Rounding to Significant Figures

How to round whole numbers, decimals greater than 1 and numbers between 0 and 1 to a required number of significant figures, which zeros must be written in the answer, and what to do when a 9 is rounded up.

Before you start

The significant figures (s.f.) of a number are counted from its first non-zero digit. Non-zero digits and the zeros between them are significant. Zeros in front of the first non-zero digit (leading zeros) are never significant, and zeros at the end of a decimal always are: $0.00405$ has 3 significant figures (4, 0 and 5) and $2.50$ has 3 (2, 5 and 0).

Zeros at the end of a whole number may or may not be significant. $4000$ can have 1, 2, 3 or 4 significant figures, according to how it was rounded.

To round a number, look at the first digit that is dropped: 5 or more rounds up, 4 or less rounds down. $3.46 \approx 3.5$ , correct to 1 decimal place (1 d.p.). The sign $\approx$ is read "is approximately equal to".

How to Round to Significant Figures

A concert at the National Stadium was watched by $38\,472$ people. A report that says "about $38\,000$ people" has kept the two digits that matter most, the 3 and the 8, and let the others go. That is rounding to 2 significant figures.

Rounding to significant figures is ordinary rounding. What is new is the way you are told where to round. You are not given a place ("to the nearest thousand", "to 2 decimal places"). You are told how many figures to keep, counted from the first non-zero digit.

Rule: rounding to a required number of significant figures

Start at the first non-zero digit and count the number of significant figures asked for. The last one you count is the last digit kept.
Look at the next digit, the one just to its right. Do not look at any digit after that.
If the next digit is 5 or more, round up: add 1 to the last digit kept. If the next digit is 4 or less, round down: leave the last digit kept as it is.
Before the decimal point, write a zero (a placeholder) in place of each digit that is dropped, so that the digits kept stay in their places. After the decimal point, leave the dropped digits out.

Check: the rounded number must be about the same size as the number you started with.

When the last digit kept is a 9 that has to be rounded up, adding 1 to it gives 10: write 0 and carry 1 to the place on its left, as you do when you round a whole number to the nearest 10 or a decimal to 1 decimal place. The sections "Zeros That Show the Accuracy" and "A Carry That Adds a Digit" show which zeros to write then.

Example 1: A whole number

Round $38\,472$ to 2 significant figures.

Solution

Count from the first non-zero digit. The first significant figure is 3 and the second is 8, so the 8 is the last digit kept.

The next digit is 4. It is less than 5, so round down: the 8 stays as it is.

The digits that are dropped (4, 7 and 2) are before the decimal point, so a zero is written in place of each one. The chart shows every digit in its place.

Place	Ten thousands	Thousands	Hundreds	Tens	Ones
Digits of $38\,472$	3	8	4	7	2
In the rounding	1st significant figure	2nd significant figure: the last digit kept	next digit: less than 5	dropped	dropped
Rounded to 2 s.f.	3	8	0	0	0

The 8 is in the thousands place, so for this number "to 2 significant figures" means "to the nearest thousand". On the number line, $38\,472$ lies between $38\,000$ and $39\,000$ , on the $38\,000$ side of the halfway mark.

The answer is not 38. The three zeros keep the 3 in the ten thousands place and the 8 in the thousands place. Without them the answer would be a thousand times too small.

$38\,472 \approx 38\,000$ (correct to 2 s.f.)

A negative number is rounded by its numerical value (the number without its sign) and keeps its sign: $-38\,472 \approx -38\,000$ (correct to 2 s.f.).

Whole Numbers: Zeros That Hold the Place

When a whole number is rounded, the zeros written in place of the dropped digits are placeholders. They keep the number the right size, but they are not significant figures. A rounded whole number that ends in zeros therefore cannot show how accurate it is, so the accuracy is written after it: "(correct to 2 s.f.)".

Example 2: One number, three accuracies

Round $75\,491$ to
(a) 1 significant figure
(b) 2 significant figures
(c) 3 significant figures

Solution

(a) The last digit kept is 7. The next digit is 5, so round up: the 7 becomes 8. Four zeros hold the places of the four digits dropped.

(b) The last digit kept is the second significant figure, 5. The next digit is 4, so round down: the 5 stays. Three zeros hold the place.

(c) The last digit kept is the third significant figure, 4. The next digit is 9, so round up: the 4 becomes 5. Two zeros hold the place.

The three answers differ from $75\,491$ by $4509$ , by 491 and by 9. Keeping more significant figures never takes the rounded number further from the original, and usually brings it closer.

(a) $75\,491 \approx 80\,000$ (correct to 1 s.f.)
(b) $75\,491 \approx 75\,000$ (correct to 2 s.f.)
(c) $75\,491 \approx 75\,500$ (correct to 3 s.f.)

Example 3: A zero that is counted

Round
(a) $60\,220$ to 2 significant figures
(b) $304\,961$ to 3 significant figures

Solution

(a) The first significant figure is 6. The zero after it lies between non-zero digits, so it is significant: it is the second significant figure and the last digit kept.

The next digit is 2, so round down. The three digits dropped are replaced by zeros.

In the answer $60\,000$ the first zero is a significant figure and the other three are placeholders. The number alone cannot show this, which is why "(correct to 2 s.f.)" is written.

(b) The significant figures are 3, 0 and 4, so the 4 is the last digit kept. The next digit is 9, so round up: the 4 becomes 5. Three zeros hold the place.

(a) $60\,220 \approx 60\,000$ (correct to 2 s.f.)
(b) $304\,961 \approx 305\,000$ (correct to 3 s.f.)

Example 4: In a news report

On one day, $136\,738$ passengers passed through an airport.
(a) A newspaper gives this number correct to 3 significant figures. What number does it print?
(b) A website gives the number as $140\,000$ . To how many significant figures has the website rounded it?

Solution

(a) The significant figures are 1, 3 and 6, so the 6 is the last digit kept. The next digit is 7, so round up: the 6 becomes 7. Three zeros hold the place.

(b) Try 1 significant figure: the last digit kept is 1 and the next digit is 3, so round down. That gives $100\,000$ , which is not the website's number.

Try 2 significant figures: the last digit kept is 3 and the next digit is 6, so round up: the 3 becomes 4. That gives $140\,000$ .

(a) The newspaper prints $137\,000$ .
(b) The website has rounded the number to 2 significant figures.

Decimals Greater Than 1

In a decimal, the digits dropped after the decimal point are simply left out. Writing zeros for them would claim an accuracy the answer does not have: $47.400$ says "correct to 5 significant figures". A zero is written after the decimal point only when it is one of the significant figures asked for.

Example 5: A whole part and a decimal part

Round
(a) $47.382$ to 3 significant figures
(b) $1234.56$ to 2 significant figures
(c) $28.046$ to 3 significant figures

Solution

(a) The significant figures are 4, 7 and 3. The next digit is 8, so round up: the 3 becomes 4. The digits dropped are after the decimal point, so they are left out.

(b) The significant figures are 1 and 2. The next digit is 3, so round down. The 3 and the 4 are before the decimal point, so a zero is written for each. The 5 and the 6 are after it, so they are left out. The answer is $1200$ , not $1200.00$ .

(c) The significant figures are 2, 8 and 0. The next digit is 4, so round down. The 0 is the third significant figure, so it must be written: $28.0$ . Writing 28 would show only 2 significant figures.

(a) $47.382 \approx 47.4$ (correct to 3 s.f.)
(b) $1234.56 \approx 1200$ (correct to 2 s.f.)
(c) $28.046 \approx 28.0$ (correct to 3 s.f.)

Numbers Between 0 and 1

In a number between 0 and 1, the zeros in front of the first non-zero digit are not significant, so the counting starts after them. They are still written in the answer, because they hold the place of every digit that follows.

Example 6: Zeros in front

Round
(a) $0.04736$ to 2 significant figures
(b) $0.008101$ to 3 significant figures
(c) $0.0030891$ to 4 significant figures

Solution

(a) The two zeros in front are not significant. The first significant figure is 4 and the second is 7. The next digit is 3, so round down. The 3 and the 6 are left out, and the zeros in front stay.

(b) The significant figures are 8, 1 and 0. The next digit is 1, so round down. The 0 is the third significant figure, so it must be written: $0.00810$ . Writing $0.0081$ would show only 2 significant figures.

(c) The significant figures are 3, 0, 8 and 9. The next digit is 1, so round down: the 9 stays as it is, and the 1 is left out.

Check (c) by counting: $0.003089$ has the significant figures 3, 0, 8 and 9, which is 4.

(a) $0.04736 \approx 0.047$ (correct to 2 s.f.)
(b) $0.008101 \approx 0.00810$ (correct to 3 s.f.)
(c) $0.0030891 \approx 0.003089$ (correct to 4 s.f.)

Zeros That Show the Accuracy

A rounded answer tells the reader two things: about how big the number is, and how accurately it is known. $2.5$ and $2.50$ have the same value, but $2.5$ shows 2 significant figures and $2.50$ shows 3. So a zero that is one of the significant figures asked for must be written, even at the end of a decimal. Example 5 had one: $28.046 \approx 28.0$ (correct to 3 s.f.), not 28.

Such zeros also turn up when a 9 is rounded up. If the last digit kept is 9 and it has to be rounded up, then $9 + 1 = 10$ : write 0 in that place and carry 1 to the place on its left, as in an addition. That 0 is written, even at the end of a decimal, when it is still one of the significant figures asked for. (The next section has the one case in which it is not one of them.)

Example 7: Zeros that show the accuracy

Round each number to 3 significant figures.
(a) $89.95$
(b) $0.03996$
(c) $1.999$

Solution

(a) The significant figures are 8, 9 and 9. The next digit is 5, so round up. Adding 1 to the last digit kept is adding 1 tenth: $89.9 + 0.1 = 90.0$ .

The answer has the 3 significant figures 9, 0 and 0. Writing 90 would be wrong: it shows 2 significant figures at most.

(b) The zeros in front are not significant. The significant figures are 3, 9 and 9, and the next digit is 6, so round up. The last digit kept is in the ten-thousandths place: $0.0399 + 0.0001 = 0.0400$ .

Both 9s have become 0 and the 3 has become 4. The answer has the 3 significant figures 4, 0 and 0. Writing $0.04$ would be wrong: it shows only 1 significant figure.

Place	Ones	Decimal point	Tenths	Hundredths	Thousandths	Ten-thousandths	Hundred-thousandths
Digits of $0.03996$	0	.	0	3	9	9	6
In the rounding	not significant		not significant	1st significant figure	2nd significant figure	3rd significant figure: the last digit kept	next digit: 5 or more
Rounded to 3 s.f.	0	.	0	4	0	0

(c) The significant figures are 1, 9 and 9. The next digit is 9, so round up: $1.99 + 0.01 = 2.00$ . Both zeros are written. Neither 2 nor $2.0$ shows 3 significant figures.

(a) $89.95 \approx 90.0$ (correct to 3 s.f.)
(b) $0.03996 \approx 0.0400$ (correct to 3 s.f.)
(c) $1.999 \approx 2.00$ (correct to 3 s.f.)

A Carry That Adds a Digit

Sometimes the carry passes through every digit that is kept and makes a new first significant figure: a 1, one place to the left of the old one. The significant figures then have to be counted again from it.

Example 8: A carry that adds a digit

Round
(a) $9962$ to 2 significant figures
(b) $0.99999$ to 2 significant figures
(c) $99.96$ to 3 significant figures
(d) $0.0996$ to 2 significant figures

Solution

(a) The significant figures are 9 and 9. The second 9 is in the hundreds place, and it is the last digit kept. The next digit is 6, so round up.

Adding 1 to the last digit kept is adding 1 hundred: $9900 + 100 = 10\,000$ . Both 9s have become 0, and a new digit, 1, has appeared in front. On the number line, $9962$ lies between $9900$ and $10\,000$ , past the halfway mark.

In the answer $10\,000$ , the 2 significant figures are the 1 and the first 0. The last three zeros hold the place. Written alone, $10\,000$ does not show which of its zeros are significant, which is why the accuracy is written beside it. (Correct to 1 significant figure, $9962$ is $10\,000$ as well.)

(b) The zero in front is not significant. The significant figures are 9 and 9, and the next digit is 9, so round up: $0.99 + 0.01 = 1.00$ .

The carry has made a new first significant figure, the 1 in the ones place. Count 2 significant figures again, starting from it: the 1 and one 0. The answer is $1.0$ . Writing $1.00$ would show 3 significant figures, and writing 1 would show only 1.

(c) The significant figures are 9, 9 and 9, and the next digit is 6, so round up: $99.9 + 0.1 = 100.0$ . Count 3 significant figures from the new first digit: 1, 0 and 0. The answer is 100, with no decimal part. Both of its zeros are significant here, although the number alone cannot show it.

(d) The zeros in front are not significant. The significant figures are 9 and 9, and the next digit is 6, so round up. The last digit kept is in the thousandths place: $0.099 + 0.001 = 0.100$ .

The carry has made a new first significant figure, the 1 in the tenths place, one place to the left of the old one. No digit has been written in front of the number, which still begins with "0.", but the significant figures are counted again all the same. Count 2, starting from the 1: the 1 and one 0. The answer is $0.10$ , not $0.100$ , which would show 3 significant figures.

(a) $9962 \approx 10\,000$ (correct to 2 s.f.)
(b) $0.99999 \approx 1.0$ (correct to 2 s.f.)
(c) $99.96 \approx 100$ (correct to 3 s.f.)
(d) $0.0996 \approx 0.10$ (correct to 2 s.f.)

Rule: after a carry, count again

If the carry makes a new first significant figure (a 1 one place to the left of the old one), count the significant figures again from it. Write only as many as the question asks for, and after them only the zeros needed to hold the place.

Common Mistakes

Leaving out the zeros that hold the place. $38\,472$ correct to 2 significant figures is $38\,000$ , not 38. A rounded number is always about the same size as the original.
Leaving out a zero that is a significant figure. Correct to 3 significant figures, $28.046$ is $28.0$ , not 28; $89.95$ is $90.0$ , not 90; and $0.03996$ is $0.0400$ , not $0.04$ . Count the significant figures in your answer.
Counting the zeros in front. $0.008101$ correct to 3 significant figures is $0.00810$ , not $0.008$ . The counting starts at the 8.
Writing zeros for digits dropped after the decimal point. $47.382$ correct to 3 significant figures is $47.4$ , not $47.400$ .
Looking past the next digit. $0.2449$ correct to 2 significant figures is $0.24$ , because the next digit is 4. Do not round the 4 up to 5 first.
Keeping too many zeros after a carry. $0.99999$ correct to 2 significant figures is $1.0$ , not $1.00$ . When the carry makes a new first significant figure, count the significant figures again from it.

Practice

Work each question on paper before you open its solution.

Question 1

Round
(a) $4618$ to 1 significant figure
(b) $26\,849$ to 2 significant figures
(c) $70\,049$ to 3 significant figures
(d) $2\,807\,356$ to 3 significant figures

ShowHide solution

(a) The last digit kept is 4. The next digit is 6, so round up: the 4 becomes 5. Three zeros hold the place.

(b) The significant figures are 2 and 6. The next digit is 8, so round up: the 6 becomes 7. Three zeros hold the place.

(c) The significant figures are 7, 0 and 0: the zeros lie between non-zero digits, so they are counted. The next digit is 4, so round down. Two zeros hold the place.

(d) The significant figures are 2, 8 and 0. The next digit is 7, so round up: the 0 becomes 1. Four zeros hold the place.

(a) $4618 \approx 5000$ (correct to 1 s.f.)
(b) $26\,849 \approx 27\,000$ (correct to 2 s.f.)
(c) $70\,049 \approx 70\,000$ (correct to 3 s.f.)
(d) $2\,807\,356 \approx 2\,810\,000$ (correct to 3 s.f.)

Question 2

Round
(a) $5.763$ to 2 significant figures
(b) $20.078$ to 3 significant figures
(c) $8271.4$ to 2 significant figures
(d) $906.03$ to 4 significant figures

ShowHide solution

(a) The significant figures are 5 and 7. The next digit is 6, so round up: the 7 becomes 8. The digits after it are left out.

(b) The significant figures are 2, 0 and 0. The next digit is 7, so round up: the second 0 becomes 1.

(c) The significant figures are 8 and 2. The next digit is 7, so round up: the 2 becomes 3. The 7 and the 1 are before the decimal point, so a zero is written for each. The 4 is after it and is left out.

(d) The significant figures are 9, 0, 6 and 0. The next digit is 3, so round down. The 0 after the decimal point is the fourth significant figure, so it is written: $906.0$ .

(a) $5.763 \approx 5.8$ (correct to 2 s.f.)
(b) $20.078 \approx 20.1$ (correct to 3 s.f.)
(c) $8271.4 \approx 8300$ (correct to 2 s.f.)
(d) $906.03 \approx 906.0$ (correct to 4 s.f.)

Question 3

Round
(a) $0.070185$ to 3 significant figures
(b) $0.00462$ to 1 significant figure
(c) $0.00025049$ to 2 significant figures
(d) $0.5049$ to 1 significant figure

ShowHide solution

(a) The two zeros in front are not significant. The significant figures are 7, 0 and 1. The next digit is 8, so round up: the 1 becomes 2.

(b) The first significant figure is 4. The next digit is 6, so round up: the 4 becomes 5. The zeros in front stay.

(d) The first significant figure is 5. The next digit is 0, so round down. The 4 and the 9 further along are not looked at.

(a) $0.070185 \approx 0.0702$ (correct to 3 s.f.)
(b) $0.00462 \approx 0.005$ (correct to 1 s.f.)
(c) $0.00025049 \approx 0.00025$ (correct to 2 s.f.)
(d) $0.5049 \approx 0.5$ (correct to 1 s.f.)

Question 4

Round $508.274$ to
(a) 1 significant figure
(b) 2 significant figures
(c) 3 significant figures
(d) 4 significant figures
(e) 5 significant figures

ShowHide solution

(a) The last digit kept is 5. The next digit is 0, so round down. The 0 and the 8 are before the decimal point, so a zero is written for each. The digits after the decimal point are left out.

(b) The significant figures are 5 and 0. The next digit is 8, so round up: the 0 becomes 1. A zero is written for the 8.

(c) The significant figures are 5, 0 and 8. The next digit is 2, so round down. Every digit dropped is after the decimal point, so no zero is written for it.

(d) The significant figures are 5, 0, 8 and 2. The next digit is 7, so round up: the 2 becomes 3.

(e) The significant figures are 5, 0, 8, 2 and 7. The next digit is 4, so round down.

(a) $508.274 \approx 500$ (correct to 1 s.f.)
(b) $508.274 \approx 510$ (correct to 2 s.f.)
(c) $508.274 \approx 508$ (correct to 3 s.f.)
(d) $508.274 \approx 508.3$ (correct to 4 s.f.)
(e) $508.274 \approx 508.27$ (correct to 5 s.f.)

Question 5

Round
(a) $-6481$ to 2 significant figures
(b) $-15.062$ to 3 significant figures
(c) $-0.07253$ to 1 significant figure

ShowHide solution

Round the numerical value of each number and keep the negative sign.

(a) The significant figures of $6481$ are 6 and 4. The next digit is 8, so round up: the 4 becomes 5. Two zeros hold the place.

(b) The significant figures of $15.062$ are 1, 5 and 0. The next digit is 6, so round up: the 0 becomes 1.

(a) $-6481 \approx -6500$ (correct to 2 s.f.)
(b) $-15.062 \approx -15.1$ (correct to 3 s.f.)
(c) $-0.07253 \approx -0.07$ (correct to 1 s.f.)

Question 6

Each of these pupils has made a mistake. Say what the mistake is and give the correct answer.
(a) Mei Ling says that $7306$ correct to 2 significant figures is 73.
(b) Ravi says that $52.347$ correct to 3 significant figures is $52.300$ .
(c) Aisyah says that $0.06049$ correct to 2 significant figures is $0.061$ .

ShowHide solution

(a) Mei Ling kept the right digits, 7 and 3, and rounded down correctly, because the next digit is 0. But she left out the zeros that hold the place. The 0 and the 6 are before the decimal point, so a zero is written for each.

(b) Ravi rounded down correctly, because the next digit is 4. But he wrote zeros for the two digits dropped after the decimal point. They are left out: $52.300$ would claim 5 significant figures.

(c) Aisyah looked past the next digit. She rounded the 4 up to 5 because of the 9 after it, and then rounded the 0 up to 1. The significant figures are 6 and 0, and the next digit is 4, so round down. The 0 is the second significant figure, so it is written.

(a) $7306 \approx 7300$ (correct to 2 s.f.)
(b) $52.347 \approx 52.3$ (correct to 3 s.f.)
(c) $0.06049 \approx 0.060$ (correct to 2 s.f.)

Question 7

A town has $263\,548$ residents.
(a) Three reports give the number of residents as $300\,000$ , as $260\,000$ and as $264\,000$ . To how many significant figures has each report rounded it?
(b) Write the number of residents correct to 4 significant figures.

ShowHide solution

(a) Try 1 significant figure: the last digit kept is 2 and the next digit is 6, so round up: the 2 becomes 3. That gives $300\,000$ .

Try 2 significant figures: they are 2 and 6, and the next digit is 3, so round down. That gives $260\,000$ .

Try 3 significant figures: they are 2, 6 and 3, and the next digit is 5, so round up: the 3 becomes 4. That gives $264\,000$ .

(b) The significant figures are 2, 6, 3 and 5. The next digit is 4, so round down. Two zeros hold the place.

(a) $300\,000$ is correct to 1 significant figure, $260\,000$ to 2 significant figures and $264\,000$ to 3 significant figures.
(b) The town has about $263\,500$ residents (correct to 4 s.f.).

Question 8

The computer on Farah's bicycle shows that she has cycled 1286.45 km this year. Write this distance correct to
(a) 2 significant figures
(b) 3 significant figures
(c) 5 significant figures

ShowHide solution

(a) The significant figures are 1 and 2. The next digit is 8, so round up: the 2 becomes 3. The 8 and the 6 are before the decimal point, so a zero is written for each. The digits after the decimal point are left out.

(b) The significant figures are 1, 2 and 8. The next digit is 6, so round up: the 8 becomes 9. A zero is written for the 6.

(a) The distance is 1300 km (correct to 2 s.f.).
(b) The distance is 1290 km (correct to 3 s.f.).
(c) The distance is 1286.5 km (correct to 5 s.f.).

Question 9

A lorry carries $12\,750$ kg of sand.
(a) Write this mass correct to 1 significant figure, to 2 significant figures and to 3 significant figures.
(b) How far is each answer from the mass of the sand?

ShowHide solution

(a) 1 significant figure: the last digit kept is 1 and the next digit is 2, so round down. That gives $10\,000$ .

2 significant figures: they are 1 and 2, and the next digit is 7, so round up: the 2 becomes 3. That gives $13\,000$ .

3 significant figures: they are 1, 2 and 7, and the next digit is 5, so round up: the 7 becomes 8. That gives $12\,800$ .

(b) $12\,750 - 10\,000 = 2750$ , $13\,000 - 12\,750 = 250$ and $12\,800 - 12\,750 = 50$ . Each extra significant figure brings the answer closer to the mass of the sand.

(a) The mass is $10\,000$ kg (correct to 1 s.f.), $13\,000$ kg (correct to 2 s.f.) and $12\,800$ kg (correct to 3 s.f.).
(b) The three answers are 2750 kg, 250 kg and 50 kg from the mass of the sand.

Question 10

A science book gives the thickness of a strand of hair as 0.007284 cm and the mass of a grain of rice as 0.02046 g.
(a) Write the thickness of the hair correct to 2 significant figures.
(b) Write the mass of the grain of rice correct to 2 significant figures.
(c) Write the mass of the grain of rice correct to 3 significant figures.

ShowHide solution

(a) The zeros in front are not significant. The significant figures are 7 and 2. The next digit is 8, so round up: the 2 becomes 3. The zeros in front stay.

(b) The significant figures are 2 and 0. The next digit is 4, so round down. The 0 is the second significant figure, so it must be written: $0.020$ . Writing $0.02$ would show only 1 significant figure.

(a) The hair is about 0.0073 cm thick (correct to 2 s.f.).
(b) The mass of the grain of rice is about 0.020 g (correct to 2 s.f.).
(c) The mass of the grain of rice is about 0.0205 g (correct to 3 s.f.).

Question 11

One school has 1449 pupils and another has 1450 pupils.
(a) Round each number to 2 significant figures.
(b) The two schools differ by 1 pupil. By how much do the two rounded numbers differ? Explain why the difference is so large.

ShowHide solution

(a) In both numbers the significant figures are 1 and 4, and the last digit kept is in the hundreds place.

In 1449 the next digit is 4, so round down. That gives 1400. The 9 at the end is not looked at.

In 1450 the next digit is 5, so round up: the 4 becomes 5. That gives 1500.

(b) $1500 - 1400 = 100$ . The number 1450 is exactly halfway between 1400 and 1500, and a number that is halfway is rounded up. The number 1449 is just below halfway, so it is rounded down.

(a) $1449 \approx 1400$ and $1450 \approx 1500$ (both correct to 2 s.f.)
(b) The rounded numbers differ by 100. 1450 is at the halfway mark between 1400 and 1500 and is rounded up, while 1449 is just below the halfway mark and is rounded down.

Question 12

In each of these, a 9 is rounded up. Round
(a) $3998$ to 2 significant figures
(b) $19.98$ to 3 significant figures
(c) $0.0030896$ to 4 significant figures
(d) $0.00997$ to 2 significant figures
(e) $304.96$ to 4 significant figures
(f) $9951$ to 2 significant figures
(g) $99.97$ to 3 significant figures

ShowHide solution

(a) The significant figures are 3 and 9. The next digit is 9, so round up. The last digit kept is in the hundreds place: $3900 + 100 = 4000$ . In this answer the 4 and the first 0 are the 2 significant figures.

(b) The significant figures are 1, 9 and 9. The next digit is 8, so round up: $19.9 + 0.1 = 20.0$ . The answer needs 3 significant figures, so the 0 after the decimal point is written.

(c) The significant figures are 3, 0, 8 and 9. The next digit is 6, so round up: $0.003089 + 0.000001 = 0.003090$ . The last 0 is the fourth significant figure, so it is written.

(d) The zeros in front are not significant. The significant figures are 9 and 9. The next digit is 7, so round up: $0.0099 + 0.0001 = 0.0100$ . The carry has made a new first significant figure, the 1 in the hundredths place. Count 2 significant figures from it: the 1 and one 0. The answer is $0.010$ .

(e) The significant figures are 3, 0, 4 and 9. The next digit is 6, so round up: $304.9 + 0.1 = 305.0$ . The 0 is the fourth significant figure, so it is written.

(f) The significant figures are 9 and 9. The next digit is 5, so round up. The last digit kept is in the hundreds place: $9900 + 100 = 10\,000$ . The carry has put a new digit, 1, in front. In the answer the 1 and the first 0 are the 2 significant figures, and the last three zeros hold the place.

(g) The significant figures are 9, 9 and 9. The next digit is 7, so round up: $99.9 + 0.1 = 100.0$ . The carry has made a new first significant figure, the 1 in the hundreds place. Count 3 significant figures from it: 1, 0 and 0. The answer is 100, with no decimal part.

(a) $3998 \approx 4000$ (correct to 2 s.f.)
(b) $19.98 \approx 20.0$ (correct to 3 s.f.)
(c) $0.0030896 \approx 0.003090$ (correct to 4 s.f.)
(d) $0.00997 \approx 0.010$ (correct to 2 s.f.)
(e) $304.96 \approx 305.0$ (correct to 4 s.f.)
(f) $9951 \approx 10\,000$ (correct to 2 s.f.)
(g) $99.97 \approx 100$ (correct to 3 s.f.)

Question 13

Each of these pupils has made a mistake. Say what the mistake is and give the correct answer.
(a) Siti says that $9981$ correct to 2 significant figures is $1000$ .
(b) Jun Wei says that $0.9996$ correct to 3 significant figures is $1.000$ .
(c) Kumar says that $6.997$ correct to 3 significant figures is 7.

ShowHide solution

(a) The significant figures are 9 and 9, and the next digit is 8, so Siti was right to round up. But she lost a zero. The last digit kept is in the hundreds place, and $9900 + 100 = 10\,000$ : the carry puts a new digit, 1, in front of the four places the number already had, so the answer has five digits. A rounded number is always about the same size as the original, and $1000$ is only about a tenth of $9981$ .

(b) Jun Wei rounded up correctly, $0.999 + 0.001 = 1.000$ , but he did not count again. The carry has made a new first significant figure, the 1 in the ones place. Count 3 significant figures from it: 1, 0 and 0. His answer, $1.000$ , shows 4 significant figures.

(c) Kumar rounded up correctly, $6.99 + 0.01 = 7.00$ , but he left out the two zeros. They are the second and third significant figures and must be written.

(a) $9981 \approx 10\,000$ (correct to 2 s.f.)
(b) $0.9996 \approx 1.00$ (correct to 3 s.f.)
(c) $6.997 \approx 7.00$ (correct to 3 s.f.)

Question 14

The number $0.046973$ has been rounded. To how many significant figures has it been rounded to give
(a) $0.05$
(b) $0.047$
(c) $0.0470$

ShowHide solution

The significant figures of $0.046973$ are 4, 6, 9, 7 and 3.

(a) $0.05$ has 1 significant figure. Check: the first is 4 and the next digit is 6, so the 4 rounds up to 5.

(b) $0.047$ has 2 significant figures. Check: they are 4 and 6, and the next digit is 9, so the 6 rounds up to 7.

(c) $0.0470$ has 3 significant figures, because the zero at the end of a decimal is significant. Check: they are 4, 6 and 9, and the next digit is 7, so round up: $0.0469 + 0.0001 = 0.0470$ .

(a) 1 significant figure
(b) 2 significant figures
(c) 3 significant figures

Question 15

A marathon is a race of 42.195 km. Write this distance correct to
(a) 2 significant figures
(b) 3 significant figures
(c) 4 significant figures

ShowHide solution

(a) The significant figures are 4 and 2. The next digit is 1, so round down. The digits after the decimal point are left out.

(b) The significant figures are 4, 2 and 1. The next digit is 9, so round up: the 1 becomes 2.

(c) The significant figures are 4, 2, 1 and 9. The next digit is 5, so round up: $42.19 + 0.01 = 42.20$ . The 0 is the fourth significant figure, so it is written.

(a) The distance is 42 km (correct to 2 s.f.).
(b) The distance is 42.2 km (correct to 3 s.f.).
(c) The distance is 42.20 km (correct to 4 s.f.).

Question 16

The pupils of a school collected $18\,964$ drink cans for recycling.
(a) The school newsletter gives the number correct to 2 significant figures. What number does it give?
(b) The principal gives the number correct to 3 significant figures. What number does she give?
(c) What do you notice about the two numbers? Which zeros are significant in each?

ShowHide solution

(a) The significant figures are 1 and 8. The next digit is 9, so round up: the 8 becomes 9. Three zeros hold the place.

(b) The significant figures are 1, 8 and 9. The next digit is 6, so round up. The last digit kept is in the hundreds place: $18\,900 + 100 = 19\,000$ .

(c) The two numbers are the same. In (a) the significant figures are 1 and 9, and all three zeros hold the place. In (b) the significant figures are 1, 9 and the first 0, and the last two zeros hold the place.

(a) The newsletter gives $19\,000$ (correct to 2 s.f.).
(b) The principal gives $19\,000$ (correct to 3 s.f.).
(c) Both numbers are $19\,000$ . The first zero is significant only in (b), so the number alone cannot show the accuracy.

Question 17

A whole number, correct to 2 significant figures, is $4700$ . (Think of the halfway marks, as in Question 11.) Find
(a) the smallest whole number it could be
(b) the largest whole number it could be

ShowHide solution

The second significant figure of $4700$ is in the hundreds place, so the number was rounded to the nearest hundred.

(a) Halfway between $4600$ and $4700$ is $4650$ . Its next digit is 5, so it rounds up to $4700$ , and $4649$ rounds down to $4600$ . The smallest whole number is $4650$ .

(b) Halfway between $4700$ and $4800$ is $4750$ , which rounds up to $4800$ . The whole number just below it, $4749$ , rounds down to $4700$ .

(a) The smallest whole number it could be is $4650$ .
(b) The largest whole number it could be is $4749$ .

Key Points

Count the significant figures from the first non-zero digit. Look only at the next digit: 5 or more rounds up, 4 or less rounds down.
Before the decimal point, write a zero for each digit dropped: $38\,472 \approx 38\,000$ (correct to 2 s.f.). After the decimal point, leave the dropped digits out: $47.382 \approx 47.4$ (correct to 3 s.f.).
The zeros in front of a number between 0 and 1 are not counted, but they are written: $0.008101 \approx 0.00810$ (correct to 3 s.f.).
A zero that is one of the significant figures asked for must be written: $28.046 \approx 28.0$ , $89.95 \approx 90.0$ and $0.03996 \approx 0.0400$ (all correct to 3 s.f.).
When a 9 is rounded up, write 0 and carry 1. When the carry makes a new first significant figure, count the significant figures again from it: $0.99999 \approx 1.0$ and $9962 \approx 10\,000$ (both correct to 2 s.f.).
A rounded whole number that ends in zeros cannot show its own accuracy, so write "(correct to 2 s.f.)" after it.
A negative number is rounded by its numerical value and keeps its sign: $-38\,472 \approx -38\,000$ (correct to 2 s.f.).