What is N10: Fractions and decimals?

A terminating decimal has finitely many decimal places, so write it over the matching power of 10 and simplify: 0.375=375/1000=3/8. Convert a fraction to a decimal by dividing its numerator by its denominator.

How many marks are N10 questions worth?

In verified past-paper appearances, N10 questions on Edexcel GCSE Maths 1MA1 have been worth 1–4 marks.

Which past exam papers has N10 appeared in?

N10 has appeared in 19 verified past-paper questions indexed on this page, with direct links to the official question paper and mark scheme for each.

N10

Work interchangeably with terminating decimals and their corresponding fractions (such as 3.5 and 7/2 or 0.375 or 3/8); change recurring decimals into their corresponding fractions and vice versa

Fractions and decimals

Worked answers, methods and verified real exam appearances for N10 on Edexcel GCSE Maths 1MA1.

Explanation

A terminating decimal has finitely many decimal places, so write it over the matching power of $10$ and simplify: $0.375=\dfrac{375}{1000}=\dfrac38$ . Convert a fraction to a decimal by dividing its numerator by its denominator.
On Higher tier, recurring decimals are also converted algebraically: name the decimal $x$ , multiply by a power of $10$ so the repeating blocks align, subtract to remove the recurring part, and solve for $x$ .
A recurring dot or bar must cover the whole repeating block.
Foundation questions use terminating decimal–fraction equivalents only.
Examiners expect the resulting fraction in its simplest form.

Worked example

Higher tier: Convert $0.\dot2\dot7=0.272727\ldots$ to a fraction.

1.Let $x=0.272727\ldots$ , so $100x=27.272727\ldots$ .
2.Subtract: $100x-x=27$ , giving $99x=27$ .
3.Solve and simplify: $x=\dfrac{27}{99}=\dfrac3{11}$ .

Answer: $\dfrac3{11}$ .

Common mistakes

Don't write $0.375$ as $\dfrac{375}{100}$ or use the wrong place-value denominator.
Don't stop before simplifying the numerator and denominator fully.
Don't multiply by $10$ when the repeating block has two digits for a recurring decimal.

Exam tip

For a recurring decimal, align identical recurring tails before subtracting so they cancel exactly.

Worked practice

Tier 1 · Easy

Write

0.375

as a fraction in its simplest form.

(1)

(Total for Question 1 is 1 mark)

Mark scheme

Mark scheme for question 1
Question	Answer	Mark	Mark scheme
1	$\dfrac{3}{8}$	1	$0.375=\dfrac{375}{1000}$ . Divide numerator and denominator by $125$ to obtain $\dfrac{3}{8}$ .

Tier 2 · Standard

Write

\dfrac{37}{80}

as a decimal.

(1)

(Total for Question 2 is 1 mark)

Mark scheme

Mark scheme for question 2
Question	Answer	Mark	Mark scheme
2	$0.4625$	1	Make the denominator a power of $10$ : $\dfrac{37}{80}=\dfrac{4625}{10000}=0.4625$ .

Tier 3 · Hard

Which is greater,

0.56

\dfrac{9}{16}

? Work out the difference as a fraction.

(3)

(Total for Question 3 is 3 marks)

Mark scheme

Mark scheme for question 3
Question	Answer	Mark	Mark scheme
3	$\dfrac{9}{16}$ is greater. The difference is $\dfrac{1}{400}$ .	3	$\dfrac{9}{16}=0.5625$ , so it is greater than $0.56$ . The difference is $0.5625-0.56=0.0025=\dfrac{25}{10000}=\dfrac{1}{400}$ .

Tier 1 · Easy

Write

2\dfrac{9}{25}

as a decimal.

(1)

(Total for Question 4 is 1 mark)

Mark scheme

Mark scheme for question 4
Question	Answer	Mark	Mark scheme
4	$2.36$	1	$\dfrac{9}{25}=\dfrac{36}{100}=0.36$ , so $2\dfrac{9}{25}=2.36$ .

Tier 2 · Standard

A bottle is

0.74

full. Write the fraction of the bottle that is empty in its simplest form.

(2)

(Total for Question 5 is 2 marks)

Mark scheme

Mark scheme for question 5
Question	Answer	Mark	Mark scheme
5	$\dfrac{13}{50}$	2	$0.74=\dfrac{74}{100}=\dfrac{37}{50}$ , so the empty fraction is $1-\dfrac{37}{50}=\dfrac{13}{50}$ .

Tier 3 · Hard

Work out

0.875+\dfrac{3}{20}

. Write the result as a fully simplified fraction.

(3)

(Total for Question 6 is 3 marks)

Mark scheme

Mark scheme for question 6
Question	Answer	Mark	Mark scheme
6	$\dfrac{41}{40}$ (or $1\dfrac{1}{40}$ )	3	$0.875=\dfrac78=\dfrac{35}{40}$ and $\dfrac3{20}=\dfrac6{40}$ . Therefore the sum is $\dfrac{35}{40}+\dfrac6{40}=\dfrac{41}{40}$ .

Tier 2 · Standard

Two of these three values are equal:

0.55

\dfrac{11}{20}

and

\dfrac{14}{25}

. Write down the value that is not equal to the other two. Show your working.

(2)

(Total for Question 7 is 2 marks)

Mark scheme

Mark scheme for question 7
Question	Answer	Mark	Mark scheme
7	$\dfrac{14}{25}$	2	$\dfrac{11}{20}=\dfrac{55}{100}=0.55$ , while $\dfrac{14}{25}=\dfrac{56}{100}=0.56$ . Therefore $\dfrac{14}{25}$ is the value that is not equal to the other two.

Tier 3 · Hard

A number is added to

0.36

to make

\dfrac78

. Work out the number. Write it as a fully simplified fraction and as a decimal.

(3)

(Total for Question 8 is 3 marks)

Mark scheme

Mark scheme for question 8
Question	Answer	Mark	Mark scheme
8	$\dfrac{103}{200}=0.515$	3	$\dfrac78=0.875$ , so the number is $0.875-0.36=0.515$ . As a fraction, $0.515=\dfrac{515}{1000}=\dfrac{103}{200}$ .

Tier 3 · Hard

Higher only:

x=0.\dot{1}\dot{8}

and

y=0.2\dot{7}

. Work out

x+y

as a fully simplified fraction. You must show all your working.

(5)

(Total for Question 9 is 5 marks)

Mark scheme

Mark scheme for question 9
Question	Answer	Mark	Mark scheme
9	$\dfrac{91}{198}$	5	For $x=0.1818\ldots$ , subtracting $x$ from $100x$ gives $99x=18$ , so $x=\dfrac2{11}$ . For $y=0.2777\ldots$ , subtracting $10y$ from $100y$ gives $90y=25$ , so $y=\dfrac5{18}$ . Hence $x+y=\dfrac2{11}+\dfrac5{18}=\dfrac{36+55}{198}=\dfrac{91}{198}$ .

Q10

Tier 3 · Hard

Higher only: Write

\dfrac{29}{24}

as a recurring decimal. Show enough working to identify the recurring digit.

(3)

(Total for Question 10 is 3 marks)

Mark scheme

Mark scheme for question 10
Question	Answer	Mark	Mark scheme
10	$1.208\dot{3}$ (or $1.208333\ldots$ )	3	Long division gives $29\div24=1.208333\ldots$ . After the first $3$ , the remainder repeats, so every following digit is also $3$ . Therefore the recurring decimal is $1.208\dot3$ .

Verified exam appearances

Series	Paper	Question	Marks	Calculator	Tier	Links
2024-11	3H	Q14	3	Allowed	Higher	QP MS
2021-11	1H	Q13	1	Non-calculator	Higher	QP MS
2024-06	2F	Q3	1	Allowed	Foundation	QP MS
2019-06	2F	Q1	1	Allowed	Foundation	QP MS
2022-06	1F	Q5	1	Non-calculator	Foundation	QP MS
2022-11	2F	Q12	4	Allowed	Foundation	QP MS
2023-06	2F	Q2	1	Allowed	Foundation	QP MS
2023-11	2F	Q2	1	Allowed	Foundation	QP MS
2021-11	2F	Q15	2	Allowed	Foundation	QP MS
2021-11	1F	Q3	1	Non-calculator	Foundation	QP MS
2022-06	1H	Q12	3	Non-calculator	Higher	QP MS
2019-11	1H	Q15	3	Non-calculator	Higher	QP MS
2022-11	1F	Q1	1	Non-calculator	Foundation	QP MS
2022-11	3F	Q3	1	Allowed	Foundation	QP MS
2023-11	2H	Q13	3	Allowed	Higher	QP MS
2019-11	3F	Q3	1	Allowed	Foundation	QP MS
2023-06	3H	Q20	3	Allowed	Higher	QP MS
2022-11	2H	Q14	3	Allowed	Higher	QP MS
2024-06	1H	Q18	3	Non-calculator	Higher	QP MS

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