Rounding Errors and Why We Round at the End

What a rounding error is, how rounding too early in a calculation spoils the final answer, and how to keep the full value so that you round only once, at the end.

Before you start

To round a number, look at the digit after the last one you keep. If it is 5 or more, round up. If it is less than 5, the last digit you keep stays as it is. So $3.14159 \approx 3.14$ (correct to 2 decimal places) and $3.14159 \approx 3.142$ (correct to 4 significant figures).

Significant figures are counted from the first non-zero digit: $0.03996 \approx 0.0400$ (correct to 3 significant figures). Unless a question says otherwise, an answer that is not exact is given correct to 3 significant figures.

The sign $\approx$ means "is approximately equal to".

What Is a Rounding Error?

A rounded number is easier to read and to use, but it is no longer the exact number. When 7.38 is rounded to 7.4, the value moves by 0.02. That small difference is called a rounding error. The word "error" here does not mean a mistake: it is the amount by which the rounded value is off.

Definition

A rounding error is the difference between a rounded value and the exact value.

To find it, subtract the smaller of the two values from the larger.

A rounding error is never more than half a unit of the place rounded to: at most 0.05 when rounding to 1 decimal place, at most 0.5 when rounding to the nearest whole number, and at most 50 when rounding to the nearest hundred.

Example 1: Finding a rounding error

Round each number as stated, and find the rounding error.
(a) 7.38, to 1 decimal place
(b) $12\,449$ , to the nearest hundred
(c) 0.0465, to 2 significant figures

Solution

(a) The digit after the first decimal place is 8, so round up: $7.38 \approx 7.4$ . The rounding error is $7.4 - 7.38 = 0.02$ .

(b) The digit after the hundreds place is 4, so the hundreds digit stays as it is: $12\,449 \approx 12\,400$ . The rounding error is $12\,449 - 12\,400 = 49$ .

(c) The first two significant figures are 4 and 6. The digit after them is 5, so round up: $0.0465 \approx 0.047$ . The rounding error is $0.047 - 0.0465 = 0.0005$ .

In (c) the last digit kept is in the thousandths place, and 0.0005 is half of 0.001. This is the largest error that rounding to the thousandths place can make.

(a) $7.38 \approx 7.4$ , and the rounding error is 0.02.
(b) $12\,449 \approx 12\,400$ , and the rounding error is 49.
(c) $0.0465 \approx 0.047$ , and the rounding error is 0.0005.

One rounding error on its own is small. The trouble starts when a rounded number is used in a further calculation.

Rounding Too Early

Many calculations take more than one step. The answer to one step that is used in the next step is called an intermediate value. If you round an intermediate value, its rounding error goes into the next step with it, and the next step can make the error bigger.

Example 2: Rounded early and rounded at the end

Mrs Tan's car travels 438 km on 35 litres of petrol. At the same rate, how far can it travel on a full tank of 52 litres? Give the answer correct to 3 significant figures.

Solution

Distance for 1 litre $= 438 \div 35 = 12.514285\ldots$ km. This is an intermediate value: it is not the answer, and it is needed for the next step.

Rounded early. Suppose the intermediate value is rounded to 3 significant figures, 12.5 km, before it is used: $12.5 \times 52 = 650$ km.

Rounded at the end. Keep the whole value in the calculator and multiply it by 52: $12.514285\ldots \times 52 = 650.742\ldots$ km, which is 651 km correct to 3 significant figures.

The two answers are different, and only the second is right. A check in one calculation: $438 \times 52 \div 35 = 650.742\ldots$

Why did the early answer go wrong? Rounding $12.514285\ldots$ to 12.5 made a rounding error of about 0.014 km. Multiplying by 52 made the error 52 times as large: $0.014285\ldots \times 52 = 0.742\ldots$ km, which is enough to change the third significant figure.

The car can travel 651 km on a full tank (correct to 3 significant figures). Rounding early gives 650 km, which is wrong.

How a rounding error grows

When a rounded number is multiplied by another number, its error is multiplied by that number too. In Example 2 an error of about 0.014 km became an error of about 0.74 km.
When a rounded number greater than 1 is squared or cubed, its error becomes larger as well (Example 4 below).
When rounded numbers are added, their errors are added too. They may partly cancel, or they may pile up: $2.4 + 3.4 + 1.4 = 7.2$ , but with each number rounded to the nearest whole number first, the sum comes to $2 + 3 + 1 = 6$ .
Every step that follows works on the error as well as on the number, so the earlier the rounding, the further the final answer can drift.

Rule: round once, at the end

In a calculation with more than one step, do not round an intermediate value. Calculate with every digit the calculator holds, and round only the final answer, to the accuracy the question asks for.

Rounding first is right when all you want is an estimate: $\sqrt{83} \times \sqrt[3]{130} \approx 9 \times 5 = 45$ . An estimate is a quick check of the size of an answer. It is not the answer: the calculator gives $46.151\ldots$ , which is 46.2 correct to 3 significant figures.

Keeping the Full Value

There are two things to get right: what you do on the calculator, and what you write in your working.

On the calculator

Carry on from the answer on the display. After $438 \div 35 =$ , press $\times$ 52 $=$ straight away. The calculator multiplies the whole value it holds, not just the digits you copied down.
Or use the Ans key, which holds the last answer in full, or store the value in the calculator's memory. Check how your own calculator does this.
Or key in the whole calculation in one go, with brackets where they are needed.

In your working

You need not copy out every digit. Write each intermediate value to at least one more significant figure than the answer needs: to at least 4 significant figures when the final answer is to be correct to 3 significant figures. Go on calculating with the full value in the calculator. Round the final answer once, and say how: "(correct to 3 significant figures)".

Example 3: A square root in the working

A square mat has an area of 2 m². Find its perimeter, correct to 3 significant figures.

Solution

The side of the square is $\sqrt{2}$ m, and the calculator shows $\sqrt{2}$ as 1.414213562. The perimeter is 4 times the side. The table shows the perimeter worked out with three different values for the side.

Value used for the side	4 × side	Correct to 3 significant figures
1.41 (3 significant figures)	$4 \times 1.41 = 5.64$	5.64, which is wrong
1.414 (4 significant figures)	$4 \times 1.414 = 5.656$	5.66
1.414213562 (the calculator value)	$4 \times 1.414213562 = 5.656854\ldots$	5.66

With the side rounded to 3 significant figures, the perimeter comes to 5.64 m, which is wrong in the third significant figure. The other two values both give 5.66 m.

Set the working out like this.
Side $= \sqrt{2} \approx 1.414$ m (correct to 4 significant figures)
Perimeter $= 4 \times \sqrt{2} = 5.6568\ldots \approx 5.66$ m (correct to 3 significant figures)

The perimeter of the mat is 5.66 m (correct to 3 significant figures).

Is one extra figure always enough?

Nearly always, but not always. Six of these mats laid in a row make a strip $6 \times \sqrt{2}$ m long. With 1.414 for $\sqrt{2}$ , the length is $6 \times 1.414 = 8.484$ , which rounds to 8.48 m. With the full value, it is $6 \times 1.414213562 = 8.485281\ldots$ , which rounds to 8.49 m. So 4 significant figures is the least to write down, and the full calculator value is what to calculate with.

Example 4: When part (b) uses the answer to part (a)

A cube has a volume of 50 cm³.
(a) Find the length of a side of the cube, correct to 3 significant figures.
(b) Find the area of one face of the cube, correct to 3 significant figures.

Solution

(a) Side $= \sqrt[3]{50} = 3.684031\ldots$ cm, which is 3.68 cm correct to 3 significant figures.

(b) A face of a cube is a square, so its area is side $\times$ side. Use the full value from part (a), not the rounded answer 3.68: $3.684031\ldots \times 3.684031\ldots = 13.572088\ldots$ cm², which is 13.6 cm² correct to 3 significant figures.

With the rounded answer to part (a), the area would come to $3.68 \times 3.68 = 13.5424$ , or 13.5 cm², which is wrong. Squaring turned an error of about 0.004 cm in the side into an error of about 0.03 cm² in the area.

Part (a) asked for a rounded answer, and 3.68 cm is the right answer to part (a). For part (b) it is still an intermediate value, so it must not be used in its rounded form.

(a) A side of the cube is 3.68 cm long (correct to 3 significant figures).
(b) The area of one face is 13.6 cm² (correct to 3 significant figures).

Common Mistakes

Using a rounded answer in the next step. In Example 4, 3.68 cm is the answer to part (a), but part (b) needs $3.684031\ldots$ Using 3.68 gives 13.5 cm², and the answer is 13.6 cm².
Writing an intermediate value to the accuracy of the final answer. Three significant figures in the working are too few for an answer correct to 3 significant figures: $4 \times 1.41 = 5.64$ , but $4 \times \sqrt{2} \approx 5.66$ (correct to 3 significant figures).
Trusting early rounding because it once gave the right answer. Sometimes the two ways happen to agree. You cannot tell beforehand, so round at the end every time.
Thinking that a small rounding error stays small. In Example 2 the rounded value, 12.5 km, is off by only about 0.014 km. Multiplied by 52, the error is about 0.74 km, and the answer comes to 650 km when it is 651 km.
Calculating with the value written in the working. A value written to 4 significant figures is still a rounded value. For six mats in a row, $6 \times 1.414 = 8.484$ , which rounds to 8.48 m, but the full value gives 8.49 m. Write 1.414 in the working, and calculate with the value in the calculator.

Practice

Work each question on paper, with a calculator where you need one, before you open its solution.

Question 1

Round each number as stated, and find the rounding error.
(a) 0.673, to 2 decimal places
(b) 5962, to 2 significant figures
(c) 19.95, to 1 decimal place
(d) A length is rounded to the nearest centimetre. What is the largest rounding error there can be?

ShowHide solution

(a) The digit after the second decimal place is 3, so the 7 stays as it is: $0.673 \approx 0.67$ . The rounding error is $0.673 - 0.67 = 0.003$ .

(b) The first two significant figures are 5 and 9. The digit after them is 6, so round up: 59 hundreds become 60 hundreds, and $5962 \approx 6000$ . The rounding error is $6000 - 5962 = 38$ .

(c) The digit after the first decimal place is 5, so round up: $19.9 + 0.1 = 20.0$ , and $19.95 \approx 20.0$ . The rounding error is $20.0 - 19.95 = 0.05$ .

(d) A rounding error is never more than half a unit of the place rounded to, and half of 1 cm is 0.5 cm. For example, 12.5 cm is rounded to 13 cm, which is 0.5 cm away.

(a) $0.67$ ; the rounding error is 0.003.
(b) $6000$ ; the rounding error is 38.
(c) $20.0$ ; the rounding error is 0.05.
(d) The largest rounding error is 0.5 cm.

Question 2

Work out each calculation twice: first with the intermediate value rounded to 3 significant figures, then with the full calculator value. Give both answers correct to 3 significant figures, and say which one is right.
(a) $15 \times \sqrt{7}$
(b) $50 \div 7 \times 13$

ShowHide solution

(a) The calculator shows $\sqrt{7}$ as 2.645751311, which is 2.65 correct to 3 significant figures.

Rounded early: $15 \times 2.65 = 39.75$ , which is 39.8 correct to 3 significant figures.

Full value: $15 \times 2.645751311 = 39.686\ldots$ , which is 39.7 correct to 3 significant figures.

(b) The calculator shows $50 \div 7$ as 7.142857143, which is 7.14 correct to 3 significant figures.

Rounded early: $7.14 \times 13 = 92.82$ , which is 92.8 correct to 3 significant figures.

Full value: $7.142857143 \times 13 = 92.857\ldots$ , which is 92.9 correct to 3 significant figures.

In each part the answer from the full value is the right one.

(a) Rounded early 39.8; with the full value 39.7, which is right.
(b) Rounded early 92.8; with the full value 92.9, which is right.

Question 3

Aisha is paid $95 for 8 hours of work. She wants to know her pay for 30 hours at the same rate. Her brother works out the pay for 1 hour, rounds it to the nearest cent and multiplies by 30.
(a) What answer does her brother get?
(b) What is the correct pay for 30 hours?
(c) Explain the difference.

ShowHide solution

Pay for 1 hour $= 95 \div 8 = 11.875$ dollars.

(a) To the nearest cent, 11.875 dollars is $11.88. Her brother gets $11.88 \times 30 = 356.40$ dollars.

(b) Without rounding: $11.875 \times 30 = 356.25$ dollars. A check in one calculation: $95 \times 30 \div 8 = 356.25$

(c) Rounding $11.875 up to $11.88 made an error of half a cent, $0.005. Multiplying by 30 made the error 30 times as large: $0.005 \times 30 = 0.15$ dollars, which is 15 cents.

(a) Her brother gets $356.40.
(b) The correct pay is $356.25.
(c) His answer is 15 cents too much, because the rounding error of half a cent was multiplied by 30.

Question 4

Mei Ling's walk to school is 0.847 km long. She walks to school and back home on each of the 195 school days of a year.
(a) She rounds 0.847 km to 1 significant figure before she calculates. What total distance does she get?
(b) Find the total distance without rounding early. Give it correct to 3 significant figures.
(c) How large is the rounding error in the length of one walk, and how large is the error in her total?

ShowHide solution

She makes $2 \times 195 = 390$ walks in the year.

(a) Correct to 1 significant figure, 0.847 km is 0.8 km. Her total is $0.8 \times 390 = 312$ km.

(b) $0.847 \times 390 = 330.33$ km, which is 330 km correct to 3 significant figures.

(c) The rounding error in one walk is $0.847 - 0.8 = 0.047$ km. The error in her total is $330.33 - 312 = 18.33$ km. It is 390 times as large: $0.047 \times 390 = 18.33$

(a) She gets 312 km.
(b) The total distance is 330 km (correct to 3 significant figures).
(c) The rounding error is 0.047 km in one walk and 18.33 km in the total.

Question 5

Shu Fen has to find $6.8 \times \sqrt{50}$ correct to 3 significant figures.
(a) Estimate the answer by using 7 for 6.8 and $\sqrt{49}$ for $\sqrt{50}$ .
(b) Find the answer with a calculator.
(c) Shu Fen writes her estimate as the answer. Why is that wrong, and what is the estimate good for?

ShowHide solution

(a) $6.8 \times \sqrt{50} \approx 7 \times \sqrt{49} = 7 \times 7 = 49$

(b) Key in the whole calculation: $6.8 \times \sqrt{50} = 48.083\ldots$ , which is 48.1 correct to 3 significant figures.

(c) In an estimate the numbers are rounded before the calculation, on purpose, to make it quick. That is rounding as early as possible, so the result is only close to the answer: 49 against 48.1. The question asks for 3 significant figures, and that needs the full values.

The estimate is a check. It says that the answer should be close to 49. A calculator answer of 480.8 or 4.81 would show a slip in the keying.

(a) The estimate is 49.
(b) $6.8 \times \sqrt{50} \approx 48.1$ (correct to 3 significant figures).
(c) An estimate is made from rounded numbers, so it is not correct to 3 significant figures. It is used to check the size of the calculator answer.

Question 6

Farah runs 2.4 km in 13 minutes.
(a) Find her speed in metres per minute, correct to 3 significant figures.
(b) At the same speed, how long does she take to run 10 km? Give the answer in minutes, correct to 3 significant figures.
(c) What answer to part (b) would the rounded speed from part (a) give?

ShowHide solution

(a) 2.4 km is 2400 m. Speed $= 2400 \div 13 = 184.615384\ldots$ metres per minute, which is 185 metres per minute correct to 3 significant figures.

(b) 10 km is $10\,000$ m. Time $= 10\,000 \div 184.615384\ldots = 54.1666\ldots$ minutes, which is 54.2 minutes correct to 3 significant figures. Use the full value of the speed, which is still in the calculator.

(c) With the rounded speed: $10\,000 \div 185 = 54.054\ldots$ minutes, which is 54.1 minutes. This is wrong. The rounded speed, 185, is a little more than the true speed, so the time comes out a little too short.

(a) Her speed is 185 metres per minute (correct to 3 significant figures).
(b) She takes 54.2 minutes (correct to 3 significant figures).
(c) The rounded speed gives 54.1 minutes, which is wrong.

Question 7

Two taps fill a water tank. Tap A delivers 20 litres in 7 minutes, and tap B delivers 25 litres in 9 minutes.
(a) Find how much water each tap delivers in 1 minute, correct to 3 significant figures.
(b) Find how much water the two taps together deliver in 1 minute, correct to 3 significant figures.
(c) What answer to part (b) do the rounded answers to part (a) give? Explain the difference.

ShowHide solution

(a) Tap A: $20 \div 7 = 2.857142\ldots$ litres, which is 2.86 litres correct to 3 significant figures. Tap B: $25 \div 9 = 2.777777\ldots$ litres, which is 2.78 litres.

(b) Add the full values: $2.857142\ldots + 2.777777\ldots = 5.634920\ldots$ litres, which is 5.63 litres correct to 3 significant figures. In one calculation: $20 \div 7 + 25 \div 9 = 5.634920\ldots$

(c) The rounded answers give $2.86 + 2.78 = 5.64$ litres, which is wrong. Both values were rounded up, by about 0.0029 and by about 0.0022. When numbers are added, their rounding errors are added too: $0.0029 + 0.0022 = 0.0051$ , which was enough to change the third significant figure.

Four significant figures would not have been enough here either: $2.857 + 2.778 = 5.635$ , which rounds to 5.64. Calculate with the full values, and the answer is 5.63.

(a) Tap A delivers 2.86 litres in 1 minute and tap B delivers 2.78 litres (each correct to 3 significant figures).
(b) Together the taps deliver 5.63 litres in 1 minute (correct to 3 significant figures).
(c) The rounded answers give 5.64 litres, which is wrong. Both were rounded up, and the two rounding errors added up.

Question 8

A piece of wire 100 cm long is cut into 12 equal pieces. The pieces are joined to make the 12 edges of a cube.
(a) Find the length of an edge of the cube, correct to 3 significant figures.
(b) Find the volume of the cube, correct to 3 significant figures.
(c) What volume does the rounded answer to part (a) give?
(d) Write out the working for parts (a) and (b) as it should be set out.

ShowHide solution

(a) Edge $= 100 \div 12 = 8.333333\ldots$ cm, which is 8.33 cm correct to 3 significant figures.

(b) Volume $=$ edge $\times$ edge $\times$ edge. With the full value of the edge: $8.333333\ldots \times 8.333333\ldots \times 8.333333\ldots = 578.703\ldots$ cm³, which is 579 cm³ correct to 3 significant figures.

(c) With the rounded edge: $8.33 \times 8.33 \times 8.33 = 578.009\ldots$ cm³, which is 578 cm³. This is wrong. Cubing turned an error of about 0.003 cm in the edge into an error of about 0.7 cm³ in the volume.

(d) Set the working out like this.
(a) Edge $= 100 \div 12 = 8.3333\ldots \approx 8.33$ cm (correct to 3 significant figures)
(b) Volume $= 8.3333\ldots \times 8.3333\ldots \times 8.3333\ldots = 578.70\ldots \approx 579$ cm³ (correct to 3 significant figures)

(a) An edge of the cube is 8.33 cm long (correct to 3 significant figures).
(b) The volume of the cube is 579 cm³ (correct to 3 significant figures).
(c) The rounded edge gives 578 cm³, which is wrong.
(d) The working for part (a) shows the full value of the edge and then its answer, 8.33 cm. Part (b) is worked out with the full value, not with 8.33, and each answer is rounded once.

Question 9

Wei Jie has to find $12 \times \sqrt{7}$ correct to 3 significant figures. He writes $\sqrt{7} \approx 2.646$ and works out $12 \times 2.646$ .
(a) What answer does he get, correct to 3 significant figures?
(b) Find the answer with the full calculator value.
(c) Wei Jie kept 4 significant figures in his working. Why is his answer still wrong, and what should he do?

ShowHide solution

(a) $12 \times 2.646 = 31.752$ , which is 31.8 correct to 3 significant figures.

(b) The calculator shows $\sqrt{7}$ as 2.645751311, and $12 \times 2.645751311 = 31.749\ldots$ , which is 31.7 correct to 3 significant figures.

(c) The true value, $31.749\ldots$ , lies just below 31.75, the halfway point between 31.7 and 31.8. The rounding error in 2.646 is only about 0.00025, and 12 times that is about 0.003, but this was enough to carry the value past the halfway point.

He should write 2.646 in his working and go on calculating with the full value in the calculator.

(a) Wei Jie gets 31.8.
(b) The answer is 31.7 (correct to 3 significant figures).
(c) The true value is very close to the halfway point 31.75, so even a tiny rounding error changes the answer. He should calculate with the full calculator value.

Question 10

The circumference of a circular pond is 30 m. Use the $\pi$ key of your calculator.
(a) Find the radius of the pond, correct to 3 significant figures.
(b) Find the area of the pond, correct to 3 significant figures.
(c) What area would the rounded answer to part (a) give?

ShowHide solution

(a) Circumference $= 2 \times \pi \times$ radius, so radius $= 30 \div (2 \times \pi) = 4.774648\ldots$ m, which is 4.77 m correct to 3 significant figures. Key in the brackets.

(b) Area $= \pi \times$ radius $\times$ radius. With the full value of the radius: $\pi \times 4.774648\ldots \times 4.774648\ldots = 71.619\ldots$ m², which is 71.6 m² correct to 3 significant figures.

(c) With the rounded radius: $\pi \times 4.77 \times 4.77 = 71.480\ldots$ m², which is 71.5 m². This is wrong: the rounding error in the radius was made larger by the squaring and by the multiplying.

(a) The radius of the pond is 4.77 m (correct to 3 significant figures).
(b) The area of the pond is 71.6 m² (correct to 3 significant figures).
(c) The rounded radius gives 71.5 m², which is wrong.

Key Points

A rounding error is the difference between a rounded value and the exact value. It is never more than half a unit of the place rounded to.
A rounded intermediate value carries its error into every step that follows, and multiplying, squaring or adding can make the error larger.
Round once, at the end. Calculate with the full value in the calculator: carry on from the display, use the Ans key or the memory, or key in the whole calculation in one go.
In the working, write an intermediate value to at least 4 significant figures when the answer is to be correct to 3 significant figures.
When part (b) uses the answer to part (a), use the unrounded value from part (a).
Rounding the numbers first is right for an estimate, which checks the size of an answer. It is not the answer.