Calculating with Bounds
Find the biggest and smallest values of longer formulas made from rounded numbers, and give an answer to a suitable degree of accuracy.
In the exam for Cambridge IGCSE Maths 0580, Edexcel IGCSE Maths 4MA1, Cambridge O Level Maths 4024 · 41 real past-paper questions below
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What Changes in a Longer Formula
Calculating with bounds means a formula with two, three or more rounded numbers. You want the biggest or the smallest answer it could give.
Go one number at a time. Ask: if this number gets bigger, does the answer go up or down?
| Where the number is | For the upper bound |
| added on | its upper bound |
| taken away | its lower bound |
| multiplied in | its upper bound |
| squared or cubed (a length, a radius) | its upper bound |
| on top of a fraction | its upper bound |
| on the bottom | its lower bound |
Rule
For the lower bound of the answer, swap every choice.
Bounds of a Formula
The 3-mark question: a formula like . Here , and , all to 2 significant figures. Find the upper bound of , to 1 decimal place.
Worked example
- 1
— 9.6 to 2 s.f. can be 0.05 either way. p is on top, so its upper bound makes the answer biggest.
- 2
— q is taken away. Taking away the smallest q leaves the most, so use its lower bound.
- 3
— 0.35 to 2 s.f. can be 0.005 either way. t is on the bottom, and dividing by a smaller number gives a bigger answer.
- 4
— Put the three chosen bounds into the formula.
- 5
— Work out the top first: 9.65 − 4.05 = 5.6.
- 6
— Divide. Keep every figure on your calculator for now.
- 7
— Round only at the end, to 1 decimal place as the question asks.
Key idea
The 2-mark version is a plain difference. The lower bound of is the lower bound of minus the upper bound of .
Common mistake
Using 4.15 because it is an upper bound question. is taken away, so its lower bound makes the answer bigger.
Squares, Cubes and Exact Numbers
Density and volume questions often cube a rounded length. Find the length's bound first, then cube it.
Formula
is a rounded length, and UB means upper bound. The same works for a square, and for the lower bound: .
A metal cube has mass 52.3 g and side 1.9 cm, both to 1 decimal place. Find the upper bound of its density, to 3 s.f.
Worked example
- 1
— 1 d.p. means 0.05 either way. Mass is on top of density = mass ÷ volume, so use its upper bound.
- 2
— The volume is on the bottom. The shortest side gives the smallest volume, and a smaller bottom gives a bigger answer.
- 3
— Cube the bound, not the rounded 1.9. Volume of a cube is side × side × side.
- 4
— Divide mass by volume, keeping every figure.
- 5
— Round at the end, to 3 s.f.
Key idea
Exact numbers have no bounds: a count like 8 boxes, “exactly 2 hours”, π, or the 2 in 2a. Use them as they are. Eight boxes of 12 kg, to the nearest kg, weigh at least kg.
Common mistake
Cubing first, then taking off 0.05: , then 6.809. Bound the side first, then cube it.
A Suitable Degree of Accuracy
The 4-mark question: "work out the value to a suitable degree of accuracy". Find both bounds, then round both until they agree.
A runner covers 452 metres, to the nearest metre, in 51.3 seconds, to 1 decimal place. Find the speed.
Worked example
- 1
— To the nearest metre means half a metre either way.
- 2
— To 1 decimal place means 0.05 either way.
- 3
— The most distance in the least time gives the fastest possible speed.
- 4
— The least distance in the most time gives the slowest possible speed.
- 5
— Round both the same way. They differ, so 3 s.f. is not safe.
- 6
— Try one figure fewer. Now they agree, so the true speed must round to 8.8.
- 7
— Give the answer with the reason: both bounds round to 8.8 to 2 s.f.
Rule
Round the upper and the lower bound the same way. Your answer is the most accurate rounding where they agree.
Common mistake
Giving 8.83, or rounding only one bound. Write both bounds and say they both round to 8.8.
Bounds in a Triangle
The 4-mark question: a triangle with a rounded side and rounded angles. Use the sine rule, then choose each bound.
Formula
Here and are sides, and and are the angles opposite them.
Side cm to 1 decimal place. Angles and to the nearest degree. Find the upper bound of .
Worked example
- 1
— 9.5 to 1 d.p. can be 0.05 either way. c is on top, so use its upper bound.
- 2
— To the nearest degree means 0.5° either way. sin B is on top, and a bigger angle (below 90°) has a bigger sine.
- 3
— sin C is on the bottom. The smaller angle gives a smaller sine, and dividing by less gives more.
- 4
— Put the three chosen bounds into the sine rule.
- 5
— Work out the top and the bottom on your calculator, in degree mode.
- 6
— Divide. Keep every figure for now.
- 7
— Round only at the end, to 3 significant figures.
Key idea
For angles below 90°, a bigger angle gives a bigger sine. Cosine goes the other way: a bigger angle gives a smaller cosine.
Common mistake
Using 55.5° on the bottom. A bigger bottom makes the answer smaller.
Bounds of an Angle
Some questions want the bound of an angle found with or . Bound the fraction first: a bigger fraction gives a bigger angle.
A ramp is 1.8 m high and its sloping length is 7.4 m, both to 1 decimal place. Find the lower bound of the angle the ramp makes with the ground.
Worked example
- 1
— The height is opposite the angle and the slope is the hypotenuse. To make the angle smallest, make this fraction smallest.
- 2
— The height is on top, so use its lower bound. 1 d.p. means 0.05 either way.
- 3
— The slope is on the bottom, so use its upper bound. A bigger bottom makes the fraction smaller.
- 4
— Divide, keeping every figure.
- 5
— Use the inverse sine key to turn the fraction into an angle.
- 6
— Round at the end, to 3 s.f.
Key idea
An angle found as works like a difference. Its upper bound uses the lower bounds of x and y.
Common mistake
Making the same choice with . For cosine a bigger fraction gives a smaller angle, so swap your choices.
How Many Fit? Round Down or Up
“What is the largest number of pieces?” and “how many tins must she buy?” need a whole number at the end. Choose the bounds, then round the right way.
A rope is 95 m long, to the nearest metre. It is cut into pieces 1.6 m long, to the nearest 10 cm. Find the largest possible number of pieces.
Worked example
- 1
— To the nearest metre means half a metre either way. The longest rope gives the most pieces.
- 2
— To the nearest 10 cm means 5 cm, or 0.05 m, either way. The shortest pieces give the most pieces.
- 3
— Divide the most rope by the shortest piece.
- 4
— Round down. The 0.61 is only part of a piece, so it does not count.
Key idea
For the least number of tins of paint you must buy to be sure, use the biggest area and the least that one tin covers. Then round up: a part-used tin still has to be bought.
Common mistake
Rounding 61.61 to 62. You cannot cut 62 whole pieces, so the answer is 61.
Which Method? Read the Question
Take each rounded number, find its bounds, then read what the question wants.
| You see | You do |
| Upper bound of a formula | each number: which bound pushes it up |
| Lower bound of a formula | swap every choice |
| Lower bound of | lower e, upper f |
| A suitable degree of accuracy | both bounds, round until they agree |
| A triangle with rounded angles | bound each angle, then the sine rule |
| A cube or square of a rounded length | bound the length, then cube or square it |
| An angle from or | bound the fraction first |
| The largest number of pieces | most ÷ least, then round down |
| A time in minutes, a speed in km/h | bound the minutes, then divide by 60 |
| Write down one bound | Upper and Lower Bounds |
Key idea
Keep every figure on your calculator until the end. Round only the final answer.
Key idea
Write down every bound you use before you calculate. Each correct bound can earn a mark, even if the final answer goes wrong.
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