Find the upper and lower bounds of rounded measurements and of calculations that use them, then choose a sensible accuracy. It is Higher only, aimed at grade 7. Free questions with answers are on this page. Worked examples show each line before you try the questions.
Error intervals become calculations: the maximum of a+b uses both upper bounds; the minimum of a−b is lower a minus upper b (smallest minus largest leftover). For a÷b, maximum is upper a over lower b. Write the error interval for each quantity first, then pick the bounds the operation needs.
A truncated value has a different interval from a rounded one. 'Calculate the maximum possible…' is upper bounds; 'to a suitable degree of accuracy' may want you to truncate a result so that even the bounds would round to the same displayed number. Show which bounds you used — that is the method mark.
Lower and upper as for error intervals. For a calculation, choose the combination of bounds that makes the result as large or as small as possible: for a−b, upper is (upper a) − (lower b).
Objectives
State the lower bound of a length to the nearest centimetre.
State the upper bound.
Multiply bounds to bound an area.
Key points
Write the error interval for a value rounded to a given accuracy, e.g. 4.55≤x<4.65.
Find the bounds of a value rounded to the nearest 10, 100, or to a number of decimal places or significant figures.
Choose the correct combination of bounds for adding, subtracting, multiplying and dividing.
Use bounds in a formula such as speed, density or area.
Worked examples
Example 1
Give the lower bound of a length measured as 9 cm to the nearest cm.
2 marks
Identify the unit of rounding: nearest 1 cm.
Halve the unit to find the error limit: 1 ÷ 2 = 0.5 cm.
Subtract 0.5 from 9: 9 - 0.5 = 8.5 cm.
Answer: 8.5
Common slips
Slip: Use the reading 9 as the lower bound. The reading is rounded, so the true value can be less.
Slip: Subtract 1 cm. The maximum rounding difference is half a unit.
Slip: Add 0.5 cm. Lower bound subtracts the error limit.
Example 2
Give the upper bound of a length measured as 15 cm to the nearest cm.
2 marks
Identify unit error: 1 cm ÷ 2 = 0.5 cm.
Add 0.5 to the measurement: 15 + 0.5 = 15.5 cm.
State the upper bound is 15.5 cm.
Answer: 15.5
Common slips
Slip: Add 1 cm. Error limit is half a unit.
Slip: Keep 15. Upper bound adds the error limit.
Slip: State 16 cm. 15.5 cm is the exact upper bound.
Example 3
Give the upper bound of the area of an 8 cm by 4 cm rectangle measured to the nearest cm.
3 marks
Find upper bound of each side: 8 + 0.5 = 8.5 and 4 + 0.5 = 4.5.
Multiply the upper bounds: 8.5 × 4.5 = 38.25.
State the maximum area: 38.25 cm².
Answer: 38.25
Common slips
Slip: Multiply 8 × 4. Both dimensions must be at their upper bounds for maximum area.
Slip: Multiply 8.5 × 4. Both lengths must be increased.
Slip: State 32. The upper bound of area is 38.25 cm².
Questions and answers
GCSE Maths bounds questions and answers. Try each one, then open the answer. Easy questions come first.
Question 1 Easy2 marks
Give the lower bound of a length measured as 7 cm to the nearest cm.
Show answer
Answer: 6.5
Half unit is 0.5.
7 - 0.5 = 6.5.
M1 — Half unit is 0.5.
A1 — 7 - 0.5 = 6.5.
Question 2 Easy2 marks
Give the upper bound of a length measured as 7 cm to the nearest cm.
Show answer
Answer: 7.5
Half unit is 0.5.
7 + 0.5 = 7.5.
M1 — Half unit is 0.5.
A1 — 7 + 0.5 = 7.5.
Question 3 Medium2 marks
Give the upper bound of the area of a 10 cm by 6 cm rectangle measured to the nearest cm.
Show answer
Answer: 68.25
Upper bounds: 10.5 and 6.5.
10.5 × 6.5 = 68.25.
M1 — Upper bounds: 10.5 and 6.5.
A1 — 10.5 × 6.5 = 68.25.
Question 4 Medium2 marks
Give the lower bound of the area of a 10 cm by 6 cm rectangle measured to the nearest cm.
Show answer
Answer: 52.25
Lower bounds: 9.5 and 5.5.
9.5 × 5.5 = 52.25.
M1 — Lower bounds: 9.5 and 5.5.
A1 — 9.5 × 5.5 = 52.25.
Question 5 Hard2 marks
Distance is 12 m to the nearest metre and time is 3 s to the nearest second. Find the upper bound of speed.
Show answer
Answer: 5
Upper distance 12.5, lower time 2.5.
12.5 ÷ 2.5 = 5.
M1 — Upper distance 12.5, lower time 2.5.
A1 — 12.5 ÷ 2.5 = 5.
Question 6 Hard2 marks
Distance is 17 m to the nearest metre and time is 3 s to the nearest second. Find the upper bound of speed.
Show answer
Answer: 7
Upper distance 17.5, lower time 2.5.
17.5 ÷ 2.5 = 7.
M1 — Upper distance 17.5, lower time 2.5.
A1 — 17.5 ÷ 2.5 = 7.
Question 7 Exam-style4 marks
A car travels 240 m, measured to the nearest 10 m, in 12 s, measured to the nearest second. Work out the upper bound of the average speed of the car. Give your answer to 3 significant figures.
Show answer
Answer: 21.3
Distance to the nearest 10 m: upper bound 240 + 5 = 245 m.
Time to the nearest second: lower bound 12 − 0.5 = 11.5 s.