Integer indices include positive, zero and negative powers. For non-zero a, a^0=1 and a^-n=1/a^n; the negative index creates a reciprocal, not a negative answer.

How many marks are N7 questions worth?

In verified past-paper appearances, N7 questions on Edexcel GCSE Maths 1MA1 have been worth 1–4 marks.

Which past exam papers has N7 appeared in?

N7 has appeared in 22 verified past-paper questions indexed on this page, with direct links to the official question paper and mark scheme for each.

Calculate with roots, and with integer and fractional indices

Indices

Worked answers, methods and verified real exam appearances for N7 on Edexcel GCSE Maths 1MA1.

Explanation

Integer indices include positive, zero and negative powers. For non-zero $a$ , $a^0=1$ and $a^{-n}=\dfrac{1}{a^n}$ ; the negative index creates a reciprocal, not a negative answer.
Roots reverse powers, including odd roots of negative numbers such as $\sqrt[3]{-216}=-6$ .
Higher-tier questions also use fractional indices: $a^{1/n}=\sqrt[n]{a}$ and $a^{m/n}=\sqrt[n]{a^m}$ .
Foundation questions stay with roots and integer indices.
Choose the form that makes evaluation simplest, keep brackets around a negative base, and show the reciprocal or root step very clearly before giving the final value.

Worked example

Higher tier: Work out $81^{3/4}$ .

1.Interpret the denominator as a root: $81^{3/4}=(\sqrt[4]{81})^3$ .
2.Evaluate $\sqrt[4]{81}=3$ .
3.Calculate $3^3=27$ .

Answer: $27$ .

Common mistakes

Don't write $4^{-2}=-16$ instead of taking the reciprocal.
Don't treat $a^{1/2}$ as $a\div2$ rather than $\sqrt a$ .
Don't evaluate $(-3)^2$ and $-3^2$ as though the brackets made no difference.

Exam tip

On Foundation, show the reciprocal for a negative integer index; on Higher, rewrite a fractional index as a root before evaluating.

Worked practice

Tier 1 · Easy

Work out

4^{-2}

(1)

(Total for Question 1 is 1 mark)

Mark scheme

Mark scheme for question 1
Question	Answer	Mark	Mark scheme
1	$\dfrac{1}{16}$	1	A negative index means take the reciprocal: $4^{-2}=\dfrac{1}{4^2}=\dfrac{1}{16}$ .

Tier 2 · Standard

Work out

\sqrt[3]{-216}

(1)

(Total for Question 2 is 1 mark)

Mark scheme

Mark scheme for question 2
Question	Answer	Mark	Mark scheme
2	$-6$	1	The cube root is the number whose cube is $-216$ . Since $(-6)^3=-216$ , $\sqrt[3]{-216}=-6$ .

Tier 3 · Hard

Work out

\sqrt{144}+5^{-2}

. Give an exact answer.

(3)

(Total for Question 3 is 3 marks)

Mark scheme

Mark scheme for question 3
Question	Answer	Mark	Mark scheme
3	$\dfrac{301}{25}$	3	$\sqrt{144}=12$ and $5^{-2}=1/5^2=1/25$ . Therefore $12+\dfrac{1}{25}=\dfrac{300}{25}+\dfrac{1}{25}=\dfrac{301}{25}$ .

Tier 1 · Easy

Work out

10^0

(1)

(Total for Question 4 is 1 mark)

Mark scheme

Mark scheme for question 4
Question	Answer	Mark	Mark scheme
4	$1$	1	Any non-zero number raised to the power $0$ is $1$ , so $10^0=1$ .

Tier 2 · Standard

Work out

2^{-3}+(-2)^3

(2)

(Total for Question 5 is 2 marks)

Mark scheme

Mark scheme for question 5
Question	Answer	Mark	Mark scheme
5	$-\dfrac{63}{8}$ (or $-7\dfrac{7}{8}$ or $-7.875$ )	2	$2^{-3}=\dfrac{1}{2^3}=\dfrac18$ and $(-2)^3=-8$ . Therefore $\dfrac18-8=\dfrac18-\dfrac{64}{8}=-\dfrac{63}{8}$ .

Tier 3 · Hard

Work out

(-3)^4\times3^{-2}\div\sqrt{81}

(3)

(Total for Question 6 is 3 marks)

Mark scheme

Mark scheme for question 6
Question	Answer	Mark	Mark scheme
6	$1$	3	$(-3)^4=81$ , $3^{-2}=\dfrac19$ and $\sqrt{81}=9$ . Therefore $81\times\dfrac19\div9=9\div9=1$ .

Tier 2 · Standard

5^n=\dfrac{1}{625}

. Work out the value of

n

and then work out

(-n)^2

(3)

(Total for Question 7 is 3 marks)

Mark scheme

Mark scheme for question 7
Question	Answer	Mark	Mark scheme
7	$n=-4$ and $(-n)^2=16$	3	$625=5^4$ , so $\dfrac1{625}=5^{-4}$ and therefore $n=-4$ . This gives $(-n)^2=4^2=16$ .

Tier 3 · Hard

Work out

(-4)^3\times2^{-5}-9^0

(3)

(Total for Question 8 is 3 marks)

Mark scheme

Mark scheme for question 8
Question	Answer	Mark	Mark scheme
8	$-3$	3	$(-4)^3=-64$ , $2^{-5}=\dfrac1{32}$ and $9^0=1$ . Therefore $-64\times\dfrac1{32}-1=-2-1=-3$ .

Tier 3 · Hard

Higher only: Work out

81^{-3/4}+32^{2/5}

. Give your answer as a fully simplified fraction.

(4)

(Total for Question 9 is 4 marks)

Mark scheme

Mark scheme for question 9
Question	Answer	Mark	Mark scheme
9	$\dfrac{109}{27}$	4	$81^{1/4}=3$ , so $81^{-3/4}=3^{-3}=\dfrac1{27}$ . Also $32^{1/5}=2$ , so $32^{2/5}=2^2=4$ . Hence the total is $4+\dfrac1{27}=\dfrac{109}{27}$ .

Q10

Tier 3 · Hard

Higher only: The positive integer

n

satisfies

n^{3/2}=125

. Work out the value of

n

(3)

(Total for Question 10 is 3 marks)

Mark scheme

Mark scheme for question 10
Question	Answer	Mark	Mark scheme
10	$n=25$	3	$n^{3/2}=(\sqrt n)^3$ . Since $125=5^3$ , $\sqrt n=5$ . Squaring gives $n=25$ ; checking, $25^{3/2}=5^3=125$ .

Verified exam appearances

Series	Paper	Question	Marks	Calculator	Tier	Links
2022-11	1H	Q17	2	Non-calculator	Higher	QP MS
2019-11	3H	Q21	2	Allowed	Higher	QP MS
2022-06	2H	Q8	2	Allowed	Higher	QP MS
2024-11	1H	Q10	4	Non-calculator	Higher	QP MS
2019-11	1H	Q19	3	Non-calculator	Higher	QP MS
2023-06	1F	Q29	2	Non-calculator	Foundation	QP MS
2021-11	3H	Q12	2	Allowed	Higher	QP MS
2023-06	1H	Q11	3	Non-calculator	Higher	QP MS
2019-06	3H	Q12	1	Allowed	Higher	QP MS
2022-06	1H	Q18	4	Non-calculator	Higher	QP MS
2022-06	3H	Q9	4	Allowed	Higher	QP MS
2024-06	1H	Q20	3	Non-calculator	Higher	QP MS
2024-06	1F	Q18	1	Non-calculator	Foundation	QP MS
2021-11	1H	Q9	4	Non-calculator	Higher	QP MS
2023-11	1H	Q2	4	Non-calculator	Higher	QP MS
2019-06	1H	Q8	4	Non-calculator	Higher	QP MS
2023-11	1F	Q19	4	Non-calculator	Foundation	QP MS
2022-06	3F	Q5	1	Allowed	Foundation	QP MS
2023-11	1H	Q14	3	Non-calculator	Higher	QP MS
2023-06	1H	Q14	4	Non-calculator	Higher	QP MS
2022-11	1F	Q21	2	Non-calculator	Foundation	QP MS
2022-11	1H	Q3	2	Non-calculator	Higher	QP MS

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