Upper & Lower Bounds in IGCSE Maths
Finding bounds for rounded values, and using those bounds in calculations involving area, speed, density and more.
0580 Extended · E1.9What are bounds?
When a measurement is rounded, the true value could be slightly higher or lower than the stated value. The upper bound is the largest value that would round down to the stated value. The lower bound is the smallest value that would round up to it.
Together, these define the range of possible true values — the limits of accuracy.
For any rounded value, the bounds sit half a unit of accuracy either side:
\[ \begin{aligned} \text{LB} &= \text{value} - \tfrac{1}{2}\text{unit} \\[6pt] \text{UB} &= \text{value} + \tfrac{1}{2}\text{unit} \end{aligned} \]The "unit" depends on what the value was rounded to. Rounded to the nearest 10? The unit is 10, so you go 5 either side. Rounded to 1 decimal place? The unit is 0.1, so you go 0.05 either side.
Finding upper & lower bounds
(a) A length is 47 m, correct to the nearest metre.
(b) A mass is 3.6 kg, correct to 1 decimal place.
(c) A time is 250 seconds, correct to the nearest 10 seconds.
Find the upper and lower bounds in each case.
Part (a)
Nearest metre, so the unit is 1. Half-unit = 0.5.
\[ \begin{aligned} \text{LB} &= 46.5 \text{ m} \\[6pt] \text{UB} &= 47.5 \text{ m} \end{aligned} \]Part (b)
1 d.p., so the unit is 0.1. Half-unit = 0.05.
\[ \begin{aligned} \text{LB} &= 3.55 \text{ kg} \\[6pt] \text{UB} &= 3.65 \text{ kg} \end{aligned} \]Part (c)
Nearest 10, so the unit is 10. Half-unit = 5.
\[ \begin{aligned} \text{LB} &= 245 \text{ s} \\[6pt] \text{UB} &= 255 \text{ s} \end{aligned} \]Bounds questions sometimes ask you to express the range using inequality notation. For Example 1(a):
\[ 46.5 \leqslant \text{length} < 47.5 \]Note: the lower bound uses \(\leqslant\) (could equal), but the upper bound uses \(<\) (strictly less than), because a value of exactly 47.5 would round up to 48.
Bounds in calculations Extended only
When rounded values are used in a calculation, you need to choose the right combination of upper and lower bounds to find the extreme result. The logic depends on the operation:
To get the largest possible answer:
Addition: UB + UB.
Subtraction: UB − LB.
Multiplication: UB × UB.
Division: UB ÷ LB.
To get the smallest possible answer:
Addition: LB + LB.
Subtraction: LB − UB.
Multiplication: LB × LB.
Division: LB ÷ UB.
Multiplication: biggest inputs give biggest output.
Addition: biggest inputs give biggest output.
Division: you get the biggest answer by dividing a big number by a small number.
Subtraction: the biggest difference comes from subtracting a small number from a big number.
A rectangle has length 12.4 cm and width 7.8 cm, both measured correct to 1 decimal place.
Find the upper bound of the perimeter.
UB of length = 12.45 cm, UB of width = 7.85 cm.
Perimeter = 2(length + width). To maximise, use UB + UB:
\[ \begin{aligned} \text{UB of P} &= 2(12.45 + 7.85) \\[6pt] &= 2 \times 20.3 \\[6pt] &= 40.6 \text{ cm} \end{aligned} \]Using the same rectangle from Example 2, find the lower bound of the area.
Lower bounds: length = 12.35 cm, width = 7.75 cm.
Area = length × width, and to minimise a product we use LB × LB:
\[ \begin{aligned} \text{LB of area} &= 12.35 \times 7.75 \\[6pt] &= 95.7125 \text{ cm}^2 \end{aligned} \]Bounds with speed & density Extended only
Speed, density, and pressure all involve division, so you need to think carefully about which bound goes on top and which goes on the bottom. The rule is always the same: to maximise a fraction, make the numerator as large as possible and the denominator as small as possible.
A car travels a distance of 180 km, correct to the nearest 10 km, in a time of 2.5 hours, correct to 1 decimal place.
Find the lower bound of the speed.
Speed = distance ÷ time. To minimise speed, use the smallest distance and the largest time:
\[ \begin{aligned} \text{LB of distance} &= 175 \text{ km} \\[6pt] \text{UB of time} &= 2.55 \text{ hours} \end{aligned} \] \[ \begin{aligned} \text{LB of speed} &= \frac{175}{2.55} \\[6pt] &= 68.627... \\[6pt] &= 68.6 \text{ km/h (3 s.f.)} \end{aligned} \]A block has mass 450 g, correct to the nearest 10 g, and volume 120 cm\(^3\), correct to the nearest 10 cm\(^3\).
Find the upper bound of the density.
Density = mass ÷ volume. To maximise, use UB of mass ÷ LB of volume:
\[ \begin{aligned} \text{UB of mass} &= 455 \text{ g} \\[6pt] \text{LB of volume} &= 115 \text{ cm}^3 \end{aligned} \] \[ \begin{aligned} \text{UB of density} &= \frac{455}{115} \\[6pt] &= 3.956... \\[6pt] &= 3.96 \text{ g/cm}^3 \end{aligned} \]Exam-style question
The diagram shows a rectangle with length 18.5 cm and width 6.4 cm. Both measurements are correct to 1 decimal place.
(a) Write down the upper bound of the length. [1]
(b) Write down the lower bound of the width. [1]
(c) Calculate the upper bound of the area of the rectangle. [2]
(d) Calculate the lower bound of the perimeter of the rectangle. [2]
(e) The rectangle is cut from a sheet of material. The offcut (waste material) has a mass of 12.0 g, correct to 3 significant figures.
Calculate the upper bound of the density of the offcut if its volume is 4.8 cm\(^3\), correct to 1 decimal place. [2]
Part (a)
1 d.p. means the unit is 0.1, half-unit is 0.05:
\[ \begin{aligned} \text{UB of length} &= 18.5 + 0.05 \\[6pt] &= 18.55 \text{ cm} \end{aligned} \]Part (b)
\[ \begin{aligned} \text{LB of width} &= 6.4 - 0.05 \\[6pt] &= 6.35 \text{ cm} \end{aligned} \]Part (c)
Area = length × width. To maximise, use UB × UB:
\[ \begin{aligned} \text{UB of area} &= 18.55 \times 6.45 \\[6pt] &= 119.6475 \text{ cm}^2 \end{aligned} \]Part (d)
Perimeter = 2(length + width). To minimise, use LB + LB:
\[ \begin{aligned} \text{LB of perimeter} &= 2(18.45 + 6.35) \\[6pt] &= 2 \times 24.8 \\[6pt] &= 49.6 \text{ cm} \end{aligned} \]Part (e)
Density = mass ÷ volume. To maximise, use UB ÷ LB.
Mass 12.0 g (3 s.f.): UB = 12.05 g.
Volume 4.8 cm\(^3\) (1 d.p.): LB = 4.75 cm\(^3\).
\[ \begin{aligned} \text{UB of density} &= \frac{12.05}{4.75} \\[6pt] &= 2.536... \\[6pt] &= 2.54 \text{ g/cm}^3 \end{aligned} \]This is a typical 8-mark bounds question. Parts (a) and (b) are straightforward — don't rush them, because every later part depends on getting the bounds right. Part (e) combines bounds with a division formula (density), which is the type of question that separates A* from A.
Common mistakes
To find the upper bound of a fraction, you need UB ÷ LB (big top, small bottom). Using UB ÷ UB is the most common error on speed and density questions.
"Correct to 3 significant figures" and "correct to 1 decimal place" give different units even when the numbers look similar. 12.0 to 3 s.f. has a unit of 0.1, but 12 to the nearest whole number has a unit of 1. Read the accuracy statement carefully.
The upper bound uses a strict inequality: \(x < \text{UB}\), not \(x \leqslant \text{UB}\). A value exactly equal to the upper bound would round up, not down. The lower bound uses \(\leqslant\).
When bounds are used in a multi-step calculation, keep full precision in the intermediate values. Only round the final answer to an appropriate degree of accuracy (usually 3 s.f. unless the question specifies otherwise).
Need help with bounds?
I teach IGCSE Maths one-to-one online. Book a free chat and we can discuss strategies to improve your understanding.
Book a Free Chat