Calculating with integers, decimals and fractions - Edexcel GCSE Maths (Calculator) Exam Questions

Practise 28 exam-style questions on Calculating with integers, decimals and fractions 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

At 6 am the temperature in a town was −4.8 °C.
By 2 pm the temperature had risen by 11.3 °C.
By midnight the temperature had fallen by 9.75 °C from its value at 2 pm.

Work out the temperature at midnight.

Model answer

At 2 pm: \(-4.8 + 11.3 = 6.5\) °C
At midnight: \(6.5 - 9.75 = -3.25\) °C

Explanation

A rise means add and a fall means subtract. Starting from a negative number, adding 11.3 takes you past zero to 6.5. Then subtracting 9.75, which is more than 6.5, takes you below zero again.

Mark scheme

1 mark
M1 for \(-4.8 + 11.3\) (= 6.5) or \(11.3 - 9.75\) (= 1.55)
1 mark
A1 for −3.25 °C
cao; correct answer with no working scores full marks

Question 2

3 marks

Petrol costs 147.9p per litre.
Amir has £60 to spend on petrol.

The petrol pump measures petrol in steps of 0.1 litres.

Work out the greatest amount of petrol, in litres, that Amir can buy.

Model answer

147.9p = £1.479
\(60 \div 1.479 = 40.567...\) litres
The pump measures in 0.1 litres, and 40.6 litres would cost \(40.6 \times 1.479 = £60.05\), which is more than £60.
40.5 litres costs \(40.5 \times 1.479 = £59.90\)
Greatest amount: 40.5 litres

Explanation

Use the same units for the money: either £60 and £1.479, or 6000p and 147.9p. The calculator gives 40.567..., but rounding to the nearest 0.1 gives 40.6 litres, which Amir cannot afford. In a problem like this you must round down.

Mark scheme

1 mark
M1 for \(60 \div 1.479\) or \(6000 \div 147.9\)
correct answer with no working scores full marks
1 mark
A1 for 40.56...
accept 40.5 to 40.6 seen
1 mark
A1 for 40.5 litres
cao; do not accept 40.6 (it would cost £60.05)

Question 3

3 marks

A rectangular room is \(4\frac{3}{8}\) m long and \(3\frac{5}{6}\) m wide.

Carpet costs £18.40 per square metre.
Carpet must be bought in a whole number of square metres.

Work out the cost of buying enough carpet to cover the floor of the room.

Model answer

Area = \(4\frac{3}{8} \times 3\frac{5}{6} = \dfrac{35}{8} \times \dfrac{23}{6} = \dfrac{805}{48} = 16.77...\) m²
Carpet must be bought in whole square metres, so buy 17 m².
Cost = \(17 \times 18.40 = £312.80\)

Explanation

Find the area by multiplying the mixed numbers (change them to improper fractions or decimals). Because carpet is sold in whole square metres, the area must be rounded up, not to the nearest whole number. A common mistake is to work out \(16.77... \times 18.40 = 308.58\).

Mark scheme

1 mark
M1 for \(4\frac{3}{8} \times 3\frac{5}{6}\) (= \(\frac{805}{48}\) = 16.77...)
correct answer with no working scores full marks
1 mark
M1 for rounding up to 17 square metres
1 mark
A1 for £312.80
cao

Question 4

4 marks

The fuel tank of Tom's car is \(\dfrac{7}{12}\) full.

Tom drives 216.5 km. The fuel tank is then \(\dfrac{1}{4}\) full.

He keeps driving and uses fuel at the same rate.

How much further can Tom drive before the fuel tank is \(\dfrac{1}{16}\) full?
Give your answer to the nearest kilometre.

Model answer

Fuel used for 216.5 km: \(\dfrac{7}{12} - \dfrac{1}{4} = \dfrac{7}{12} - \dfrac{3}{12} = \dfrac{4}{12} = \dfrac{1}{3}\) of the tank
A full tank lasts \(216.5 \times 3 = 649.5\) km
Fuel left to use: \(\dfrac{1}{4} - \dfrac{1}{16} = \dfrac{3}{16}\) of the tank
Distance: \(649.5 \times \dfrac{3}{16} = 121.78...\) km
122 km (to the nearest km)

Explanation

Work out what fraction of the tank was used for the 216.5 km, then scale this up to a full tank. The remaining distance uses \(\frac{1}{4} - \frac{1}{16} = \frac{3}{16}\) of the tank, not \(\frac{1}{16}\) of it. Keep the exact value until the final rounding.

Mark scheme

1 mark
P1 for \(\dfrac{7}{12} - \dfrac{1}{4} = \dfrac{1}{3}\) of the tank used for 216.5 km
1 mark
P1 for \(\dfrac{1}{4} - \dfrac{1}{16} = \dfrac{3}{16}\) of the tank
1 mark
P1 for \(216.5 \div \dfrac{1}{3} \times \dfrac{3}{16}\) oe
e.g. 649.5 km for a full tank, then \(649.5 \times \frac{3}{16}\), or \(216.5 \times \frac{9}{16}\)
1 mark
A1 for 122 km
accept 121.78...; correct answer with no working scores full marks

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