Upper and lower bounds - Edexcel GCSE Maths (Non-calculator) Exam Questions

Practise 28 exam-style questions on Upper and lower bounds for Edexcel GCSE Maths (Non-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

A rectangular room is 4.3 m wide and 5.2 m long.
Both measurements are correct to 1 decimal place.

Work out the upper bound for the perimeter of the room.

Model answer

Upper bounds: width 4.35 m, length 5.25 m
Upper bound of perimeter \(= 2 \times (4.35 + 5.25) = 2 \times 9.6 = 19.2\) m

Explanation

The perimeter is greatest when both the width and the length are at their upper bounds. Add the upper bounds and double. Using the rounded values gives 19 m, which is not the upper bound.

Mark scheme

1 mark
M1 for using 4.35 and 5.25, e.g. \(2 \times (4.35 + 5.25)\)
correct answer with no working scores full marks
1 mark
A1 for 19.2 (m)
cao

Question 2

2 marks

A lift can carry a maximum load of 600 kg.
8 people get into the lift.
Each person has a mass of 72 kg, correct to the nearest kg.

Could the total mass of the 8 people be more than the maximum load? You must show how you get your answer.

Model answer

Upper bound of one person's mass = 72.5 kg
Upper bound of total mass \(= 8 \times 72.5 = 580\) kg
580 kg < 600 kg, so the total mass cannot be more than the maximum load.

Explanation

To check whether the load could be too heavy, use the heaviest possible mass of each person: the upper bound. Even then the total is only 580 kg, which is less than 600 kg.

Mark scheme

1 mark
M1 for \(8 \times 72.5\) (= 580)
1 mark
C1 for No, with 580 kg compared with 600 kg

Question 3

3 marks

\(v = u + at\)

\(u = 12\), correct to the nearest integer.
\(a = 2.5\), correct to 1 decimal place.
\(t = 4\), correct to the nearest integer.

Work out the upper bound for \(v\).

Model answer

Upper bounds: \(u = 12.5\), \(a = 2.55\), \(t = 4.5\)
\(2.55 \times 4.5 = 11.475\)
Upper bound of \(v = 12.5 + 11.475 = 23.975\)

Explanation

All the quantities are positive and are added or multiplied, so \(v\) is greatest when every value is at its upper bound. Note that \(a\) is given to 1 decimal place, so its upper bound is 2.55, not 3.

Mark scheme

1 mark
B1 for at least two of 12.5, 2.55 and 4.5 seen
correct answer with no working scores full marks
1 mark
M1 for \(12.5 + 2.55 \times 4.5\)
1 mark
A1 for 23.975
cao

Question 4

4 marks

\(x = 3.4\), correct to 1 decimal place.

By considering bounds, work out the value of \(x^{2}\) to a suitable degree of accuracy. You must give a reason for your answer.

Model answer

Lower bound of \(x^{2} = 3.35^{2} = 11.2225\)
Upper bound of \(x^{2} = 3.45^{2} = 11.9025\)
To 2 s.f. the bounds are 11 and 12, which are different.
To 1 s.f. both bounds are 10.
So \(x^{2} = 10\) to a suitable degree of accuracy (1 s.f.).

Explanation

A suitable degree of accuracy is the most accurate rounding that the lower and upper bounds agree on. Square both bounds (\(335^{2} = 112\,225\) and \(345^{2} = 119\,025\)), then round both to 2 s.f., then 1 s.f., until they match.

Mark scheme

1 mark
B1 for 3.35 and 3.45
1 mark
M1 for \(3.35^{2}\) or \(3.45^{2}\)
1 mark
A1 for 11.2225 and 11.9025
1 mark
C1 for 10, with a reason, e.g. both bounds round to 10 to 1 significant figure (but to 11 and 12 to 2 significant figures)

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