Powers and roots - Edexcel GCSE Maths (Calculator) Exam Questions

Practise 28 exam-style questions on Powers and roots 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

Work out \(\sqrt{18.4^2 - 9.6^2}\)

Give your answer correct to 3 significant figures.

Model answer

\(18.4^2 = 338.56\) and \(9.6^2 = 92.16\)
\(338.56 - 92.16 = 246.4\)
\(\sqrt{246.4} = 15.697...\)
15.7 (3 s.f.)

Explanation

Work out the squares first, then subtract, then take the square root of the result. The square root does not split over a subtraction: \(\sqrt{18.4^2 - 9.6^2}\) is not \(18.4 - 9.6 = 8.8\).

Mark scheme

1 mark
M1 for 338.56 − 92.16 or 246.4 seen
1 mark
A1 for 15.7
awrt 15.7 (15.697...); correct answer with no working scores full marks

Question 2

3 marks

Find the smallest positive integer \(n\) such that \(1.5^n > 100\)

Model answer

\(1.5^{10} = 57.66...\)
\(1.5^{11} = 86.49...\) (less than 100)
\(1.5^{12} = 129.74...\) (more than 100)
So \(n = 12\)

Explanation

Use your calculator to try powers of 1.5, increasing \(n\) until the value first goes over 100. Show the two powers either side of 100 to prove that 12 is the smallest. Repeatedly multiplying by 1.5 on the calculator (ANS × 1.5) is quick.

Mark scheme

1 mark
M1 for a correct trial of a power of 1.5, e.g. \(1.5^{10} = 57.66...\)
correct answer with no working scores full marks
1 mark
M1 for \(1.5^{11} = 86.49...\) and \(1.5^{12} = 129.74...\)
1 mark
A1 for 12
cao; dependent on seeing both trials

Question 3

3 marks

\(N\) is a power of 2
\(N\) is also a square number and a cube number.
\(N\) is between 100 000 and 10 000 000

Find \(N\).

Model answer

\(N = 2^k\). For a square number \(k\) must be even; for a cube number \(k\) must be a multiple of 3. So \(k\) is a multiple of 6.
\(2^{12} = 4096\) (too small)
\(2^{18} = 262\,144\)
\(2^{24} = 16\,777\,216\) (too big)
\(N = 262\,144\) (\(= 512^2 = 64^3\))

Explanation

A power of a prime is a square when the power is even and a cube when the power is a multiple of 3, so here the power must be a multiple of 6. Then check which of \(2^6, 2^{12}, 2^{18}, ...\) lies in the range. Check: \(\sqrt{262\,144} = 512\) and \(\sqrt[3]{262\,144} = 64\).

Mark scheme

1 mark
M1 for recognising that the power of 2 must be a multiple of both 2 and 3, i.e. a multiple of 6
or listing powers of 2 that are square and cube numbers, e.g. 64, 4096; correct answer with no working scores full marks
1 mark
M1 for testing \(2^{12} = 4096\), \(2^{18} = 262\,144\) or \(2^{24} = 16\,777\,216\)
1 mark
A1 for 262 144
cao

Question 4

5 marks

A cube has edges of length 4.6 cm.

Ellie makes a new cube with twice the volume of this cube.
Fred makes a new cube with twice the surface area of this cube.

Ellie says, "The edges of my cube will be longer than the edges of Fred's cube."

Is Ellie correct?
You must show how you get your answer.

Model answer

Ellie: volume = \(4.6^3 = 97.336\) cm³, doubled = 194.672 cm³
Edge = \(\sqrt[3]{194.672} = 5.7956...\) cm
Fred: surface area = \(6 \times 4.6^2 = 126.96\) cm², doubled = 253.92 cm²
Area of one face = \(253.92 \div 6 = 42.32\) cm², edge = \(\sqrt{42.32} = 6.5053...\) cm
5.80 cm < 6.51 cm, so Ellie's edges are shorter. Ellie is not correct.

Explanation

Doubling the volume multiplies the edge by \(\sqrt[3]{2} \approx 1.26\), but doubling the surface area multiplies it by \(\sqrt{2} \approx 1.41\). Work out each new edge with the correct inverse: a cube root for volume and a square root for area (after dividing by the 6 faces).

Mark scheme

1 mark
P1 for the volume \(4.6^3\) (= 97.336) and doubling it (= 194.672)
1 mark
P1 for Ellie's edge, \(\sqrt[3]{194.672}\) (= 5.795...)
1 mark
P1 for the surface area \(6 \times 4.6^2\) (= 126.96) and doubling it (= 253.92)
1 mark
P1 for Fred's edge, \(\sqrt{253.92 \div 6}\) (= 6.505...)
1 mark
C1 for No, with correct edges, e.g. 5.80 cm < 6.51 cm
accept 5.79 to 5.80 and 6.50 to 6.51; dependent on P1 P1 P1 P1

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