Integer and fractional indices - Edexcel GCSE Maths (Non-calculator) Exam Questions

Practise 37 exam-style questions on Integer and fractional indices 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

Work out the value of \(27^{\frac{2}{3}}\)

Model answer

\(27^{\frac{2}{3}} = \left(\sqrt[3]{27}\right)^2 = 3^2 = 9\)

Explanation

In a fractional index \(\frac{m}{n}\), the denominator \(n\) is the root and the numerator \(m\) is the power. Taking the root first keeps the numbers small. A common error is to multiply 27 by \(\frac{2}{3}\) to get 18.

Mark scheme

1 mark
M1 for \(\sqrt[3]{27} = 3\) or \((\sqrt[3]{27})^2\) or \(\sqrt[3]{729}\)
correct answer with no working scores full marks
1 mark
A1 for 9
cao

Question 2

2 marks

Work out \(\sqrt[3]{-64} + 16^{\frac{1}{4}} \times 2^{-1}\)

Model answer

\(\sqrt[3]{-64} = -4\) because \((-4)^3 = -64\)
\(16^{\frac{1}{4}} = 2\) because \(2^4 = 16\)
\(2^{-1} = \dfrac{1}{2}\)
\(-4 + 2 \times \dfrac{1}{2} = -4 + 1 = -3\)

Explanation

Evaluate each power or root, then use the order of operations: multiply before adding. A negative number has a real cube root, because a negative number cubed is negative.

Mark scheme

1 mark
M1 for two of \(\sqrt[3]{-64} = -4\), \(16^{\frac{1}{4}} = 2\), \(2^{-1} = \dfrac{1}{2}\)
correct answer with no working scores full marks
1 mark
A1 for −3
cao

Question 3

3 marks

\(n\) is a positive integer.

Prove that \(2^{n+2} + 2^n\) is a multiple of 5.

Model answer

\(2^{n+2} = 2^n \times 2^2 = 4 \times 2^n\)
\(2^{n+2} + 2^n = 4 \times 2^n + 2^n = 5 \times 2^n\)
\(2^n\) is an integer, so \(5 \times 2^n\) is a multiple of 5.

Explanation

Use \(a^{m+n} = a^m \times a^n\) to write both terms in terms of \(2^n\), then take out the common factor. Checking a few values of \(n\) suggests the result but does not prove it for every \(n\).

Mark scheme

1 mark
M1 for \(2^{n+2} = 2^n \times 2^2\) (= \(4 \times 2^n\))
1 mark
M1 for \(2^n(4 + 1)\) or \(5 \times 2^n\)
1 mark
C1 for a conclusion that \(5 \times 2^n\) is a multiple of 5 (as \(2^n\) is an integer)
numerical examples only score 0

Question 4

4 marks

Work out the value of
\[\left(\dfrac{27}{8}\right)^{-\frac{2}{3}} \times \left(\dfrac{4}{9}\right)^{-\frac{3}{2}}\]
Give your answer as a fraction in its simplest form.

Model answer

\(\left(\dfrac{27}{8}\right)^{-\frac{2}{3}} = \left(\dfrac{8}{27}\right)^{\frac{2}{3}} = \left(\dfrac{2}{3}\right)^2 = \dfrac{4}{9}\)
\(\left(\dfrac{4}{9}\right)^{-\frac{3}{2}} = \left(\dfrac{9}{4}\right)^{\frac{3}{2}} = \left(\dfrac{3}{2}\right)^3 = \dfrac{27}{8}\)
\(\dfrac{4}{9} \times \dfrac{27}{8} = \dfrac{108}{72} = \dfrac{3}{2}\)

Explanation

For each fraction: the minus sign turns it upside down, the denominator of the index gives the root (of top and bottom), and the numerator gives the power. Cancelling before multiplying: \(\frac{4}{9} \times \frac{27}{8} = \frac{1 \times 3}{1 \times 2}\).

Mark scheme

1 mark
M1 for using the reciprocal for a negative index, e.g. \(\left(\dfrac{8}{27}\right)^{\frac{2}{3}}\) or \(\left(\dfrac{9}{4}\right)^{\frac{3}{2}}\)
correct answer with no working scores full marks
1 mark
M1 for \(\left(\dfrac{27}{8}\right)^{-\frac{2}{3}} = \dfrac{4}{9}\)
1 mark
M1 for \(\left(\dfrac{4}{9}\right)^{-\frac{3}{2}} = \dfrac{27}{8}\)
1 mark
A1 for \(\dfrac{3}{2}\)
oe, e.g. \(1\frac{1}{2}\)

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