Exact calculation with fractions, surds and π - Edexcel GCSE Maths (Non-calculator) Exam Questions

Practise 37 exam-style questions on Exact calculation with fractions, surds and \\\\u03c0 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.

37 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

Simplify \(\sqrt{8} + \sqrt{18}\)

Give your answer in the form \(a\sqrt{2}\).

Model answer

\(\sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2}\)
\(\sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2}\)
\(2\sqrt{2} + 3\sqrt{2} = 5\sqrt{2}\)

Explanation

Surds can only be added when they are multiples of the same surd, like collecting like terms. Simplify each one first so both are multiples of \(\sqrt{2}\). Note that \(\sqrt{8} + \sqrt{18}\) is not \(\sqrt{26}\).

Mark scheme

1 mark
M1 for \(\sqrt{8} = 2\sqrt{2}\) or \(\sqrt{18} = 3\sqrt{2}\)
correct answer with no working scores full marks
1 mark
A1 for \(5\sqrt{2}\)
cao; \(\sqrt{26}\) scores 0

Question 2

2 marks

The area of a circle is \(\dfrac{49}{4}\pi\) cm².

Work out the circumference of the circle. Give your answer in terms of \(\pi\).

Model answer

\(\pi r^2 = \dfrac{49}{4}\pi\), so \(r^2 = \dfrac{49}{4}\)
\(r = \dfrac{7}{2}\) cm
Circumference \(= 2\pi r = 2 \times \dfrac{7}{2} \times \pi = 7\pi\) cm

Explanation

Divide the area by \(\pi\) to get \(r^2\), then square root the top and bottom of the fraction separately: \(\sqrt{49} = 7\) and \(\sqrt{4} = 2\). Keeping fractions avoids rounding.

Mark scheme

1 mark
M1 for \(r^2 = \dfrac{49}{4}\) leading to \(r = \dfrac{7}{2}\)
correct answer with no working scores full marks
1 mark
A1 for \(7\pi\) (cm)
cao

Question 3

3 marks

\((a + \sqrt{b})^2 = 14 + 6\sqrt{5}\), where \(a\) and \(b\) are positive integers.

Find the value of \(a\) and the value of \(b\).

Model answer

\((a + \sqrt{b})^2 = a^2 + 2a\sqrt{b} + b\)
Compare with \(14 + 6\sqrt{5}\):
Rational parts: \(a^2 + b = 14\)
Surd parts: \(2a\sqrt{b} = 6\sqrt{5}\), so \(a\sqrt{b} = 3\sqrt{5}\)
Try \(b = 5\), \(a = 3\): \(a^2 + b = 9 + 5 = 14\) ✓
\(a = 3\), \(b = 5\)

Explanation

Expand the bracket, then match the whole-number part and the surd part on each side. The surd part suggests \(b = 5\) and \(a = 3\), and the whole-number part confirms it. (The other option, \(a = 1\) and \(b = 45\), gives \(a^2 + b = 46\), which does not work.)

Mark scheme

1 mark
M1 for expanding: \(a^2 + 2a\sqrt{b} + b\)
correct answer with no working scores full marks
1 mark
M1 for \(a^2 + b = 14\) and \(2a\sqrt{b} = 6\sqrt{5}\)
1 mark
A1 for \(a = 3\) and \(b = 5\)
cao; correct values with a check \((3 + \sqrt{5})^2 = 14 + 6\sqrt{5}\) score full marks

Question 4

4 marks

A water trough is a prism. Its cross-section is a semicircle of diameter 40 cm.
The trough is 1.5 m long.

Work out the volume of water the trough holds when full. Give your answer in litres, in terms of \(\pi\).
(1 litre = 1000 cm³)

Model answer

Radius = 40 ÷ 2 = 20 cm
Area of cross-section \(= \frac{1}{2} \times \pi \times 20^2 = 200\pi\) cm²
Length = 1.5 m = 150 cm
Volume \(= 200\pi \times 150 = 30\,000\pi\) cm³
In litres: \(30\,000\pi \div 1000 = 30\pi\) litres

Explanation

Volume of a prism = area of cross-section × length. Use the radius (half the diameter) for the semicircle, and make sure all lengths are in the same unit before multiplying. Then divide by 1000 to change cm³ to litres.

Mark scheme

1 mark
P1 for the area of the cross-section: \(\frac{1}{2} \times \pi \times 20^2\) (= \(200\pi\) cm²)
correct answer with no working scores full marks
1 mark
P1 for converting the length: 1.5 m = 150 cm
or working in metres throughout
1 mark
P1 for the volume: \(200\pi \times 150\) (= \(30\,000\pi\) cm³)
ft their area and length
1 mark
A1 for \(30\pi\) (litres)
cao

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