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WJEC GCSE Maths

Topic questions Number Standard form

Standard form

24 questions · 72 marks · about 1.5 hours

Standard form questions

WJEC 3300 Standard form 24 questions

Unit 1: Non-calculator June 2024 Intermediate tier, Question 16, also on the Higher paper 3 marks

(a)

Express 0·0057 in standard form.

[1]
Reveal mark scheme Mark scheme for question 16(a)
(b)
Question 16, shown as an image

Give your answer in standard form.

[2]
Reveal mark scheme Mark scheme for question 16(b)

Unit 1: Non-calculator June 2024 Higher tier, Question 7, also on the Intermediate paper 3 marks

(a)

Express 0·0057 in standard form.

[1]
Reveal mark scheme Mark scheme for question 7(a)
(b)
Question 7, shown as an image

Give your answer in standard form.

[2]
Reveal mark scheme Mark scheme for question 7(b)

Unit 1: Non-calculator June 2023 Intermediate tier, Question 19, also on the Higher paper 4 marks

(a)

Express 0·0076 in standard form.

[1]
Reveal mark scheme Mark scheme for question 19(a)
(b)

Calculate the value of (3 × 1017) × (2 × 10−12).

Give your answer in standard form.

[1]
Reveal mark scheme Mark scheme for question 19(b)
(c)

Calculate the value of (2·3 × 104) + (5 × 103).

Give your answer in standard form.

[2]
Reveal mark scheme Mark scheme for question 19(c)

Unit 1: Non-calculator June 2023 Higher tier, Question 9, also on the Intermediate paper 4 marks

(a)

Express 0·0076 in standard form.

[1]
Reveal mark scheme Mark scheme for question 9(a)
(b)

Calculate the value of (3 × 1017) × (2 × 10−12).

Give your answer in standard form.

[1]
Reveal mark scheme Mark scheme for question 9(b)
(c)

Calculate the value of (2·3 × 104) + (5 × 103).

Give your answer in standard form.

[2]
Reveal mark scheme Mark scheme for question 9(c)

Unit 1: Non-calculator November 2021 Intermediate tier, Question 15, also on the Higher paper 3 marks

(a)

Express 0·0058 in standard form.

[1]
Reveal mark scheme Mark scheme for question 15(a)
(b)
Question 15, shown as an image

Give your answer in standard form.

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

Unit 1: Non-calculator November 2021 Higher tier, Question 6, also on the Intermediate paper 3 marks

(a)

Express 0·0058 in standard form.

[1]
Reveal mark scheme Mark scheme for question 6(a)
(b)
Question 6, shown as an image

Give your answer in standard form.

[2]
Reveal mark scheme Mark scheme for question 6(b)

Unit 2: Calculator-allowed November 2020 Intermediate tier, Question 16, also on the Higher paper 2 marks

Calculate the value of (3·2 × 107) × (8·3 × 10−2).

Give your answer in standard form.

[2]
Reveal mark scheme Mark scheme for question 16

Unit 2: Calculator-allowed November 2020 Higher tier, Question 7, also on the Intermediate paper 2 marks

Calculate the value of (3·2 × 107) × (8·3 × 10−2).

Give your answer in standard form.

[2]
Reveal mark scheme Mark scheme for question 7

Unit 1: Non-calculator June 2019 Intermediate tier, Question 16, also on the Higher paper 5 marks

(a)(i)

A mass is written as 430 kg, correct to the nearest 10 kg.

Circle the least possible value of this mass.

420 kg 425 kg 429·5 kg 426 kg 424·9 kg

[1]
Reveal mark scheme Mark scheme for question 16(a)(i)
(a)
(a)(ii)

A time period is written as 22 seconds, correct to the nearest second.

Circle the least possible value of this time period.

22 s 20 s 21 s 21·5 s 21·4 s

[1]
Reveal mark scheme Mark scheme for question 16(a)(ii)
(a)(iii)

A population is written as 85 people, correct to the nearest five people.

Circle the least possible value of this population.

83 people 81 people 84 people 82 people 80 people

[1]
Reveal mark scheme Mark scheme for question 16(a)(iii)
(b)
Question 16, shown as an image
[2]
Reveal mark scheme Mark scheme for question 16(b)

Unit 1: Non-calculator June 2019 Higher tier, Question 7, also on the Intermediate paper 5 marks

(a)
(a)(i)

A mass is written as 430 kg, correct to the nearest 10 kg.

Circle the least possible value of this mass.

420 kg 425 kg 429·5 kg 426 kg 424·9 kg

[1]
Reveal mark scheme Mark scheme for question 7(a)(i)
(a)(ii)

A time period is written as 22 seconds, correct to the nearest second.

Circle the least possible value of this time period.

22 s 20 s 21 s 21·5 s 21·4 s

[1]
Reveal mark scheme Mark scheme for question 7(a)(ii)
(a)(iii)

A population is written as 85 people, correct to the nearest five people.

Circle the least possible value of this population.

83 people 81 people 84 people 82 people 80 people

[1]
Reveal mark scheme Mark scheme for question 7(a)(iii)
(b)
Question 7, shown as an image
[2]
Reveal mark scheme Mark scheme for question 7(b)

Unit 1: Non-calculator June 2018 Intermediate tier, Question 14, also on the Higher paper 4 marks

(a)
Question 14, shown as an image
[2]
Reveal mark scheme Mark scheme for question 14(a)
(b)
Question 14, shown as an image

Give your answer in standard form.

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

Unit 1: Non-calculator June 2018 Higher tier, Question 5, also on the Intermediate paper 4 marks

(a)
Question 5, shown as an image
[2]
Reveal mark scheme Mark scheme for question 5(a)
(b)
Question 5, shown as an image

Give your answer in standard form.

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

Unit 1: Non-calculator November 2017 Intermediate tier, Question 13, also on the Higher paper 4 marks

(a)

Express 0·000 42 in standard form.

[1]
Reveal mark scheme Mark scheme for question 13(a)
(b)
Question 13, shown as an image

Give your answer in standard form.

[1]
Reveal mark scheme Mark scheme for question 13(b)
(c)
Question 13, shown as an image
[2]
Reveal mark scheme Mark scheme for question 13(c)

Unit 1: Non-calculator November 2017 Higher tier, Question 5, also on the Intermediate paper 4 marks

(a)

Express 0·000 42 in standard form.

[1]
Reveal mark scheme Mark scheme for question 5(a)
(b)
Question 5, shown as an image

Give your answer in standard form.

[1]
Reveal mark scheme Mark scheme for question 5(b)
(c)
Question 5, shown as an image
[2]
Reveal mark scheme Mark scheme for question 5(c)

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