A length given as 17 m to the nearest metre could really be anything from 16.5 m up to (but not including) 17.5 m. Find these bounds and (Extended) use them to find the bounds of perimeters, areas and speeds.
The bounds are half a rounding unit below and above the given value: for 17 m to the nearest metre, \(16.5 \le L < 17.5\). (Extended) For a maximum, combine the values that make the answer biggest: when dividing, that is the upper bound over the LOWER bound.
Fig. 1: lengths that round to 17 m to the nearest metre. The filled circle includes 16.5; the open circle excludes 17.5, which would round up to 18
Section 01
Half a unit either way.
In the lab, set the rounded value and the accuracy, then drag the true value along the line. Watch the rounded value stay the same until you cross a bound, and notice that the upper bound itself already rounds up.
Find the rounding unit (1 m, 0.1 cm, 100 people, ...), then halve it.
Lower bound = value \(-\) half a unit; upper bound = value \(+\) half a unit.
Write the error interval as lower bound \(\le x <\) upper bound.
Worked example Nearest metre
A length \(L\) is 17 m, correct to the nearest metre. Write down the upper and lower bounds and the error interval.
The unit is 1 m, so half a unit is 0.5 m.
Lower bound \(17 - 0.5 = 16.5\) m; upper bound \(17 + 0.5 = 17.5\) m.
16.5 rounds to 17, so it is included; 17.5 rounds to 18, so it is not (Fig. 1).
Answer: Lower 16.5 m, upper 17.5 m; \(16.5 \le L < 17.5\)
Worked example Decimal places and significant figures
Find the bounds of (a) 4.7 cm, correct to 1 decimal place (b) 3400, correct to the nearest 100 (c) 2.4 kg, correct to 2 significant figures.
(a) Unit 0.1, half 0.05: \(4.65\) and \(4.75\).
(b) Unit 100, half 50: \(3350\) and \(3450\).
(c) The second significant figure is in the tenths, so the unit is 0.1 and half is 0.05: \(2.35\) and \(2.45\).
Answer: (a) 4.65 cm, 4.75 cm (b) 3350, 3450 (c) 2.35 kg, 2.45 kg
Examiner trap
The upper bound of 17 m is 17.5 m, not 17.4 or 17.49. Cambridge writes the upper bound as 17.5, even though 17.5 itself is excluded from the interval.
Section 02
Bounds in context.
Read the accuracy from the wording: 'to the nearest 5 kg', 'to the nearest tenth of a second', 'to 1 decimal place'. The half unit always comes from that accuracy, not from the size of the number.
Worked example Nearest 5
A sack of rice has mass \(m\) = 60 kg, correct to the nearest 5 kg. Write the error interval for \(m\).
The unit is 5 kg, so half a unit is 2.5 kg.
\(60 - 2.5 = 57.5\) and \(60 + 2.5 = 62.5\).
Answer: \(57.5 \le m < 62.5\)
Your turn
A sprinter's time is 9.8 s, correct to the nearest tenth of a second. Write the error interval for the time \(t\).
Unit 0.1 s, half 0.05 s: \(9.75 \le t < 9.85\).
Examiner trap
60 kg to the nearest 5 kg is NOT \(59.5 \le m < 60.5\). The half unit is half of 5, not half of 1.
Stage 1 · Learn
Check your understanding
Four quick checks. Each correct answer earns XP and records this skill on your profile.
Quick check+10 XP
A length is 17 m, correct to the nearest metre. What is its upper bound?
Quick check+10 XP
A length is 4.7 cm, correct to 1 decimal place. What is its lower bound?
Quick check+10 XP
A crowd is 3400 people, correct to the nearest 100. What is the lower bound?
Quick check+10 XP
(Extended) A rectangle is 6 cm by 4 cm, each to the nearest centimetre. What is the upper bound of its area?
Section 03
Bounds of calculations.
Extended. When rounded values are combined, choose the bound of each one that pushes the answer the way you want.
Calculation
Maximum
Minimum
\(a + b\)
\(a_{\max} + b_{\max}\)
\(a_{\min} + b_{\min}\)
\(a - b\)
\(a_{\max} - b_{\min}\)
\(a_{\min} - b_{\max}\)
\(a \times b\)
\(a_{\max} \times b_{\max}\)
\(a_{\min} \times b_{\min}\)
\(a \div b\)
\(a_{\max} \div b_{\min}\)
\(a_{\min} \div b_{\max}\)
Worked example Perimeter and area (Extended)
A rectangle measures 8 cm by 5 cm, each correct to the nearest centimetre. Calculate the upper bound of (a) the perimeter (b) the area.
A runner covers 100 m, correct to the nearest metre, in 12.4 s, correct to the nearest 0.1 s. Find the lower bound of the average speed.
Speed = distance \(\div\) time. For the smallest speed, use the smallest distance and the LARGEST time.
Distance bounds 99.5 and 100.5 m; time bounds 12.35 and 12.45 s.
\(99.5 \div 12.45 = 7.9919\dots\)
Answer: 7.99 m/s (3 s.f.)
Your turn
(Extended) Two rods are 15 cm and 6 cm long, each to the nearest centimetre. Find the largest possible difference in their lengths.
Largest minus smallest: \(15.5 - 5.5 = 10\) cm.
Examiner trap
For the lower bound of a speed, do NOT use both lower bounds: \(99.5 \div 12.35 = 8.06\) is bigger than the true minimum. Dividing by the larger time gives the smaller answer.
Stage 2 · Exam
Exam-style questions
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Finish the four checks above to unlock the exam questions
Exam style+20 XP
The mass \(m\) of a sack is 60 kg, correct to the nearest 5 kg. Which is the error interval for \(m\)?
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A time is 12.4 s, correct to the nearest 0.1 s. Which of these could NOT be the actual time?
Exam style+20 XP
(Extended) A runner covers 100 m, correct to the nearest metre, in 12.4 s, correct to the nearest 0.1 s. Find the lower bound of the average speed.
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(Extended) A rectangle is 8 cm by 5 cm, each correct to the nearest centimetre. Find the upper bound of its perimeter.
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