Practise 20 exam-style questions on Fractions and percentages as operators 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.
20 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
Sophie earns £2346.50 each month. She pays 28% of this on rent.
How much of her monthly earnings does Sophie have left after she has paid her rent?
Model answer
She has 100% − 28% = 72% left. \(2346.50 \times 0.72 = 1689.48\) £1689.48
Explanation
Paying 28% leaves 72% of her earnings, so multiply by 0.72 in one step. You could also find 28% (£657.02) and subtract it. Do not give £657.02 as the answer - that is the rent, not the amount left.
Mark scheme
1 mark
M1 for \(2346.50 \times 0.72\), or for \(2346.50 \times 0.28\) (= 657.02) and subtracting from 2346.50
oe
1 mark
A1 for £1689.48
cao; correct answer with no working scores full marks
Question 2
3 marks
Ella wants to increase £68.40 by 7.5%. Here is her working.
\(68.40 \times 1.75 = 119.70\) The new amount is £119.70.
Ella's answer is wrong. Explain the mistake she has made and work out the correct new amount.
Model answer
Ella multiplied by 1.75, which increases an amount by 75%. For an increase of 7.5% the multiplier is 100% + 7.5% = 107.5% = 1.075. \(68.40 \times 1.075 = 73.53\) The correct new amount is £73.53.
Explanation
To write a percentage as a decimal divide by 100: 7.5% = 0.075, so the multiplier for a 7.5% increase is 1 + 0.075 = 1.075. Ella used 0.75 instead of 0.075. A quick estimate would have caught it: increasing about £70 by under 10% cannot give nearly £120.
Mark scheme
1 mark
C1 for explaining that 1.75 is the multiplier for an increase of 75% (not 7.5%), or that she should have multiplied by 1.075
oe, e.g. "7.5% is 0.075 not 0.75"
1 mark
M1 for \(68.40 \times 1.075\), or for \(68.40 \times 0.075\) (= 5.13) and adding to 68.40
oe
1 mark
A1 for £73.53
cao
Question 3
3 marks
A ball is dropped from a height of 3.2 m. Each time the ball bounces, it rises to \(\dfrac{3}{5}\) of the height it fell from.
After which bounce does the ball first rise to a height of less than 10 cm? You must show your working.
Model answer
Heights after each bounce (multiply by 0.6 each time): 1st: 1.92 m, 2nd: 1.152 m, 3rd: 0.6912 m, 4th: 0.41472 m, 5th: 0.248832 m 6th: \(3.2 \times 0.6^6 = 0.149299...\) m = 14.9 cm (not less than 10 cm) 7th: \(3.2 \times 0.6^7 = 0.089579...\) m = 8.96 cm (less than 10 cm) The ball first rises to less than 10 cm after the 7th bounce.
Explanation
Each bounce multiplies the height by \(\dfrac{3}{5}\) = 0.6, so after n bounces the height is \(3.2 \times 0.6^n\) metres. Work in one unit: 10 cm is 0.1 m. Show the bounce just before and just after the height drops below 0.1 m to justify your answer.
Mark scheme
1 mark
M1 for \(3.2 \times 0.6^n\) for any \(n \ge 2\), or for repeated multiplication by \(\dfrac{3}{5}\) (1.92, 1.152, 0.6912, ...)
oe in cm (320, 192, 115.2, ...)
1 mark
M1 for the heights after the 6th bounce, 0.149... m (14.9 cm), and after the 7th bounce, 0.0895... m (8.96 cm)
accept 0.149 to 0.15 and 0.089 to 0.09 (or in cm)
1 mark
A1 for the 7th bounce
supported by both values; answer only scores 0
Question 4
4 marks
The length of a cuboid is increased by 12%. The width of the cuboid is decreased by \(\dfrac{1}{8}\). The height of the cuboid is changed so that the volume of the cuboid stays the same.
Work out the percentage change in the height. Give your answer correct to 1 decimal place and state whether it is an increase or a decrease.
Model answer
Volume = length × width × height New length = 1.12 × length, new width = 0.875 × width \(1.12 \times 0.875 = 0.98\) For the volume to stay the same, the height must be multiplied by \(\dfrac{1}{0.98} = 1.020408...\) This is an increase of 2.0408...%, so the height increases by 2.0% (1 d.p.).
Explanation
Volume is the product of the three dimensions, so the multipliers multiply: \(1.12 \times 0.875 \times m = 1\). Solving gives \(m = \dfrac{1}{0.98}\), an increase of about 2.0%. A common mistake is to add the percentages (+12% and −12.5%) and say the height must increase by 0.5%.
Mark scheme
1 mark
P1 for the multipliers 1.12 and \(\dfrac{7}{8}\) (or 0.875)
oe; correct answer with no working scores full marks
1 mark
P1 for \(1.12 \times 0.875\) (= 0.98)
oe, e.g. using lengths 1, 1, 1 or other values
1 mark
P1 for \(1 \div 0.98\) (= 1.0204...)
oe
1 mark
A1 for an increase of 2.0%
awrt 2.0 (2.0408...); must state increase
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