Rounding to Decimal Places

Rounding a number to 1, 2 or 3 decimal places and to the nearest whole number: the rule, a digit of exactly 5, the zero that must be kept, a carry across the decimal point, why a number is rounded in one step, and how a negative decimal is rounded.

Before you start

In a decimal, the places after the decimal point are the tenths, the hundredths, the thousandths and the ten-thousandths. In $8.4627$ the 4 is in the tenths place, the 6 in the hundredths place, the 2 in the thousandths place and the 7 in the ten-thousandths place.

To round a whole number to the nearest 10, look at the ones digit: 5 or more rounds up, less than 5 rounds down. So $3847 \approx 3850$ and $3842 \approx 3840$ , each correct to the nearest 10. The sign $\approx$ is read "is approximately equal to".

Decimal Places

A decimal place is the position of a digit after the decimal point. The first decimal place is the tenths place, the second is the hundredths place, the third is the thousandths place, and so on. The number of decimal places a number is written with is the number of digits after its decimal point: $8.4627$ has 4 decimal places, $12.50$ has 2, and the whole number 7 has none. "Decimal places" is shortened to d.p.

Rounding to	means rounding to the nearest	The last digit kept is the
the nearest whole number	one, or $1$	ones digit
1 decimal place (1 d.p.)	tenth, or $0.1$	tenths digit
2 decimal places (2 d.p.)	hundredth, or $0.01$	hundredths digit
3 decimal places (3 d.p.)	thousandth, or $0.001$	thousandths digit

A question may use either wording: "to the nearest hundredth" and "to 2 decimal places" ask for the same thing. Rounding to 4 or more decimal places goes on in the same way.

Example 1: Naming decimal places

(a) Write down the digit in the second decimal place of $30.5082$ .
(b) Write down the digit in the thousandths place of $30.5082$ .
(c) How many decimal places does $30.5082$ have?
(d) How many decimal places do $6.20$ and 45 have?

Solution

(a) The digits after the decimal point are 5, 0, 8 and 2, in that order. The second of them is 0.

(b) The thousandths place is the third decimal place. The digit there is 8.

(d) $6.20$ has two digits after the decimal point, the 2 and the 0: a zero written after the decimal point counts as a decimal place. 45 has no decimal point, so it has no decimal places.

(a) 0
(b) 8
(c) $30.5082$ has 4 decimal places.
(d) $6.20$ has 2 decimal places and 45 has none.

Rounding on the Number Line

To round a number to 1 decimal place is to replace it by the nearest number that has one decimal place. $8.4627$ lies between two such numbers that are next to each other, $8.4$ and $8.5$ . The number halfway between them, $8.45$ , is called the halfway point, and it decides which is nearer. A number to the left of the halfway point is nearer the lower one, and a number to the right of it is nearer the upper one. Rounding to 2 or 3 decimal places is the same, between two numbers of 2 or 3 decimal places.

The words correct to say how a value was rounded: " $8.5$ , correct to 1 decimal place". A question that says "give your answer correct to 2 decimal places" is asking you to round your answer to 2 decimal places.

Example 2: One number to 1, 2 and 3 decimal places

Use a number line to round $8.4627$ to
(a) 1 decimal place
(b) 2 decimal places
(c) 3 decimal places.

Solution

(a) The numbers with one decimal place on either side of $8.4627$ are $8.4$ and $8.5$ . Halfway between them is $8.45$ . Since $8.4627$ is greater than $8.45$ , it is to the right of halfway, so it is nearer $8.5$ .

(b) With two decimal places, $8.4627$ lies between $8.46$ and $8.47$ . Halfway between them is $8.465$ . Since $8.4627$ is less than $8.465$ , it is to the left of halfway, so it is nearer $8.46$ .

(c) With three decimal places, $8.4627$ lies between $8.462$ and $8.463$ . Halfway between them is $8.4625$ . Since $8.4627$ is greater than $8.4625$ , it is nearer $8.463$ .

(a) $8.4627 \approx 8.5$ (correct to 1 d.p.)
(b) $8.4627 \approx 8.46$ (correct to 2 d.p.)
(c) $8.4627 \approx 8.463$ (correct to 3 d.p.)

The Rule for Rounding

You do not need to draw a number line every time. One digit tells you which side of halfway the number is on: the digit just after the last one you keep. In Example 2(a) the tenths digit is kept and the digit after it is 6. Because 6 is 5 or more, $8.4627$ is at or past the halfway point $8.45$ . In (b) the hundredths digit is kept and the digit after it is 2. Because 2 is less than 5, the number is before the halfway point $8.465$ .

Rule: rounding to a given number of decimal places

Find the last digit you keep: the tenths digit for 1 decimal place, the hundredths digit for 2 decimal places, the thousandths digit for 3 decimal places. For the nearest whole number it is the ones digit.
Look at the next digit to its right, and at that digit only.
If the next digit is 5 or more (5, 6, 7, 8 or 9), round up: add 1 to the last digit you keep. If the next digit is less than 5 (0, 1, 2, 3 or 4), round down: leave the last digit you keep as it is.
Drop every digit after the last digit you keep. Do not write zeros in their place: after the decimal point, the digits that are kept do not change their places. (A whole number rounded to the nearest 10 or 100 is different. There the zeros are needed to keep the other digits in their places: $3847 \approx 3850$ , not 385.)

Rounding down never makes the last digit smaller: it stays the same. Rounding up a 9 makes 10, so write 0 and carry 1 to the place on its left, as in addition.

Example 3: Using the rule

Round $89.451$ to the nearest
(a) tenth
(b) hundredth
(c) whole number.

Solution

(a) The nearest tenth means 1 decimal place. The last digit kept is the tenths digit, 4. The next digit is 5, which is 5 or more, so round up: the 4 becomes 5. Drop the digits 5 and 1.

(b) The nearest hundredth means 2 decimal places. The last digit kept is the hundredths digit, 5. The next digit is 1, which is less than 5, so round down: the 5 stays. Drop the 1.

(c) The last digit kept is the ones digit, 9. The next digit is the tenths digit, 4, which is less than 5, so round down: the 9 stays. Drop every digit after the decimal point.

(a) $89.451 \approx 89.5$ (correct to the nearest tenth)
(b) $89.451 \approx 89.45$ (correct to the nearest hundredth)
(c) $89.451 \approx 89$ (correct to the nearest whole number)

When the next digit is 5

A next digit of 5 with other digits after it that are not all zero puts the number past halfway. In Example 3(a), $89.451$ is greater than the halfway point $89.45$ , so it really is nearer $89.5$ . A next digit of 5 with nothing after it, or with only zeros after it, puts the number exactly halfway, equally near both neighbours. Then, so that everyone gives the same answer, we follow an agreement: a positive number exactly halfway is rounded up. Every number is positive until the section "Rounding Negative Decimals", where a negative number exactly halfway goes away from zero.

Example 4: Exactly halfway

Round
(a) $6.25$ to 1 decimal place
(b) $0.0635$ to 3 decimal places.

Solution

(a) $6.25$ lies between $6.2$ and $6.3$ , and it is the halfway point itself: $6.25 - 6.2 = 0.05$ and $6.3 - 6.25 = 0.05$ . Neither is nearer, so we use the agreement that a positive number exactly halfway is rounded up.

By the rule: the last digit kept is the tenths digit, 2. The next digit is 5, so round up: the 2 becomes 3.

(b) The last digit kept is the thousandths digit, 3. The next digit is 5, so round up: the 3 becomes 4. The zeros in front stay where they are.

(a) $6.25 \approx 6.3$ (correct to 1 d.p.)
(b) $0.0635 \approx 0.064$ (correct to 3 d.p.)

The Zero That Must Be Kept

When you round to 2 decimal places, the answer must show 2 decimal places, even when the last of them is 0. That zero is the hundredths digit of the answer. It may not be left out.

Example 5: Answers that end in zero

Round
(a) $4.03$ to 1 decimal place
(b) $2.497$ to 2 decimal places
(c) $15.0004$ to 3 decimal places.

Solution

(a) The last digit kept is the tenths digit, 0. The next digit is 3, which is less than 5, so the 0 stays. The answer is $4.0$ , with its zero. Writing 4 would be rounding to the nearest whole number.

(b) The last digit kept is the hundredths digit, 9. The next digit is 7, so round up. The 9 becomes 10: write 0 in the hundredths place and carry 1 to the tenths place, where the 4 becomes 5. In one line, $2.49 + 0.01 = 2.50$ . Keep the zero.

(c) The last digit kept is the thousandths digit, 0. The next digit is 4, so the 0 stays. All three zeros after the decimal point are written.

(a) $4.03 \approx 4.0$ (correct to 1 d.p.)
(b) $2.497 \approx 2.50$ (correct to 2 d.p.)
(c) $15.0004 \approx 15.000$ (correct to 3 d.p.)

What the zero tells the reader

$2.5$ and $2.50$ are the same value: they mark the same point on the number line. What differs is what each one tells the reader about the number that was rounded.

" $2.5$ , correct to 1 decimal place" says that the number rounds to $2.5$ , and not to $2.4$ or $2.6$ . It is somewhere from $2.45$ up to, but not including, $2.55$ .

" $2.50$ , correct to 2 decimal places" says that the number rounds to $2.50$ , and not to $2.49$ or $2.51$ . It is somewhere from $2.495$ up to, but not including, $2.505$ .

On each bar, the filled dot is a number that is included and the hollow dot is a number that is not: $2.55$ itself rounds to $2.6$ , and $2.505$ rounds to $2.51$ .

So $2.50$ is given to a higher degree of accuracy: it places the number within a much shorter stretch of the number line. It is not a different value, and it is not a closer one. Leaving out the zero throws that information away. Writing a zero that was not asked for claims an accuracy you do not have.

Example 6: The same value, a different accuracy

A laboratory balance shows the mass of a sample as 8.4695 g.
(a) Round the mass to 1, 2 and 3 decimal places.
(b) How far is each rounded value from 8.4695 g?
(c) Jun Wei says, "8.470 g is closer to the mass than 8.47 g, because it has more decimal places." Is he right?

Solution

(a) To 1 decimal place: the last digit kept is 4 and the next digit is 6, so round up to $8.5$ .

To 2 decimal places: the last digit kept is 6 and the next digit is 9, so round up to $8.47$ .

To 3 decimal places: the last digit kept is 9 and the next digit is 5, so round up. The 9 becomes 10: write 0 and carry 1 to the hundredths place, so $8.469 + 0.001 = 8.470$ . Keep the zero.

(b) $8.5 - 8.4695 = 0.0305$ , $8.47 - 8.4695 = 0.0005$ and $8.470 - 8.4695 = 0.0005$ .

(c) $8.47$ and $8.470$ are the same value, so they are equally far from $8.4695$ : each is $0.0005$ away. What the zero of 8.470 g does is tell the reader that the mass was rounded to the nearest 0.001 g, and not only to the nearest 0.01 g.

(a) 8.5 g (1 d.p.), 8.47 g (2 d.p.) and 8.470 g (3 d.p.)
(b) 0.0305 g, 0.0005 g and 0.0005 g
(c) No. 8.47 g and 8.470 g are equally close to the mass. 8.470 g is given to a higher degree of accuracy; it is not a closer value.

Keeping more decimal places often does bring the rounded value closer. In Example 2, $8.5$ , $8.46$ and $8.463$ are $0.0373$ , $0.0027$ and $0.0003$ away from $8.4627$ . It never takes the rounded value further away. But as Example 6 shows, it does not always bring it closer.

A Carry Across the Decimal Point

Rounding up adds 1 to the last digit you keep. When that digit is 9, the 1 is carried to the left, and the carry can pass the decimal point into the whole-number part. The rule is the same, and the zeros left behind after the decimal point are still written.

Example 7: When a 9 is rounded up

Round
(a) $9.96$ to 1 decimal place
(b) $0.996$ to 2 decimal places
(c) $19.998$ to 2 decimal places
(d) $59.52$ to the nearest whole number.

Solution

(a) $9.96$ lies between $9.9$ and $10.0$ . It is past the halfway point $9.95$ , so it is nearer $10.0$ .

By the rule: the last digit kept is the tenths digit, 9. The next digit is 6, so round up. The 9 becomes 10: write 0 in the tenths place and carry 1 to the ones place. The ones digit is also 9, so it becomes 10 too: write 0 and carry 1 to the tens place. In one line, $9.9 + 0.1 = 10.0$ . The answer has one decimal place, as asked: $10.0$ , not 10.

(b) The last digit kept is the hundredths digit, 9. The next digit is 6, so round up: $0.99 + 0.01 = 1.00$ . The carry runs through both 9s into the ones place. Keep both zeros.

(d) The last digit kept is the ones digit, 9. The next digit is 5, so round up: $59 + 1 = 60$ . No decimal place was asked for, so none is written: $60.0$ would claim 1 decimal place.

(a) $9.96 \approx 10.0$ (correct to 1 d.p.)
(b) $0.996 \approx 1.00$ (correct to 2 d.p.)
(c) $19.998 \approx 20.00$ (correct to 2 d.p.)
(d) $59.52 \approx 60$ (correct to the nearest whole number)

Round in One Step

Always round from the original number, in one step. Do not round to more places first and then round the result again. Each rounding moves the number a little, and two moves in the same direction can carry it across halfway.

Example 8: One step, not stages

Round
(a) $2.449$ to 1 decimal place
(b) $6.4546$ to the nearest whole number.

Solution

(a) The last digit kept is the tenths digit, 4. The next digit is 4, which is less than 5, so round down to $2.4$ . The 9 after it is not looked at.

Check on the number line: $2.449$ lies between $2.4$ and $2.5$ , just to the left of the halfway point $2.45$ . It is $0.049$ from $2.4$ and $0.051$ from $2.5$ , so $2.4$ is nearer.

The wrong way, in stages: $2.449$ to 2 decimal places is $2.45$ , and $2.45$ to 1 decimal place is $2.5$ . The first step moved the number up to $2.45$ , which is exactly halfway, and the second step moved it up again. The result, $2.5$ , is not the nearest number with one decimal place.

(b) The last digit kept is the ones digit, 6. The next digit is the tenths digit, 4, which is less than 5, so round down to 6.

The wrong way, in stages: $6.4546$ becomes $6.455$ , then $6.46$ , then $6.5$ , then 7. But $6.4546$ is less than the halfway point $6.5$ , so it is nearer 6 than 7.

(a) $2.449 \approx 2.4$ (correct to 1 d.p.)
(b) $6.4546 \approx 6$ (correct to the nearest whole number)

Rule: one step

Round the original number once, looking only at the first digit that is dropped. If you need the same number to a different number of decimal places, go back to the original number. Do not round a value that has already been rounded.

Rounding Negative Decimals

Every number so far has been positive. The numerical value of a number is the number without its sign: $-2.45$ has numerical value $2.45$ . A negative decimal is rounded by its digits in the same way. On the number line it still lies between two neighbours and goes to the nearer one. The neighbour with the larger numerical value is now on the left.

Rule: negative numbers

To round a negative number, round its numerical value by the rule, then put the negative sign back. A negative number that is exactly halfway is therefore rounded away from zero: $-2.45$ to 1 decimal place is $-2.5$ . If the result is zero, it is written without a sign: $-0.04$ to 1 decimal place is $0.0$ .

Example 9: Rounding negative decimals

Round
(a) $-3.276$ to 1 decimal place
(b) $-2.45$ to 1 decimal place
(c) $-0.698$ to 2 decimal places.

Solution

(a) The numerical value is $3.276$ . The last digit kept is 2 and the next digit is 7, so round up: $3.276 \approx 3.3$ . Put the sign back: $-3.3$ .

Check on the number line: $-3.276$ lies between $-3.3$ and $-3.2$ , to the left of the halfway point $-3.25$ , so it is nearer $-3.3$ .

(b) The numerical value is $2.45$ . The last digit kept is 4 and the next digit is 5, so round up: $2.45 \approx 2.5$ . Put the sign back: $-2.5$ . Here $-2.45$ is exactly halfway between $-2.5$ and $-2.4$ , and the answer is the one further from zero.

(c) The numerical value is $0.698$ . The last digit kept is 9 and the next digit is 8, so round up: $0.69 + 0.01 = 0.70$ . Put the sign back and keep the zero: $-0.70$ .

(a) $-3.276 \approx -3.3$ (correct to 1 d.p.)
(b) $-2.45 \approx -2.5$ (correct to 1 d.p.)
(c) $-0.698 \approx -0.70$ (correct to 2 d.p.)

Rounding up makes a positive number greater. With a negative number it is the other way round: rounding up the numerical value moves the number further from zero, to the left on the number line, so the rounded number is less than the original. Rounding down the numerical value moves it towards zero, so the rounded number is greater than the original, unless only zeros were dropped. So "halfway is rounded up" is true only of a positive number. What is true of every number is that halfway goes away from zero: to the right for a positive number and to the left for a negative one.

Example 10: A temperature below zero

The temperature inside a freezer is $-18.25$ °C.
(a) Round the temperature to 1 decimal place.
(b) Round the temperature to the nearest whole number.
(c) Is each rounded temperature higher or lower than the temperature in the freezer?

Solution

(a) The numerical value is $18.25$ . The last digit kept is the tenths digit, 2, and the next digit is 5, so round up: $18.25 \approx 18.3$ . Put the sign back: $-18.3$ .

(b) Go back to $-18.25$ . The numerical value is $18.25$ . The last digit kept is the ones digit, 8, and the next digit is 2, so round down: $18.25 \approx 18$ . Put the sign back: $-18$ .

(c) On the number line, $-18.3$ is to the left of $-18.25$ , so it is the lower temperature. $-18$ is to the right of $-18.25$ , so it is the higher temperature.

(a) The temperature is $-18.3$ °C, correct to 1 decimal place.
(b) The temperature is $-18$ °C, correct to the nearest whole number.
(c) $-18.3$ °C is lower than $-18.25$ °C, and $-18$ °C is higher.

Common Mistakes

Leaving out the zero. $9.96$ to 1 decimal place is $10.0$ , not 10, and $2.497$ to 2 decimal places is $2.50$ , not $2.5$ . The answer must show the number of decimal places asked for.
Writing zeros in place of the dropped digits. $8.4627$ to 1 decimal place is $8.5$ , not $8.5000$ . After the decimal point, the dropped digits are left out.
Rounding in stages. $2.449$ to 1 decimal place is $2.4$ . Going by way of $2.45$ gives $2.5$ , which is wrong. Round the original number once.
Making the last digit smaller when rounding down. $7.82$ to 1 decimal place is $7.8$ , not $7.7$ . Rounding down leaves the last digit kept as it is.
Thinking that $8.470$ is closer than $8.47$ . They are the same value. The zero shows a higher degree of accuracy, which is about how the number was rounded.
Dropping the sign, or rounding a negative number the wrong way at halfway. $-2.45$ to 1 decimal place is $-2.5$ , not $-2.4$ and not $2.5$ .

Practice

Work each question on paper before you open its solution.

Question 1

(a) Write down the digit in the third decimal place of $0.07360$ .
(b) How many decimal places is $0.07360$ written with?
(c) How many decimal places does "to the nearest thousandth" ask for?

ShowHide solution

(a) The digits after the decimal point are 0, 7, 3, 6 and 0, in that order. The third of them is 3.

(b) There are five digits after the decimal point. The zero at the end counts.

(a) 3
(b) 5 decimal places
(c) 3 decimal places

Question 2

Round $7.3862$ to
(a) the nearest whole number
(b) 1 decimal place
(c) 2 decimal places
(d) 3 decimal places.

ShowHide solution

(a) The last digit kept is the ones digit, 7. The next digit is 3, so round down: the 7 stays.

(b) The last digit kept is the tenths digit, 3. The next digit is 8, so round up: the 3 becomes 4.

(d) The last digit kept is the thousandths digit, 6. The next digit is 2, so round down: the 6 stays.

(a) $7.3862 \approx 7$ (correct to the nearest whole number)
(b) $7.3862 \approx 7.4$ (correct to 1 d.p.)
(c) $7.3862 \approx 7.39$ (correct to 2 d.p.)
(d) $7.3862 \approx 7.386$ (correct to 3 d.p.)

Question 3

Draw a number line from $4.3$ to $4.4$ and mark the number halfway between them. Use it to round each number to 1 decimal place.
(a) $4.35$
(b) $4.349$

ShowHide solution

The number halfway between $4.3$ and $4.4$ is $4.35$ .

(a) $4.35$ is the halfway point itself: it is $0.05$ from each end. A positive number exactly halfway is rounded up, so it goes to $4.4$ .

(b) $4.349$ is just to the left of halfway. It is $0.049$ from $4.3$ and $0.051$ from $4.4$ , so it is nearer $4.3$ .

(a) $4.35 \approx 4.4$ (correct to 1 d.p.)
(b) $4.349 \approx 4.3$ (correct to 1 d.p.)

Question 4

Round $9.00350$ to
(a) the nearest tenth
(b) 2 decimal places
(c) 3 decimal places.

ShowHide solution

(a) The nearest tenth means 1 decimal place. The last digit kept is the tenths digit, 0. The next digit is 0, so the tenths digit stays 0, and it is written.

(b) The last digit kept is the hundredths digit, 0. The next digit is 3, so the 0 stays. Both zeros are written.

(a) $9.00350 \approx 9.0$ (correct to the nearest tenth)
(b) $9.00350 \approx 9.00$ (correct to 2 d.p.)
(c) $9.00350 \approx 9.004$ (correct to 3 d.p.)

Question 5

The height of a door is measured as 2.036 m. Write the height correct to
(a) 2 decimal places
(b) 1 decimal place
(c) the nearest whole number.

ShowHide solution

(a) The last digit kept is the hundredths digit, 3. The next digit is 6, so round up: the 3 becomes 4.

(b) Go back to 2.036 m. The last digit kept is the tenths digit, 0. The next digit is 3, so round down: the 0 stays, and it is written.

(c) The last digit kept is the ones digit, 2. The next digit is 0, so round down: the 2 stays. No decimal place is written.

(a) The height is 2.04 m, correct to 2 decimal places.
(b) The height is 2.0 m, correct to 1 decimal place.
(c) The height is 2 m, correct to the nearest whole number.

Question 6

Round
(a) $0.95205$ to 1 decimal place
(b) $0.95205$ to 4 decimal places
(c) $29.996$ to 2 decimal places
(d) $99.97$ to 1 decimal place
(e) $199.5$ to the nearest whole number.

ShowHide solution

(a) The last digit kept is the tenths digit, 9. The next digit is 5, so round up. The 9 becomes 10: write 0 and carry 1 to the ones place, so $0.9 + 0.1 = 1.0$ . Keep the zero.

(b) The last digit kept is the ten-thousandths digit (the fourth decimal place), 0. The next digit is 5, so round up: the 0 becomes 1.

(c) The last digit kept is the hundredths digit, 9. The next digit is 6, so round up: $29.99 + 0.01 = 30.00$ . Keep both zeros.

(d) The last digit kept is the tenths digit, 9. The next digit is 7, so round up: $99.9 + 0.1 = 100.0$ . Keep the zero after the decimal point.

(e) The last digit kept is the ones digit, 9. The next digit is 5, so round up: $199 + 1 = 200$ . No decimal place is written.

(a) $0.95205 \approx 1.0$ (correct to 1 d.p.)
(b) $0.95205 \approx 0.9521$ (correct to 4 d.p.)
(c) $29.996 \approx 30.00$ (correct to 2 d.p.)
(d) $99.97 \approx 100.0$ (correct to 1 d.p.)
(e) $199.5 \approx 200$ (correct to the nearest whole number)

Question 7

An electronic balance shows the mass of a bag of flour as 0.9953 kg. Round the mass to
(a) 3 decimal places
(b) 2 decimal places
(c) 1 decimal place
(d) the nearest whole number.

ShowHide solution

(a) The last digit kept is the thousandths digit, 5. The next digit is 3, so round down: the 5 stays.

(b) The last digit kept is the hundredths digit, 9. The next digit is 5, so round up: $0.99 + 0.01 = 1.00$ . The carry passes the decimal point. Keep both zeros.

(d) The last digit kept is the ones digit, 0. The next digit is 9, so round up: $0 + 1 = 1$ . No decimal place is written.

The answers to (b), (c) and (d) are the same value. Each one tells the reader a different accuracy: the nearest 0.01 kg, the nearest 0.1 kg and the nearest 1 kg.

(a) 0.995 kg (3 d.p.)
(b) 1.00 kg (2 d.p.)
(c) 1.0 kg (1 d.p.)
(d) 1 kg (nearest whole number)

Question 8

Kumar rounds $5.3472$ to 1 decimal place like this: " $5.3472$ is $5.347$ , then $5.35$ , then $5.4$ ."
(a) What has Kumar done wrong?
(b) Find the correct answer.
(c) Show by subtraction that your answer is nearer $5.3472$ than Kumar's.

ShowHide solution

(a) Kumar rounded in stages: to 3 decimal places, then to 2, then to 1. Each stage moved the number up a little, and the last stage started from $5.35$ , which is not the original number.

(b) Round $5.3472$ in one step. The last digit kept is the tenths digit, 3. The next digit is 4, which is less than 5, so round down: the 3 stays.

(a) He rounded in stages instead of rounding the original number in one step.
(b) $5.3472 \approx 5.3$ (correct to 1 d.p.)
(c) $5.3$ is $0.0472$ away from $5.3472$ and $5.4$ is $0.0528$ away, so $5.3$ is nearer.

Question 9

Use a calculator to write $\frac{5}{11}$ as a decimal. Give its value correct to
(a) 1 decimal place
(b) 2 decimal places
(c) 3 decimal places
(d) 4 decimal places.

ShowHide solution

$\frac{5}{11} = 5 \div 11 = 0.454545\ldots$ Round this decimal each time, not an answer already rounded.

(a) The last digit kept is the tenths digit, 4. The next digit is 5, so round up: the 4 becomes 5.

(b) The last digit kept is the hundredths digit, 5. The next digit is 4, so round down: the 5 stays.

(d) The last digit kept is the ten-thousandths digit (the fourth decimal place), 5. The next digit is 4, so round down: the 5 stays.

(a) $\frac{5}{11} \approx 0.5$ (correct to 1 d.p.)
(b) $\frac{5}{11} \approx 0.45$ (correct to 2 d.p.)
(c) $\frac{5}{11} \approx 0.455$ (correct to 3 d.p.)
(d) $\frac{5}{11} \approx 0.4545$ (correct to 4 d.p.)

Question 10

In a science lesson, a data logger shows that a trolley takes 2.3496 s to roll down a ramp.
(a) Round the time to 3 decimal places.
(b) Round the time to 2 decimal places.
(c) Round the time to 1 decimal place.
(d) The time of a second trolley is shown only as 2.35 s, correct to 2 decimal places. Daniel says, "Then its time is 2.4 s, correct to 1 decimal place." Explain why he cannot be sure.

ShowHide solution

(a) The last digit kept is the thousandths digit, 9. The next digit is 6, so round up: $2.349 + 0.001 = 2.350$ . Keep the zero.

(b) Go back to 2.3496 s. The last digit kept is the hundredths digit, 4. The next digit is 9, so round up: the 4 becomes 5.

(c) Go back to 2.3496 s again. The last digit kept is the tenths digit, 3. The next digit is 4, which is less than 5, so round down: the 3 stays.

(d) A time that is 2.35 s, correct to 2 decimal places, is somewhere from 2.345 s up to, but not including, 2.355 s. If it is less than 2.35 s, as the first trolley's 2.3496 s is, it is 2.3 s, correct to 1 decimal place. If it is 2.35 s or more, it is 2.4 s. The rounded time alone does not say which. This is why a value that has already been rounded is not rounded again.

(a) 2.350 s (3 d.p.)
(b) 2.35 s (2 d.p.)
(c) 2.3 s (1 d.p.)
(d) A time of 2.35 s (2 d.p.) could be anything from 2.345 s up to, but not including, 2.355 s. Some of these times are 2.3 s and the others are 2.4 s, correct to 1 decimal place, so he cannot be sure.

Question 11

Which of these numbers are $3.70$ when rounded to 2 decimal places?
$3.695$ , $3.6949$ , $3.7049$ , $3.705$ , $3.7$ , $3.699$

ShowHide solution

Round each number to 2 decimal places. The last digit kept is the hundredths digit, and the next digit is the thousandths digit.

$3.695$ : the next digit is 5, so round up: $3.69 + 0.01 = 3.70$ . Yes.

$3.6949$ : the next digit is 4, so round down to $3.69$ . No.

$3.7049$ : the next digit is 4, so round down to $3.70$ . Yes.

$3.705$ : the next digit is 5, so round up to $3.71$ . No.

$3.7$ : no hundredths digit and no thousandths digit are written, and each of those places holds 0, so $3.7 = 3.700$ . The last digit kept is 0 and the next digit is 0, so round down to $3.70$ . Yes. Two decimal places were asked for, so the zero is written.

$3.699$ : the next digit is 9, so round up: $3.69 + 0.01 = 3.70$ . Yes.

$3.695$ , $3.7049$ , $3.7$ and $3.699$ are $3.70$ when rounded to 2 decimal places. Each of them lies from $3.695$ up to, but not including, $3.705$ .

Question 12

In a calculation, Mei Ling finds that a strip of paper must be 6.2964 cm wide.
(a) Round this width to 1, 2 and 3 decimal places.
(b) Two of your three answers are the same value. Which two?
(c) What does 6.30 cm tell a reader that 6.3 cm does not?
(d) Write down a width that is 6.3 cm, correct to 1 decimal place, but is not 6.30 cm, correct to 2 decimal places.

ShowHide solution

(a) To 1 decimal place: the last digit kept is 2 and the next digit is 9, so round up to $6.3$ .

To 2 decimal places: the last digit kept is 9 and the next digit is 6, so round up: $6.29 + 0.01 = 6.30$ . Keep the zero.

To 3 decimal places: the last digit kept is 6 and the next digit is 4, so round down to $6.296$ .

(b) $6.3$ and $6.30$ mark the same point on the number line.

(c) 6.30 cm says that the width was rounded to the nearest 0.01 cm, so it is somewhere from 6.295 cm up to, but not including, 6.305 cm. 6.3 cm says only that the width was rounded to the nearest 0.1 cm, so it could be anywhere from 6.25 cm up to, but not including, 6.35 cm.

(d) The width must round to 6.3 cm but lie outside the shorter stretch. So it is from 6.25 cm up to, but not including, 6.295 cm, or from 6.305 cm up to, but not including, 6.35 cm. For example, 6.27 cm is 6.3 cm, correct to 1 decimal place (the next digit is 7). Correct to 2 decimal places it is 6.27 cm, not 6.30 cm.

(a) 6.3 cm (1 d.p.), 6.30 cm (2 d.p.) and 6.296 cm (3 d.p.)
(b) 6.3 cm and 6.30 cm
(c) 6.30 cm is given to a higher degree of accuracy: it is correct to the nearest 0.01 cm, where 6.3 cm is correct only to the nearest 0.1 cm.
(d) For example, 6.27 cm. Any width from 6.25 cm up to, but not including, 6.295 cm, or from 6.305 cm up to, but not including, 6.35 cm, is right.

Question 13

Round
(a) $-7.062$ to 1 decimal place
(b) $-0.385$ to 2 decimal places
(c) $-14.96$ to 1 decimal place
(d) $-19.5$ to the nearest whole number
(e) $-0.03$ to 1 decimal place.

ShowHide solution

(a) The numerical value is $7.062$ . The last digit kept is 0 and the next digit is 6, so round up to $7.1$ . Put the sign back.

(b) The numerical value is $0.385$ . The last digit kept is 8 and the next digit is 5, so round up to $0.39$ . Put the sign back.

(c) The numerical value is $14.96$ . The last digit kept is 9 and the next digit is 6, so round up: $14.9 + 0.1 = 15.0$ . Put the sign back and keep the zero.

(d) The numerical value is $19.5$ . The last digit kept is the ones digit, 9, and the next digit is 5, so round up: $19 + 1 = 20$ . Put the sign back.

(e) The numerical value is $0.03$ . The last digit kept is the tenths digit, 0, and the next digit is 3, so round down to $0.0$ . The result is zero, so no sign is written.

(a) $-7.062 \approx -7.1$ (correct to 1 d.p.)
(b) $-0.385 \approx -0.39$ (correct to 2 d.p.)
(c) $-14.96 \approx -15.0$ (correct to 1 d.p.)
(d) $-19.5 \approx -20$ (correct to the nearest whole number)
(e) $-0.03 \approx 0.0$ (correct to 1 d.p.)

Question 14

Round each number to 1 decimal place, and say whether the rounded number is greater or less than the number you started with.
(a) $5.65$
(b) $-5.65$
(c) $5.62$
(d) $-5.62$

ShowHide solution

(a) The last digit kept is the tenths digit, 6. The next digit is 5, so round up: $5.65 \approx 5.7$ . On the number line, $5.7$ is to the right of $5.65$ , so it is greater.

(b) The numerical value is $5.65$ , which rounds up to $5.7$ as in (a). Put the sign back: $-5.7$ . On the number line, $-5.7$ is to the left of $-5.65$ , so it is less. ( $-5.65$ is exactly halfway between $-5.7$ and $-5.6$ , and it has gone away from zero.)

(c) The last digit kept is the tenths digit, 6. The next digit is 2, so round down: $5.62 \approx 5.6$ , which is less than $5.62$ .

(d) The numerical value is $5.62$ , which rounds down to $5.6$ as in (c). Put the sign back: $-5.6$ . On the number line, $-5.6$ is to the right of $-5.62$ , so it is greater.

Rounding up the numerical value takes a positive number to the right and a negative number to the left. Rounding down does the opposite.

(a) $5.65 \approx 5.7$ (correct to 1 d.p.), which is greater than $5.65$
(b) $-5.65 \approx -5.7$ (correct to 1 d.p.), which is less than $-5.65$
(c) $5.62 \approx 5.6$ (correct to 1 d.p.), which is less than $5.62$
(d) $-5.62 \approx -5.6$ (correct to 1 d.p.), which is greater than $-5.62$

Question 15

In a dry month, the water level of a reservoir changes by $-0.845$ m.
(a) Round the change to 2 decimal places.
(b) Round the change to 1 decimal place.
(c) Round the change to the nearest whole number.
(d) Which of your three answers are less than $-0.845$ ?

ShowHide solution

(a) The numerical value is $0.845$ . The last digit kept is the hundredths digit, 4, and the next digit is 5, so round up: $0.845 \approx 0.85$ . Put the sign back: $-0.85$ .

(b) Go back to $-0.845$ . The last digit kept is the tenths digit, 8, and the next digit is 4, so round down: $0.845 \approx 0.8$ . Put the sign back: $-0.8$ .

(c) The last digit kept is the ones digit, 0, and the next digit is 8, so round up: $0.845 \approx 1$ . Put the sign back: $-1$ .

(d) On the number line, $-0.85$ and $-1$ are to the left of $-0.845$ , so they are less than $-0.845$ . $-0.8$ is to the right of it, so it is greater.

(a) The change is $-0.85$ m, correct to 2 decimal places.
(b) The change is $-0.8$ m, correct to 1 decimal place.
(c) The change is $-1$ m, correct to the nearest whole number.
(d) $-0.85$ and $-1$ are less than $-0.845$ .

Question 16

Which of these numbers are $-1.4$ when rounded to 1 decimal place?
$-1.35$ , $-1.45$ , $-1.349$ , $-1.449$ , $-1.40$ , $-1.351$

ShowHide solution

Round the numerical value of each number to 1 decimal place and put the sign back. The last digit kept is the tenths digit, and the next digit is the hundredths digit.

$-1.35$ : the next digit is 5, so $1.35$ rounds up to $1.4$ . The number is $-1.4$ . Yes. ( $-1.35$ is exactly halfway between $-1.4$ and $-1.3$ , and it goes away from zero.)

$-1.45$ : the next digit is 5, so $1.45$ rounds up to $1.5$ . The number is $-1.5$ . No.

$-1.349$ : the next digit is 4, so $1.349$ rounds down to $1.3$ . The number is $-1.3$ . No.

$-1.449$ : the next digit is 4, so $1.449$ rounds down to $1.4$ . The number is $-1.4$ . Yes.

$-1.40$ : this is the same value as $-1.4$ . Yes.

$-1.351$ : the next digit is 5, so $1.351$ rounds up to $1.4$ . The number is $-1.4$ . Yes.

$-1.35$ , $-1.449$ , $-1.40$ and $-1.351$ are $-1.4$ when rounded to 1 decimal place. Each of them lies between $-1.45$ and $-1.35$ : the end nearer zero, $-1.35$ , is included, and the end further from zero, $-1.45$ , is not.

Question 17

The platform of an underground station is 42.95 m below ground level. An engineer records its level as $-42.95$ m.
(a) Round the level to 1 decimal place.
(b) Round the level to the nearest whole number.
(c) Farid writes $-42.9$ m for (a). He says, "A number exactly halfway is rounded up, and $-42.9$ is higher than $-42.95$ ." What is wrong with his reasoning?
(d) Is the level in your answer to (a) higher or lower than the platform?

ShowHide solution

(a) The numerical value is $42.95$ . The last digit kept is the tenths digit, 9, and the next digit is 5, so round up. The 9 becomes 10: write 0 and carry 1 to the ones place, so $42.9 + 0.1 = 43.0$ . Put the sign back and keep the zero: $-43.0$ .

(b) Go back to $-42.95$ . The numerical value is $42.95$ . The last digit kept is the ones digit, 2, and the next digit is 9, so round up: $42 + 1 = 43$ . Put the sign back: $-43$ . No decimal place is written.

(c) "Halfway is rounded up" is the rule for a positive number. $-42.95$ is exactly halfway between $-43.0$ and $-42.9$ : it is $0.05$ from each. A negative number is rounded by its numerical value, and $42.95$ rounds up to $43.0$ , so $-42.95$ goes away from zero, to $-43.0$ . Farid moved it towards zero.

(d) On the number line, $-43.0$ is to the left of $-42.95$ , so it is the lower level: 43.0 m below ground is a little deeper than the platform.

(a) The level is $-43.0$ m, correct to 1 decimal place.
(b) The level is $-43$ m, correct to the nearest whole number.
(c) He used the rule for a positive number. A negative number exactly halfway is rounded away from zero, so the level is $-43.0$ m, not $-42.9$ m.
(d) It is lower: $-43.0$ m is $0.05$ m below $-42.95$ m.

Key Points

A decimal place is a position after the decimal point. Rounding to 1, 2 or 3 decimal places is rounding to the nearest tenth, hundredth or thousandth.
Look only at the digit after the last one you keep: 5 or more, round up; less than 5, round down. Then drop every digit after the last one you keep.
A positive number exactly halfway is rounded up: $6.25 \approx 6.3$ (1 d.p.).
Show every decimal place asked for, zeros included: $4.03 \approx 4.0$ (1 d.p.) and $2.497 \approx 2.50$ (2 d.p.).
$2.5$ and $2.50$ are the same value. $2.50$ is given to a higher degree of accuracy; it is not closer.
A 9 that is rounded up carries to the left, across the decimal point if need be: $9.96 \approx 10.0$ (1 d.p.).
Round the original number in one step: $2.449 \approx 2.4$ (1 d.p.), not $2.5$ .
Round a negative number by its numerical value and put the sign back: $-3.276 \approx -3.3$ (1 d.p.).
A negative number that is exactly halfway is rounded away from zero: $-2.45 \approx -2.5$ (1 d.p.). A result of zero is written without a sign.