Worked example · Easy
Nearest centimetre
A length is 43 cm correct to the nearest centimetre. State its bounds.
The rounding unit is 1 cm, so half a unit is 0.5 cm.
\[43-0.5=42.5,\qquad43+0.5=43.5\]\[\boxed{42.5\le L<43.5}\]Chapter 10
Limits of accuracy
Measurements that have been rounded are not exact. Learn to turn each rounded measurement into a range, then use the correct ends of those ranges to find the greatest and least possible results.
By the end of this chapter, you should be able to
- find the lower and upper bounds of a rounded measurement;
- write an error interval correctly using inequalities;
- find bounds for sums and differences;
- find bounds for products such as areas and volumes;
- find bounds for quotients such as speed, density and population density;
- decide which combination of lower and upper bounds gives the greatest or least possible result.
1
Subtopic 1
Bounds of measurement
A rounded measurement stands for a whole interval of possible true values. The lower bound (LB) is the smallest possible value in the interval; the upper bound (UB) is the boundary at the top of the interval.
The half-unit rule
If a value has been rounded to the nearest unit of size \\(u\\), then:
\[\text{LB}=\text{rounded value}-\frac{u}{2},\qquad \text{UB}=\text{rounded value}+\frac{u}{2}.\]The true value \\(x\\) lies in the error interval
\[\boxed{\text{LB}\le x<\text{UB}}.\]| Rounded to | Half-unit to subtract/add | Example |
|---|---|---|
| nearest 1 | 0.5 | 57 → 56.5 to 57.5 |
| nearest 0.1 | 0.05 | 4.7 → 4.65 to 4.75 |
| nearest 0.01 | 0.005 | 3.28 → 3.275 to 3.285 |
| nearest 10 | 5 | 340 → 335 to 345 |
| nearest 100 | 50 | 5600 → 5550 to 5650 |
Worked example · Medium
Decimal-place bound
A race time is 18.4 s correct to 1 decimal place. Find the error interval.
1 d.p. means nearest 0.1 s, so half the unit is 0.05 s.
\[18.4-0.05=18.35\]\[18.4+0.05=18.45\]\[\boxed{18.35\le t<18.45}\]Worked example · Exam-style
Significant-figure bound
A population is stated as 48 000 correct to 2 significant figures. Find its lower and upper bounds.
The second significant figure is the thousands digit, so the number was rounded to the nearest 1000.
Half of 1000 is 500.
\[48\,000-500=47\,500\]\[48\,000+500=48\,500\]\[\boxed{47\,500\le P<48\,500}\]Common mistake
Do not use the final written zero to decide the rounding unit. For significant figures, first identify the place value of the last significant digit.
Additional exam-focused worked example
Worked example · Exam-style
A distance is 480 m, correct to the nearest 10 metres. Express the true distance \(d\) as an inequality.
Step-by-step solution
Half the rounding unit is 5 m. \[\boxed{475\le d<485\text{ m}}\] The upper boundary is not included.
Practice: from confidence to exam-style
Work each question first, then select Show Answer to reveal the steps.
1Easy practice
Give the bounds of 36 cm measured to the nearest centimetre.
Show Answer
Steps: Half a centimetre is 0.5 cm: \(\boxed{35.5\le L<36.5\text{ cm}}\).
2Easy practice
A time is 8.2 s to the nearest 0.1 s. State its lower bound.
Show Answer
Steps: Half of 0.1 is 0.05. \(8.2-0.05=\boxed{8.15\text{ s}}\).
3Medium practice
Write the error interval for 6.38 kg correct to 2 decimal places.
Show Answer
Steps: Half of 0.01 is 0.005: \(\boxed{6.375\le m<6.385\text{ kg}}\).
4Medium practice
A length rounds to 320 mm to the nearest 10 mm. What is the greatest boundary value?
Show Answer
Steps: \[320+5=\boxed{325\text{ mm}}\] This is the upper boundary, not an included possible exact length.
5Exam-style practice
A race time is shown as 14.6 s, correct to 1 decimal place. Can its actual time be 14.65 s? Explain.
Show Answer
Steps: The interval is \(14.55\le t<14.65\). The upper endpoint rounds to 14.7 s, so \(\boxed{\text{no}}\).
2
Subtopic 2
Sums and differences of measurements
For addition, the largest total comes from adding the largest possible values and the smallest total comes from adding the smallest possible values. Subtraction is different because making the number being subtracted larger makes the result smaller.
Sum
\[\mathrm{LB}(A+B)=\mathrm{LB}(A)+\mathrm{LB}(B)\]\[\mathrm{UB}(A+B)=\mathrm{UB}(A)+\mathrm{UB}(B)\]Difference \\(A-B\\)
\[\mathrm{LB}=\mathrm{LB}(A)-\mathrm{UB}(B)\]\[\mathrm{UB}=\mathrm{UB}(A)-\mathrm{LB}(B)\]Worked example · Easy
Upper and lower bounds of a sum
Two lengths are 12.4 cm and 8.7 cm, each correct to the nearest 0.1 cm. Find the bounds of their total.
First length: \\(12.35\le a<12.45\\).
Second length: \\(8.65\le b<8.75\\).
\[\mathrm{LB}(a+b)=12.35+8.65=21.00\]\[\mathrm{UB}(a+b)=12.45+8.75=21.20\]So \\(\boxed{21.00\le a+b<21.20}\\).
Worked example · Medium
Bounds of a difference
A longer ribbon is 35.2 cm and a shorter ribbon is 12.6 cm, both correct to the nearest millimetre. Find the lower and upper bounds of their difference.
Nearest millimetre = nearest 0.1 cm.
Long ribbon: LB 35.15, UB 35.25.
Short ribbon: LB 12.55, UB 12.65.
Smallest difference:
\[35.15-12.65=22.50\text{ cm}\]Largest difference:
\[35.25-12.55=22.70\text{ cm}\]Therefore \\(\boxed{22.50\le d<22.70}\\).
Worked example · Exam-style
Will an object definitely fit?
A shelf opening is 60 cm wide, correct to the nearest centimetre. A box is 595 mm wide, correct to the nearest millimetre. Will the box definitely fit?
Opening lower bound:
\[60\text{ cm}\Rightarrow59.5\text{ cm}=595\text{ mm}\]Box upper bound:
\[595\text{ mm}\Rightarrow595.5\text{ mm}\]In the worst case the opening can be 595 mm while the box can be almost 595.5 mm.
So the box is not guaranteed to fit. \\(\boxed{\text{No}}\\)
Think “worst case”
For a “definitely fits?” question, compare the largest possible object with the smallest possible opening.
Additional exam-focused worked example
Worked example · Exam-style
Two planks are 2.4 m and 1.7 m long, each to the nearest 0.1 m. Find the greatest possible difference between their lengths.
Step-by-step solution
The first length is below 2.45 m; the second is at least 1.65 m. Max difference approaches \[2.45-1.65=\boxed{0.80\text{ m}}\]
Practice: from confidence to exam-style
Work each question first, then select Show Answer to reveal the steps.
1Easy practice
Lengths 12 cm and 9 cm are each correct to the nearest cm. Find the lower bound of their sum.
Show Answer
Steps: \[11.5+8.5=\boxed{20\text{ cm}}\]
2Easy practice
Two masses are 3.4 kg and 2.1 kg, each to 1 d.p. Find the upper bound of their total.
Show Answer
Steps: \[3.45+2.15=\boxed{5.60\text{ kg}}\] (an upper boundary).
3Medium practice
Two lengths are 18 cm and 11 cm, each to the nearest cm. Find the lower bound of their difference.
Show Answer
Steps: \[17.5-11.5=\boxed{6\text{ cm}}\] Minimum first minus maximum second.
4Medium practice
Times of 6.8 s and 4.3 s are to 1 d.p. Find the upper bound of their difference.
Show Answer
Steps: \[6.85-4.25=\boxed{2.60\text{ s}}\]
5Exam-style practice
Three timber lengths are 1.25, 2.36 and 0.94 m, each to nearest cm. Find the lower bound of their combined length.
Show Answer
Steps: Each was rounded to 0.01 m, so subtract 0.005 m from each: \[1.245+2.355+0.935=\boxed{4.535\text{ m}}\]
3
Subtopic 3
Multiplying and dividing measurements
For positive measurements, products are straightforward: use upper bounds together for the greatest product and lower bounds together for the least product. Quotients need opposite bounds in the denominator.
Product \\(AB\\)
\[\mathrm{LB}=\mathrm{LB}(A)\mathrm{LB}(B)\]\[\mathrm{UB}=\mathrm{UB}(A)\mathrm{UB}(B)\]Quotient \\(\frac AB\\)
\[\mathrm{LB}=\frac{\mathrm{LB}(A)}{\mathrm{UB}(B)}\]\[\mathrm{UB}=\frac{\mathrm{UB}(A)}{\mathrm{LB}(B)}\]Worked example · Easy
Bounds of an area
A rectangle is 8.0 cm by 5.0 cm, each measurement correct to the nearest 0.1 cm. Find the area bounds.
Width: \\(7.95\le w<8.05\\). Height: \\(4.95\le h<5.05\\).
\[A_{\min}=7.95\times4.95=39.3525\text{ cm}^2\]\[A_{\max}=8.05\times5.05=40.6525\text{ cm}^2\]So \\(\boxed{39.3525\le A<40.6525}\\).
Worked example · Medium
Bounds of average speed
A cyclist travels 14.2 km correct to 3 significant figures in 46 minutes correct to the nearest minute. Find the speed bounds in km/h to 3 significant figures.
Distance: \\(14.15\le d<14.25\\) km.
Time: \\(45.5\le t<46.5\\) minutes.
Fastest speed uses greatest distance and shortest time:
\[v_{\max}=\frac{14.25}{45.5}\times60\approx18.8\text{ km/h}\]Slowest speed uses shortest distance and longest time:
\[v_{\min}=\frac{14.15}{46.5}\times60\approx18.3\text{ km/h}\]\(\boxed{18.3\le v<18.8\text{ km/h}}\) to 3 s.f.
Worked example · Exam-style
Population density
A town has population 108 000 to the nearest 1000 and area 129 km² to the nearest km². Find the lower and upper bounds of its population density to 3 significant figures.
Population: LB 107 500, UB 108 500.
Area: LB 128.5, UB 129.5.
Lower density:
\[\frac{107500}{129.5}\approx830\text{ people/km}^2\]Upper density:
\[\frac{108500}{128.5}\approx844\text{ people/km}^2\]So the density is approximately \\(\boxed{830\le D<844}\\) people/km².
Universal exam strategy
- Write the LB and UB for every measured value.
- Write the formula you are using.
- Ask: what makes this formula as large as possible? Use those bounds.
- Ask: what makes it as small as possible? Use those bounds.
- Round only at the end unless the question instructs otherwise.
Additional exam-focused worked example
Worked example · Exam-style
A car travels 125 km (nearest km) in 2.5 h (nearest 0.1 h). Find an upper bound for its average speed.
Step-by-step solution
The distance is below 125.5 km and the time is at least 2.45 h. For the largest speed use UB(distance)/LB(time): \[\frac{125.5}{2.45}=\boxed{51.22\text{ km/h}}\text{ (approx.)}\]
Practice: from confidence to exam-style
Work each question first, then select Show Answer to reveal the steps.
1Easy practice
A rectangle is 5 cm by 4 cm, both to nearest cm. Find its smallest possible area.
Show Answer
Steps: \[A_{\min}=4.5\times3.5=\boxed{15.75\text{ cm}^2}\]
2Easy practice
A rectangle is 5 cm by 4 cm, both to nearest cm. Find its upper area bound.
Show Answer
Steps: \[A_{\max}\to5.5\times4.5=\boxed{24.75\text{ cm}^2}\]
3Medium practice
A rectangle measures 8.0 cm by 3.0 cm (nearest 0.1 cm). Give the lower bound of its area.
Show Answer
Steps: \[7.95\times2.95=\boxed{23.4525\text{ cm}^2}\]
4Medium practice
A journey covers 100 km (nearest km) in 2 h (nearest hour). Find the lower bound of average speed.
Show Answer
Steps: Smallest distance: 99.5 km. Largest time boundary: 2.5 h. \(\dfrac{99.5}{2.5}=\boxed{39.8\text{ km/h}}\).
5Exam-style practice
A cuboid is 3.2 cm × 4.5 cm × 2.6 cm, each correct to 1 d.p. Find its lower volume bound.
Show Answer
Steps: Use all three lower bounds: \[V_{\min}=3.15\times4.45\times2.55=\boxed{35.744625\text{ cm}^3}\]
Chapter 10 practice
The first questions build confidence with error intervals. The later questions require you to decide which combination of bounds gives the largest or smallest possible result.
A. Bounds of individual measurements
1Nearest centimetre[2 marks]
A length is 71 cm correct to the nearest centimetre. State its lower and upper bounds.
Show Answer
Half of 1 cm is 0.5 cm.
\[\boxed{70.5\le L<71.5}\]2Nearest tenth[2 marks]
A mass is 6.3 kg correct to 1 decimal place. Write its error interval.
Show Answer
Half of 0.1 is 0.05.
\[\boxed{6.25\le m<6.35}\]3Nearest hundredth[2 marks]
A time is 12.48 s correct to 2 decimal places. State its bounds.
Show Answer
Half of 0.01 is 0.005.
\[\boxed{12.475\le t<12.485}\]4Significant figures[3 marks]
A distance is 7300 m correct to 2 significant figures. Find its bounds.
Show Answer
The second significant figure is the hundreds digit, so the value was rounded to the nearest 100 m. Half of 100 is 50.
\[\boxed{7250\le d<7350}\]B. Sums and differences
5Total length[4 marks]
Lengths 7.2 m and 4.5 m are each correct to the nearest 0.1 m. Find the bounds of their sum.
Show Answer
First length: 7.15 to 7.25. Second: 4.45 to 4.55.
\[\mathrm{LB}=7.15+4.45=11.60\]\[\mathrm{UB}=7.25+4.55=11.80\]\(\boxed{11.60\le S<11.80}\\).
6Difference[4 marks]
Two rods measure 18.6 cm and 11.4 cm to the nearest 0.1 cm. Find the lower bound of the difference.
Show Answer
For the smallest difference, use the lower bound of the longer rod and the upper bound of the shorter rod.
\[18.55-11.45=\boxed{7.10\text{ cm}}\]7Perimeter bounds[5 marks]
A rectangle is 12 cm by 5 cm, both measurements correct to the nearest centimetre. Find the lower and upper bounds of its perimeter.
Show Answer
Length: 11.5 to 12.5; width: 4.5 to 5.5.
\[P_{\min}=2(11.5+4.5)=32\text{ cm}\]\[P_{\max}=2(12.5+5.5)=36\text{ cm}\]\(\boxed{32\le P<36}\\).
8Definitely fits?[5 marks]
A cupboard is 79.6 cm wide correct to the nearest 0.1 cm. A recess is 80 cm wide correct to the nearest centimetre. Is the cupboard guaranteed to fit?
Show Answer
Cupboard UB = 79.65 cm.
Recess LB = 79.5 cm.
Since 79.65 > 79.5, there are possible true measurements for which the cupboard is wider than the recess.
\(\boxed{\text{No, it is not guaranteed to fit.}}\\)
C. Products and quotients
9Area bounds[5 marks]
A rectangle measures 6.2 m by 3.8 m, both correct to 1 decimal place. Find the lower and upper bounds of its area.
Show Answer
Dimensions: 6.15 to 6.25 and 3.75 to 3.85.
\[A_{\min}=6.15\times3.75=23.0625\]\[A_{\max}=6.25\times3.85=24.0625\]\(\boxed{23.0625\le A<24.0625\text{ m}^2}\\).
10Smallest total width[4 marks]
Each of 10 identical pencils has width 8 mm correct to the nearest millimetre. What is the smallest possible total width?
Show Answer
One pencil LB = 7.5 mm.
\[10\times7.5=\boxed{75\text{ mm}}\]11Speed bounds[6 marks]
A runner covers 100 m correct to the nearest metre in 12.4 s correct to 1 decimal place. Find the lower and upper bounds of the average speed.
Show Answer
Distance: 99.5 to 100.5 m. Time: 12.35 to 12.45 s.
\[v_{\min}=\frac{99.5}{12.45}\approx7.99\text{ m/s}\]\[v_{\max}=\frac{100.5}{12.35}\approx8.14\text{ m/s}\]So approximately \\(\boxed{7.99\le v<8.14\text{ m/s}}\\).
12Density reasoning[6 marks]
A block has mass 520 g correct to the nearest 10 g and volume 65 cm³ correct to the nearest cm³. Find the greatest possible density.
Show Answer
Mass UB = 525 g. Volume LB = 64.5 cm³.
\[\rho_{\max}=\frac{525}{64.5}\approx\boxed{8.14\text{ g/cm}^3}\]D. Exam-style challenge
13Volume bounds[6 marks]
A cuboid measures 8.6 cm by 4.1 cm by 2.4 cm, each correct to the nearest 0.1 cm. Find the largest possible volume.
Show Answer
Use all upper bounds:
\[8.65\times4.15\times2.45\approx\boxed{87.96\text{ cm}^3}\]14Minimum width from area[6 marks]
A rectangle has area 210 cm² correct to 3 significant figures and length 17.8 cm correct to the nearest millimetre. Find the minimum possible width.
Show Answer
210 to 3 s.f. is to the nearest 1 cm², so area LB = 209.5 cm².
17.8 cm to nearest mm means length UB = 17.85 cm.
Minimum width uses smallest area and greatest length:
\[w_{\min}=\frac{209.5}{17.85}\approx\boxed{11.7\text{ cm}}\]15Compare race times[6 marks]
Runner A is recorded as 9.9 s correct to 1 d.p. Runner B is recorded as 9.93 s correct to 2 d.p. Can you be certain which runner was faster?
Show Answer
A: \\(9.85\le t_A<9.95\\).
B: \\(9.925\le t_B<9.935\\).
The intervals overlap, for example A could have run 9.90 s while B ran 9.93 s, but A could also have run 9.94 s while B ran 9.93 s.
\(\boxed{\text{No, the rounded data are not enough to be certain.}}\\)
16Tank capacity[7 marks]
A tank measures 80 cm by 75 cm by 90 cm, each correct to the nearest centimetre. Could its true capacity be exactly 550 litres? (1000 cm³ = 1 litre.)
Show Answer
Lower volume:
\[79.5\times74.5\times89.5\approx530\,105\text{ cm}^3\approx530\text{ L}\]Upper volume:
\[80.5\times75.5\times90.5\approx550\,207\text{ cm}^3\approx550.2\text{ L}\]Since 550 L lies between the lower and upper possible capacities, \\(\boxed{\text{yes, it could}}\\).
17Formula with a square[7 marks]
\(P=\frac{V^2}{R}\), where \\(V=6\\) and \\(R=1\\), each correct to the nearest whole number. Find the greatest possible value of \\(P\\).
Show Answer
V has UB 6.5. R has LB 0.5.
To maximise \\(P\\), maximise the numerator and minimise the denominator:
\[P_{\max}=\frac{6.5^2}{0.5}=\frac{42.25}{0.5}=\boxed{84.5}\]18Explain the bound choice[6 marks]
A density is calculated from \\(\rho=\frac{m}{V}\\). Explain, without calculating, which bounds of mass and volume are needed for a) the greatest density and b) the least density.
Show Answer
a) Greatest density: make the numerator as large as possible and denominator as small as possible, so use \\(\boxed{\mathrm{UB}(m)/\mathrm{LB}(V)}\\).
b) Least density: make the numerator as small as possible and denominator as large as possible, so use \\(\boxed{\mathrm{LB}(m)/\mathrm{UB}(V)}\\).
Chapter 10 high-grade checklist
Single measurement
I identify the rounding unit and add/subtract half of it.
Error interval
I write LB ≤ value < UB and remember that the upper bound is not included.
Sums
Smallest total uses all lower bounds; largest total uses all upper bounds.
Differences
Smallest difference uses LB − UB; largest difference uses UB − LB.
Products
For positive measurements, multiply lower bounds for the minimum and upper bounds for the maximum.
Quotients
For the greatest quotient I use UB numerator ÷ LB denominator, and reverse this for the least.