inverse proportion the product \(nt\) is constant. Then \(t = \frac{90}{n}\), and "less than 10 minutes" needs \(n > 9\). 9 pipes take exactly 10 minutes, which is not less than 10, so 10 pipes are needed.
Mark scheme
1 mark
M1 for \(nt\) worked out for every pair (all 90)
1 mark
C1 for a conclusion that \(nt\) is constant (90), so \(t\) is inversely proportional to \(n\)
1 mark
M1 for \(t = \frac{90}{n}\) and \(\frac{90}{n} < 10\), or \(n > 9\)
oe, e.g. 9 pipes take exactly 10 minutes
1 mark
A1 for 10 pipes
cao
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