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WJEC GCSE Maths

Topic questions Pure mathematics Substitution into formulae

Substitution into formulae

38 questions · 172 marks · about 4 hours

Substitution into formulae questions

TopicSubstitution into formulae

WJEC 3300 Substitution into formulae 38 questions

Unit 2: Calculator-allowed June 2025 Foundation tier, Question 11, also on the Intermediate paper 3 marks

Use the formula 18 x + 5 y = t to find the value of y when x = 2∙5 and t = 60.

[3]
Reveal mark scheme Mark scheme for question 11

Unit 2: Calculator-allowed June 2025 Intermediate tier, Question 2, also on the Foundation paper 3 marks

Use the formula 18 x + 5 y = t to find the value of y when x = 2∙5 and t = 60.

[3]
Reveal mark scheme Mark scheme for question 2

Unit 2: Calculator-allowed June 2024 Foundation tier, Question 15, also on the Intermediate paper 3 marks

Use the formula v = u + at to find the value of t when v = 51·3, u = 2·3 and a = 9·8.

[3]
Reveal mark scheme Mark scheme for question 15

Unit 2: Calculator-allowed June 2024 Intermediate tier, Question 7, also on the Foundation paper 3 marks

Use the formula v = u + at to find the value of t when v = 51·3, u = 2·3 and a = 9·8.

[3]
Reveal mark scheme Mark scheme for question 7

Unit 1: Non-calculator November 2023 Intermediate tier, Question 1 6 marks

(a)

Write down the next two numbers in the following sequence.

[2]
Reveal mark scheme Mark scheme for question 1(a)
(b)
Figure for question 1
[2]
Reveal mark scheme Mark scheme for question 1(b)
(c)

Simplify the expression 5y + 7m − 3y −10m.

[2]
Reveal mark scheme Mark scheme for question 1(c)

Unit 2: Calculator-allowed November 2023 Intermediate tier, Question 3 5 marks

(a)
Figure for question 3
[2]
Reveal mark scheme Mark scheme for question 3(a)
(b)

A number machine is shown below.

Figure for question 3
(b)(i)

Calculate the OUTPUT when the INPUT is 1·5.

[1]
Reveal mark scheme Mark scheme for question 3(b)(i)
(b)(ii)

Write down an expression for the OUTPUT when the INPUT is n.

[2]
Reveal mark scheme Mark scheme for question 3(b)(ii)

Unit 2: Calculator-allowed November 2023 Intermediate tier, Question 13, also on the Higher paper 4 marks

The table below shows some of the values of y = x2 + 4x + 5 for values of x from –4 to 1.

x–4–3–2–101
y = x2 + 4x + 55225
(a)

Complete the table by finding the value of y for x maths symbol –2 and for x maths symbol 1.

[2]
Reveal mark scheme Mark scheme for question 13(a)
(b)

On the graph paper below, draw the graph of y = x2 + 4x + 5 for values of x

from –4 to 1.

Figure for question 13
[2]
Reveal mark scheme Mark scheme for question 13(b)

Unit 2: Calculator-allowed November 2023 Higher tier, Question 1, also on the Intermediate paper 4 marks

The table below shows some of the values of y = x2 + 4x + 5 for values of x from –4 to 1.

x–4–3–2–101
y = x2 + 4x + 55225
(a)

Complete the table by finding the value of y for x maths symbol –2 and for x maths symbol 1.

[2]
Reveal mark scheme Mark scheme for question 1(a)
(b)

On the graph paper below, draw the graph of y = x2 + 4x + 5 for values of x

from –4 to 1.

Figure for question 1
[2]
Reveal mark scheme Mark scheme for question 1(b)

Unit 2: Calculator-allowed June 2023 Intermediate tier, Question 4 4 marks

(a)

Write down the next two numbers in the following sequence.

Question 4, shown as an image
[2]
Reveal mark scheme Mark scheme for question 4(a)
(b)

f maths symbol 3g maths symbol 2h.

Calculate the value of f when g = 9·3 and h = −13·6.

[2]
Reveal mark scheme Mark scheme for question 4(b)

Unit 2: Calculator-allowed November 2022 Foundation tier, Question 5 7 marks

(a)

Simplify 3a maths symbol 2a maths symbol a.

[1]
Reveal mark scheme Mark scheme for question 5(a)
(b)
(b)(i)
Figure for question 5
[1]
Reveal mark scheme Mark scheme for question 5(b)(i)
(b)(ii)

How many dots will there be in Diagram 6?

[2]
Reveal mark scheme Mark scheme for question 5(b)(ii)
(c)
Figure for question 5
[3]
Reveal mark scheme Mark scheme for question 5(c)

Unit 2: Calculator-allowed November 2022 Higher tier, Question 3 5 marks

A rectangle has a width of 4x maths symbol10.

(a)
Figure for question 3
[1]
Reveal mark scheme Mark scheme for question 3(a)
(b)

The perimeter of the rectangle is 14x maths symbol 4.

Figure for question 3

Find the length of the rectangle in terms of x.

[4]
Reveal mark scheme Mark scheme for question 3(b)

Unit 1: Non-calculator June 2022 Foundation tier, Question 15, also on the Intermediate paper 4 marks

(a)

Find a whole number value of n, so that 7n maths symbol 9 is a multiple of 4.

You must show all your working.

When n =........ , 7n maths symbol 9 is a multiple of 4.

[2]
Reveal mark scheme Mark scheme for question 15(a)
(b)

Find a whole number value of n, so that 3n maths symbol 5 is a prime number.

You must show all your working.

When n =........ , 3n maths symbol 5 is a prime number.

[2]
Reveal mark scheme Mark scheme for question 15(b)

Unit 1: Non-calculator June 2022 Intermediate tier, Question 5, also on the Foundation paper 4 marks

(a)

Find a whole number value of n, so that 7n maths symbol 9 is a multiple of 4.

You must show all your working.

When n =........ , 7n maths symbol 9 is a multiple of 4.

[2]
Reveal mark scheme Mark scheme for question 5(a)
(b)

Find a whole number value of n, so that 3n maths symbol 5 is a prime number.

You must show all your working.

When n =........ , 3n maths symbol 5 is a prime number.

[2]
Reveal mark scheme Mark scheme for question 5(b)

Unit 1: Non-calculator June 2022 Intermediate tier, Question 11 5 marks

The table below shows some of the values of y maths symbol x2 maths symbol x maths symbol 4 for values of x from –3 to 3.

x–3–2–10123
y x2 x 42–2– 428
(a)

Complete the table by finding the values of y for x = –1 and for x = 1.

[2]
Reveal mark scheme Mark scheme for question 11(a)
(b)

On the graph paper opposite, draw the graph of y maths symbol x2 maths symbol x maths symbol 4 for values of x from

–3 to 3.

[2]
Reveal mark scheme Mark scheme for question 11(b)
(c)
Question 11, shown as an image
Figure for question 11
[1]
Reveal mark scheme Mark scheme for question 11(c)

Unit 1: Non-calculator November 2021 Intermediate tier, Question 14, also on the Higher paper 5 marks

(a)

Rearrange the following formula to make k the subject.

p = 3k + 2

[2]
Reveal mark scheme Mark scheme for question 14(a)
(b)

Does the midpoint of the straight line joining points (3, 15) and (7, 19) lie on the line y = 3x + 2?

You must show all your working.

[3]
Reveal mark scheme Mark scheme for question 14(b)

Unit 2: Calculator-allowed November 2021 Intermediate tier, Question 1 4 marks

(a)

Solve 7 x – 3 = 11.

[2]
Reveal mark scheme Mark scheme for question 1(a)
(b)

Find the value of 3f + 2g when f = 5·8 and g = −3·7.

[2]
Reveal mark scheme Mark scheme for question 1(b)

Unit 1: Non-calculator November 2021 Higher tier, Question 3 4 marks

Complete the table below.

Draw the graph of y = 7 − x2 for values of x between −2 and 4.

Use the graph paper below.

x−2−101234
y = 7 − x23763−9
Figure for question 3
[4]
Reveal mark scheme Mark scheme for question 3

Unit 1: Non-calculator November 2021 Higher tier, Question 5, also on the Intermediate paper 5 marks

(a)

Rearrange the following formula to make k the subject.

p = 3k + 2

[2]
Reveal mark scheme Mark scheme for question 5(a)
(b)

Does the midpoint of the straight line joining points (3, 15) and (7, 19) lie on the line y = 3x + 2?

You must show all your working.

[3]
Reveal mark scheme Mark scheme for question 5(b)

Unit 2: Calculator-allowed November 2020 Foundation tier, Question 11, also on the Intermediate paper 4 marks

(a)

Write down the next two numbers in the following sequence.

[2]
Reveal mark scheme Mark scheme for question 11(a)
(b)

Use the formula x = 4a + 3b to find the value of x when a = 7·2 and b = − 4·6.

[2]
Reveal mark scheme Mark scheme for question 11(b)

Unit 1: Non-calculator November 2020 Intermediate tier, Question 14, also on the Higher paper 6 marks

The table below shows some of the values of y = x2 − 4x − 3 for values of x from –2 to 5.

x−2–1012345
y = x2 − 4x − 32−3−6−6−32
(a)

Complete the table by finding the value of y for x = −2 and the value of y for x = 2.

[2]
Reveal mark scheme Mark scheme for question 14(a)
(b)

On the graph paper opposite, draw the graph of y = x2 − 4x − 3 for values of x from

–2 to 5.

[2]
Reveal mark scheme Mark scheme for question 14(b)
(c)

Draw the line y = 1 on the graph paper.

Write down the values of x where the line y = 1 cuts the curve y = x2 − 4x − 3.

Values of x are........ and........

Figure for question 14
[2]
Reveal mark scheme Mark scheme for question 14(c)

Unit 2: Calculator-allowed November 2020 Intermediate tier, Question 2, also on the Foundation paper 4 marks

(a)

Write down the next two numbers in the following sequence.

[2]
Reveal mark scheme Mark scheme for question 2(a)
(b)

Use the formula x = 4a + 3b to find the value of x when a = 7·2 and b = − 4·6.

[2]
Reveal mark scheme Mark scheme for question 2(b)

Unit 1: Non-calculator November 2020 Higher tier, Question 3, also on the Intermediate paper 6 marks

The table below shows some of the values of y = x2 − 4x − 3 for values of x from –2 to 5.

x−2–1012345
y = x2 − 4x − 32−3−6−6−32
(a)

Complete the table by finding the value of y for x = −2 and the value of y for x = 2.

[2]
Reveal mark scheme Mark scheme for question 3(a)
(b)

On the graph paper opposite, draw the graph of y = x2 − 4x − 3 for values of x from

–2 to 5.

[2]
Reveal mark scheme Mark scheme for question 3(b)
(c)

Draw the line y = 1 on the graph paper.

Write down the values of x where the line y = 1 cuts the curve y = x2 − 4x − 3.

Values of x are........ and........

Figure for question 3
[2]
Reveal mark scheme Mark scheme for question 3(c)

Unit 1: Non-calculator November 2019 Foundation tier, Question 13 5 marks

(a)

Calculate the value of 3 x + 4 y when x = −6 and y = 5.

[2]
Reveal mark scheme Mark scheme for question 13(a)
(b)

Simplify the expression 9g – 4f – 3g – 5f.

[2]
Reveal mark scheme Mark scheme for question 13(b)
(c)

Expand 4(3x – 5).

[1]
Reveal mark scheme Mark scheme for question 13(c)

Unit 1: Non-calculator November 2019 Intermediate tier, Question 3 7 marks

(a)

Calculate the value of 3 x + 4 y when x = −6 and y = 5.

[2]
Reveal mark scheme Mark scheme for question 3(a)
(b)

Simplify the expression 9g – 4f – 3g – 5f.

[2]
Reveal mark scheme Mark scheme for question 3(b)
(c)

Solve the equation 3m − 7 = 8.

[2]
Reveal mark scheme Mark scheme for question 3(c)
(d)

Expand 4(3x – 5).

[1]
Reveal mark scheme Mark scheme for question 3(d)

Unit 2: Calculator-allowed June 2019 Foundation tier, Question 12, also on the Intermediate paper 5 marks

(a)

Find the value of 5 f + 7 g when f = 3·8 and g = −2·6.

[2]
Reveal mark scheme Mark scheme for question 12(a)
(b)

Solve the following equation.

Give your answer correct to 1 decimal place.

7 x – 4 = 12

[3]
Reveal mark scheme Mark scheme for question 12(b)

Unit 2: Calculator-allowed June 2019 Intermediate tier, Question 2, also on the Foundation paper 5 marks

(a)

Find the value of 5 f + 7 g when f = 3·8 and g = −2·6.

[2]
Reveal mark scheme Mark scheme for question 2(a)
(b)

Solve the following equation.

Give your answer correct to 1 decimal place.

7 x – 4 = 12

[3]
Reveal mark scheme Mark scheme for question 2(b)

Unit 1: Non-calculator November 2018 Foundation tier, Question 14, also on the Intermediate paper 6 marks

(a)

Simplify the expression 15 x – 2 y – 7 x – 4 y.

[2]
Reveal mark scheme Mark scheme for question 14(a)
(b)

Solve the equation 2m – 7 = 12.

[2]
Reveal mark scheme Mark scheme for question 14(b)
(c)

Calculate the value of 5f + 3g when f = −4 and g = 7.

[2]
Reveal mark scheme Mark scheme for question 14(c)

Unit 1: Non-calculator November 2018 Intermediate tier, Question 3, also on the Foundation paper 6 marks

(a)

Simplify the expression 15 x – 2 y – 7 x – 4 y.

[2]
Reveal mark scheme Mark scheme for question 3(a)
(b)

Solve the equation 2m – 7 = 12.

[2]
Reveal mark scheme Mark scheme for question 3(b)
(c)

Calculate the value of 5f + 3g when f = −4 and g = 7.

[2]
Reveal mark scheme Mark scheme for question 3(c)

Unit 2: Calculator-allowed June 2018 Foundation tier, Question 12 6 marks

(a)

Solve.maths symbol 7

[1]
Reveal mark scheme Mark scheme for question 12(a)
(b)

Simplify 3f + 7g + f – 4g.

[2]
Reveal mark scheme Mark scheme for question 12(b)
(c)

Use the formula 5p + 2q = t to find the value of q when p = 4 and t = 24.6.

[3]
Reveal mark scheme Mark scheme for question 12(c)

Unit 2: Calculator-allowed June 2018 Intermediate tier, Question 1 6 marks

(a)
Figure for question 1
[1]
Reveal mark scheme Mark scheme for question 1(a)
(b)

Simplify 3f + 7g + f – 4g.

[2]
Reveal mark scheme Mark scheme for question 1(b)
(c)

Use the formula 5p + 2q = t to find the value of q when p = 4 and t = 24.6.

[3]
Reveal mark scheme Mark scheme for question 1(c)

Unit 1: Non-calculator June 2017 Foundation tier, Question 11 4 marks

(a)

Write down the next two numbers in the following sequence.

Question 11, shown as an image
[2]
Reveal mark scheme Mark scheme for question 11(a)
(b)

Find the value of 2x + 7y when x = –3 and y = 10.

[2]
Reveal mark scheme Mark scheme for question 11(b)

Unit 1: Non-calculator June 2017 Intermediate tier, Question 1 6 marks

(a)

Write down the next two numbers in the following sequence.

Question 1, shown as an image
[2]
Reveal mark scheme Mark scheme for question 1(a)
(b)

Find the value of 2x + 7y when x = –3 and y = 10.

[2]
Reveal mark scheme Mark scheme for question 1(b)
(c)

Simplify the expression 8k + 3m – 2k – 8m.

[2]
Reveal mark scheme Mark scheme for question 1(c)

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