Rounding and error intervals - Edexcel GCSE Maths (Calculator) Exam Questions

Practise 28 exam-style questions on Rounding and error intervals 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
\[ \sqrt{\dfrac{6.8^2 + 14.3}{2.9}} \]
Give your answer correct to 3 significant figures.

Model answer

\(6.8^2 + 14.3 = 46.24 + 14.3 = 60.54\)
\(60.54 \div 2.9 = 20.875...\)
\(\sqrt{20.875...} = 4.56901...\)
Answer: 4.57 (3 s.f.)

Explanation

Work out the numerator, divide by 2.9, and only then take the square root, keeping the full calculator value at each stage. Round only at the very end: the fourth significant figure is 9, so 4.56 rounds up to 4.57.

Mark scheme

1 mark
M1 for \(\dfrac{6.8^2 + 14.3}{2.9} = 20.87...\) or for 4.569...
accept 20.8 to 20.9
1 mark
A1 for 4.57
awrt 4.57 (4.5690...); correct answer with no working scores full marks

Question 2

2 marks

A number, x, is rounded to 1 decimal place.
The result is 7.2

Which of these numbers could be x?

7.14 7.15 7.249 7.25 7.2499

Model answer

The error interval for x is \(7.15 \le x < 7.25\).
7.14 is too small (it rounds to 7.1).
7.25 is not included (it rounds to 7.3).
x could be 7.15, 7.249 or 7.2499.

Explanation

Write down the error interval first: half of 0.1 is 0.05, so \(7.15 \le x < 7.25\). Then test each number against it. 7.15 is included because it rounds up to 7.2, and anything just below 7.25 (such as 7.2499) still rounds to 7.2.

Mark scheme

1 mark
B1 for at least two of 7.15, 7.249, 7.2499 with no incorrect numbers, or all three with one incorrect number
correct answer with no working scores full marks
1 mark
B1 for 7.15, 7.249 and 7.2499 only
cao

Question 3

3 marks

A number, x, is rounded to 1 decimal place. The result is 3.4
The same number, x, is truncated to 1 decimal place. The result is also 3.4

Write down the error interval for x.

Model answer

Rounding to 1 d.p. gives 3.4: \(3.35 \le x < 3.45\)
Truncating to 1 d.p. gives 3.4: \(3.4 \le x < 3.5\)
x must satisfy both:
\(3.4 \le x < 3.45\)

Explanation

Write the two error intervals separately, then find where they overlap. The largest of the two lower limits (3.4) and the smallest of the two upper limits (3.45) give the answer. A number such as 3.38 rounds to 3.4 but truncates to 3.3, so it is not allowed.

Mark scheme

1 mark
B1 for the rounding interval \(3.35 \le x < 3.45\)
correct answer with no working scores full marks
1 mark
B1 for the truncation interval \(3.4 \le x < 3.5\)
1 mark
B1 for \(3.4 \le x < 3.45\)
cao; the inequality signs must be correct

Question 4

4 marks

A spherical balloon has a volume of 2.8 litres.

Volume of a sphere \(= \dfrac{4}{3}\pi r^3\)

Work out the diameter of the balloon in centimetres.
Give your answer correct to 3 significant figures.

Model answer

2.8 litres = 2800 cm³
\(\dfrac{4}{3}\pi r^3 = 2800\)
\(r^3 = \dfrac{3 \times 2800}{4\pi} = 668.450...\)
\(r = \sqrt[3]{668.450...} = 8.74359...\) cm
Diameter = \(2 \times 8.74359... = 17.487...\) cm = 17.5 cm (3 s.f.)

Explanation

Change litres into cm³ (1 litre = 1000 cm³) so the radius comes out in centimetres. Rearrange the formula to make \(r^3\) the subject, take the cube root, and double it for the diameter. Keep full calculator values: rounding r to 8.7 first gives 17.4, the wrong answer.

Mark scheme

1 mark
M1 for 2.8 litres = 2800 cm³
correct answer with no working scores full marks
1 mark
M1 for \(r^3 = \dfrac{3 \times 2800}{4\pi}\) (= 668.4...)
oe
1 mark
M1 for \(\sqrt[3]{668.4...}\) (= 8.743...)
oe
1 mark
A1 for 17.5 cm
awrt 17.5 (17.487...); do not accept 8.74 (the radius)

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