Rounding Money and Measurements

Rounding an amount of money to the nearest cent, 5 cents, 10 cents or dollar and a measurement with its unit, and giving the answer to a word problem correct to the accuracy the question states.

Before you start

To round a number to a given number of decimal places (d.p.), find the last digit you keep: the tenths digit for 1 decimal place, the hundredths digit for 2, the thousandths digit for 3, and the ones digit for the nearest whole number. Then look at the next digit only. 5 or more, round up: add 1 to the last digit you keep. Less than 5, round down: leave it as it is. Drop the digits after it.

So $8.4627 \approx 8.5$ (correct to 1 d.p.) and $8.4627 \approx 8.46$ (correct to 2 d.p.). The sign $\approx$ is read "is approximately equal to", and the words "correct to" say how a value was rounded.

Show every decimal place asked for, zeros included: $2.497 \approx 2.50$ (correct to 2 d.p.) and $9.96 \approx 10.0$ (correct to 1 d.p.). The zero tells the reader how the number was rounded.

Round the original number in one step: $2.449 \approx 2.4$ (correct to 1 d.p.). Rounding it to $2.45$ first and then to $2.5$ is wrong.

A negative decimal is rounded by its numerical value, which is the number without its sign, and the sign is then put back: $-3.276 \approx -3.3$ (correct to 1 d.p.). So a negative number that is exactly halfway goes away from zero: $-2.45 \approx -2.5$ (correct to 1 d.p.).

Money

An amount of money is rounded by the same rule. What is new is how the accuracy is named, in cents and dollars, and how the answer is written.

Rule: rounding money

One cent is $0.01$ of a dollar. So for an amount in dollars, the nearest cent is 2 decimal places, the nearest 10 cents is 1 decimal place and the nearest dollar is the nearest whole number.

An amount in dollars and cents is always written with two decimal places: $12.40, never $12.4. So $12.38 to the nearest 10 cents is written $12.40. To the nearest dollar it is $12.

The nearest 5 cents is not a decimal place. Find the two amounts ending in 0 or 5 cents on either side and take the nearer one. An amount exactly halfway between the two, such as $3.725, goes up, as a halfway number always does. For an amount in whole cents, a last digit of 1 or 2 goes down to 0, a last digit of 3 or 4 goes up to 5, a last digit of 6 or 7 goes down to 5, and a last digit of 8 or 9 goes up to the next 10 cents. A question will say when it is used, for example when a shop rounds a cash payment to the nearest 5 cents.

Why money is written with 2 decimal places

There are 100 cents in a dollar, so the two decimal places of an amount in dollars are its cents: $4.50 is 4 dollars and 50 cents. Written as $4.5, it could be misread as 4 dollars and 5 cents, which is $4.05. So when a calculator shows $4.5$ for an amount in dollars, write $4.50, not $4.5.

This is the way money is written. It does not by itself say how an amount was rounded, so say that in words: "$12.40, correct to the nearest 10 cents". An amount in whole dollars may be written without its cents: $12.

Example 1: Rounding money

Ribbon costs $1.35 per metre. Siti buys 2.75 m of it.
(a) Find the cost, correct to the nearest cent.
(b) Siti pays in cash, and the shop rounds a cash payment to the nearest 5 cents. How much does she pay?
(c) Find the cost, correct to the nearest dollar.

Solution

(a) $1.35 \times 2.75 = 3.7125$ , so the exact cost is $3.7125. The nearest cent is 2 decimal places. The last digit kept is the hundredths digit, 1. The next digit is 2, so round down: $3.71.

(b) The amounts ending in 0 or 5 cents on either side of $3.7125 are $3.70 and $3.75. Halfway between them is $3.725. Since $3.7125$ is less than $3.725$ , the cost is nearer $3.70.

(c) Round the original $3.7125, not $3.71 or $3.70. The last digit kept is the ones digit, 3. The next digit is 7, so round up: $4.

(a) The cost is $3.71, correct to the nearest cent.
(b) Siti pays $3.70.
(c) The cost is $4, correct to the nearest dollar.

Example 2: Amounts that end in zero

(a) Three identical notebooks cost $13.50 altogether. Find the cost of one notebook.
(b) Prawns cost $19.99 per kilogram. Mdm Lim buys 0.2 kg. Find the cost, correct to the nearest cent.
(c) Find the cost of the prawns, correct to the nearest dollar.

Solution

(a) $13.50 \div 3 = 4.5$ . The division is exact, so nothing is rounded. As an amount of money, $4.5$ dollars is written $4.50: 4 dollars and 50 cents.

(b) $19.99 \times 0.2 = 3.998$ , so the exact cost is $3.998. The nearest cent is 2 decimal places. The last digit kept is the hundredths digit, 9. The next digit is 8, so round up: $3.99 + 0.01 = 4.00$ . Both zeros are written.

(a) One notebook costs $4.50.
(b) The cost is $4.00, correct to the nearest cent.
(c) The cost is $4, correct to the nearest dollar.

Measurements

A rounded measurement keeps its unit, and the unit decides which decimal place is meant.

Rule: rounding a measurement

A measurement is rounded like any other number and keeps its unit: 12.847 s is 12.8 s, correct to 1 decimal place.

When the accuracy is named by a smaller unit, change that unit to the unit of the measurement first: 1 cm $= 0.01$ m, so for a length in metres the nearest centimetre is 2 decimal places.

The accuracy may also be given as $0.1$ , $0.01$ or $0.001$ of the unit, which are a tenth, a hundredth and a thousandth of it: the nearest 0.1 kg is 1 decimal place of a mass in kilograms.

The table gives the accuracies you will meet most often.

The measurement is in	The question says "to the nearest"	Round to
metres	centimetre (1 cm $= 0.01$ m)	2 decimal places
metres	millimetre (1 mm $= 0.001$ m)	3 decimal places
centimetres	millimetre (1 mm $= 0.1$ cm)	1 decimal place
kilograms	gram (1 g $= 0.001$ kg)	3 decimal places
kilograms	0.1 kg	1 decimal place
seconds	0.01 s	2 decimal places

Example 3: Rounding a length

A plank 2.5 m long is cut into 7 pieces of equal length. Find the length of each piece in metres, correct to
(a) 2 decimal places
(b) the nearest millimetre.

Solution

$2.5 \div 7 = 0.357142\ldots$ The decimal does not end, so the length has to be rounded.

(a) The last digit kept is the hundredths digit, 5. The next digit is 7, so round up: the 5 becomes 6.

(b) 1 mm $= 0.001$ m, so the nearest millimetre is 3 decimal places of a length in metres. Go back to $0.357142\ldots$ , not to the answer of (a). The last digit kept is the thousandths digit, 7. The next digit is 1, so round down: the 7 stays.

(a) Each piece is 0.36 m long, correct to 2 decimal places.
(b) Each piece is 0.357 m long, correct to the nearest millimetre. That is 357 mm.

Example 4: The nearest 0.1 kg and the nearest 0.01 s

(a) A baby has a mass of 6.847 kg. Round the mass to the nearest 0.1 kg.
(b) An electronic timer shows a sprinter's time as 11.396 s. Round the time to the nearest 0.01 s.
(c) Round the sprinter's time to the nearest second.

Solution

(a) $0.1$ is one tenth, so the nearest 0.1 kg is 1 decimal place. The last digit kept is the tenths digit, 8. The next digit is 4, so round down: the 8 stays.

(b) $0.01$ is one hundredth, so the nearest 0.01 s is 2 decimal places. The last digit kept is the hundredths digit, 9. The next digit is 6, so round up: $11.39 + 0.01 = 11.40$ . Keep the zero: it shows that the time is correct to the nearest 0.01 s.

(c) Go back to 11.396 s. The nearest second is the nearest whole number. The last digit kept is the ones digit, 1. The next digit is 3, so round down: the 1 stays.

(a) The mass is 6.8 kg, correct to the nearest 0.1 kg.
(b) The time is 11.40 s, correct to the nearest 0.01 s.
(c) The time is 11 s, correct to the nearest second.

Changing units: change first, round last

Sometimes the answer is wanted in a different unit from the one you worked in. Change the unit first, then round.

The order matters because "1 decimal place" is a different accuracy in each unit. For a length in metres it is the nearest 0.1 m, which is the nearest 10 cm. For a length in centimetres it is the nearest 0.1 cm, which is the nearest millimetre. A length rounded to 1 decimal place in metres and then changed to centimetres is correct only to the nearest 10 cm.

An accuracy named by a unit, such as "the nearest centimetre", is the same accuracy in every unit: 2 decimal places of a length in metres, or the nearest whole number of a length in centimetres. Then the rounding may be done in either unit, as long as it is done once, from the value before rounding.

Example 5: Changing units and rounding

A wire 10 m long is cut into 13 pieces of equal length. Find the length of each piece
(a) in centimetres, correct to 1 decimal place
(b) correct to the nearest centimetre.

Solution

$10 \div 13 = 0.769230\ldots$ , so each piece is $0.769230\ldots$ m long.

(a) Change the unit first. 1 m $= 100$ cm, so the length is $0.769230\ldots \times 100 = 76.9230\ldots$ cm. Now round. The last digit kept is the tenths digit, 9. The next digit is 2, so round down: the 9 stays.

The wrong order: rounding first gives 0.8 m, correct to 1 decimal place, and 0.8 m $= 80$ cm. But 80 cm is the length correct only to the nearest 10 cm. It is more than 3 cm longer than a piece really is.

(b) In centimetres: round $76.9230\ldots$ to the nearest whole number. The last digit kept is the ones digit, 6. The next digit is 9, so round up: the 6 becomes 7.

In metres: 1 cm $= 0.01$ m, so round $0.769230\ldots$ to 2 decimal places. The last digit kept is the hundredths digit, 6. The next digit is 9, so round up: 0.77 m, which is 77 cm. The two ways agree, because "the nearest centimetre" is the same accuracy in both units.

(a) Each piece is 76.9 cm long, correct to 1 decimal place.
(b) Each piece is 77 cm long, correct to the nearest centimetre.

Word Problems: Calculate First, Round Last

A word problem may end "give your answer correct to 2 decimal places" or "correct to the nearest cent". These words say how to round the final answer, not the numbers you work with.

Rule: calculate first, round once, at the end

Do the whole calculation with the numbers as they are given, and keep every digit your calculator shows.
Round once, at the end, to the accuracy the question states. Write the answer with its unit, and say how it was rounded.
If a later part of the question needs an earlier answer, use that answer as it was before rounding.

Example 6: The calculation first, the rounding last

A 4 kg bag of rice is packed equally into 9 small bags.
(a) Find the mass of rice in each small bag in kilograms, correct to 2 decimal places.
(b) Find the total mass of rice in 7 of the small bags in kilograms, correct to 2 decimal places.

Solution

(a) $4 \div 9 = 0.4444\ldots$ The last digit kept is the hundredths digit, 4. The next digit is 4, so round down: the 4 stays.

(b) Use the mass before it was rounded: $7 \times 4 \div 9 = 28 \div 9 = 3.1111\ldots$ The last digit kept is the hundredths digit, 1. The next digit is 1, so round down: the 1 stays.

The wrong way: using the rounded answer of (a) gives $7 \times 0.44 = 3.08$ , which is not 3.11. The little that was dropped in (a) has been dropped 7 times.

(a) Each small bag holds 0.44 kg of rice, correct to 2 decimal places.
(b) The 7 small bags hold 3.11 kg of rice, correct to 2 decimal places.

Why an early rounding makes the error grow, and by how much, is the subject of the lesson on rounding errors.

Common Mistakes

Writing $4.5 for $4.50. An amount in dollars and cents is written with two decimal places, so the zero is written.
Treating the nearest 5 cents as a decimal place. $6.43 to the nearest 5 cents is $6.45, not $6.40. Take the nearer of the two amounts ending in 0 or 5 cents.
Leaving out the zero or the unit of a measurement. 11.396 s to the nearest 0.01 s is 11.40 s. It is not 11.4 s, and it is not 11.40 with no unit.
Rounding first and changing the unit afterwards. A length of $0.769230\ldots$ m is 76.9 cm, correct to 1 decimal place, not 80 cm. Change the unit first.
Rounding in the middle of a calculation. Round only the final answer, and use the values before rounding in every later part.

Practice

Work each question on paper before you open its solution.

Question 1

(a) Round $18.476 to the nearest cent.
(b) Round $7.996 to the nearest cent.
(c) Round $36.25 to the nearest 10 cents.
(d) Round $249.50 to the nearest dollar.
(e) A calculator shows $23.6$ for a cost in dollars. Write the cost as an amount of money.

ShowHide solution

(a) The nearest cent is 2 decimal places. The last digit kept is the hundredths digit, 7. The next digit is 6, so round up: the 7 becomes 8.

(b) The last digit kept is the hundredths digit, 9. The next digit is 6, so round up: $7.99 + 0.01 = 8.00$ . Both zeros are written.

(c) The nearest 10 cents is 1 decimal place. The last digit kept is the tenths digit, 2. The next digit is 5, so round up: the 2 becomes 3. That is $36.3$ dollars, which is written with two decimal places as $36.30.

(d) The nearest dollar is the nearest whole number. The last digit kept is the ones digit, 9. The next digit is 5, so round up: $249 + 1 = 250$ .

(e) $23.6$ dollars is 23 dollars and 60 cents. Nothing is rounded: the zero is written because the two decimal places are the cents.

(a) $18.48
(b) $8.00
(c) $36.30
(d) $250
(e) $23.60

Question 2

Round
(a) 4.2738 m to the nearest centimetre
(b) 0.8496 m to the nearest millimetre
(c) 52.96 kg to the nearest 0.1 kg
(d) 9.995 s to the nearest 0.01 s
(e) 2.4035 kg to the nearest gram.

ShowHide solution

(a) 1 cm $= 0.01$ m, so round to 2 decimal places. The last digit kept is the hundredths digit, 7. The next digit is 3, so round down: the 7 stays.

(b) 1 mm $= 0.001$ m, so round to 3 decimal places. The last digit kept is the thousandths digit, 9. The next digit is 6, so round up: $0.849 + 0.001 = 0.850$ . Keep the zero. The length is 850 mm.

(c) The nearest 0.1 kg is 1 decimal place. The last digit kept is the tenths digit, 9. The next digit is 6, so round up: $52.9 + 0.1 = 53.0$ . Keep the zero.

(d) The nearest 0.01 s is 2 decimal places. The last digit kept is the hundredths digit, 9. The next digit is 5, so round up: $9.99 + 0.01 = 10.00$ . Keep both zeros.

(e) 1 g $= 0.001$ kg, so round to 3 decimal places. The last digit kept is the thousandths digit, 3. The next digit is 5, so round up: the 3 becomes 4.

(a) 4.27 m
(b) 0.850 m
(c) 53.0 kg
(d) 10.00 s
(e) 2.404 kg

Question 3

Four friends share the cost of a $57.50 birthday cake equally.
(a) Find each friend's share, correct to the nearest cent.
(b) Find each friend's share, correct to the nearest dollar.

ShowHide solution

(a) $57.50 \div 4 = 14.375$ , so each exact share is $14.375. The nearest cent is 2 decimal places. The last digit kept is the hundredths digit, 7. The next digit is 5, so round up: $14.38. (The share is exactly halfway between $14.37 and $14.38, and a positive number exactly halfway is rounded up.)

(b) Round the exact share, $14.375, not $14.38. The last digit kept is the ones digit, 4. The next digit is 3, so round down: $14.

(a) Each share is $14.38, correct to the nearest cent.
(b) Each share is $14, correct to the nearest dollar.

Question 4

A shop rounds a cash payment to the nearest 5 cents.
(a) How much is paid in cash for a bill of $6.42, a bill of $6.43 and a bill of $19.98?
(b) Grapes cost $7.90 per kilogram at the shop. Mr Tan buys 0.65 kg of grapes and pays in cash. How much does he pay?

ShowHide solution

(a) $6.42 lies between $6.40 and $6.45. It is 2 cents from $6.40 and 3 cents from $6.45, so it is rounded to $6.40.

$6.43 is 3 cents from $6.40 and 2 cents from $6.45, so it is rounded to $6.45.

$19.98 lies between $19.95 and $20.00. It is 3 cents from $19.95 and 2 cents from $20.00, so it is rounded to $20.00.

(b) $7.90 \times 0.65 = 5.135$ , so the exact cost is $5.135. The amounts ending in 0 or 5 cents on either side of it are $5.10 and $5.15. Halfway between them is $5.125. Since $5.135$ is greater than $5.125$ , the cost is nearer $5.15.

(a) $6.40, $6.45 and $20.00
(b) Mr Tan pays $5.15.

Question 5

A piece of string 2.3 m long is cut into 6 pieces of equal length. Find the length of each piece
(a) in metres, correct to the nearest centimetre
(b) in metres, correct to the nearest millimetre
(c) in centimetres, correct to 1 decimal place.

ShowHide solution

$2.3 \div 6 = 0.383333\ldots$ , so each piece is $0.383333\ldots$ m long. Round this value each time.

(a) 1 cm $= 0.01$ m, so round to 2 decimal places. The last digit kept is the hundredths digit, 8. The next digit is 3, so round down: the 8 stays.

(b) 1 mm $= 0.001$ m, so round to 3 decimal places. The last digit kept is the thousandths digit, 3. The next digit is 3, so round down: the 3 stays.

(c) Change the unit first: $0.383333\ldots \times 100 = 38.3333\ldots$ , so each piece is $38.3333\ldots$ cm long. The last digit kept is the tenths digit, 3. The next digit is 3, so round down: the 3 stays. (Rounding first would give 0.4 m, which is 40 cm: correct only to the nearest 10 cm.)

(a) Each piece is 0.38 m long, correct to the nearest centimetre.
(b) Each piece is 0.383 m long, correct to the nearest millimetre.
(c) Each piece is 38.3 cm long, correct to 1 decimal place.

Question 6

Three durians have masses of 1.85 kg, 2.4 kg and 2.06 kg. Find their average mass in kilograms, correct to
(a) 1 decimal place
(b) 2 decimal places
(c) the nearest gram.

ShowHide solution

Total mass $= 1.85 + 2.4 + 2.06 = 6.31$ kg

Average mass $= 6.31 \div 3 = 2.10333\ldots$ kg. Round this decimal each time.

(a) The last digit kept is the tenths digit, 1. The next digit is 0, so round down: the 1 stays.

(b) The last digit kept is the hundredths digit, 0. The next digit is 3, so the 0 stays, and it is written.

(c) 1 g $= 0.001$ kg, so the nearest gram is 3 decimal places of a mass in kilograms. The last digit kept is the thousandths digit, 3. The next digit is 3, so round down: the 3 stays.

(a) The average mass is 2.1 kg, correct to 1 decimal place.
(b) The average mass is 2.10 kg, correct to 2 decimal places.
(c) The average mass is 2.103 kg, correct to the nearest gram.

Question 7

A machine caps 11 bottles in 25 s, working at a steady rate.
(a) Find the time it takes to cap one bottle, correct to the nearest 0.01 s.
(b) Find the time it takes to cap 500 bottles, correct to the nearest second.
(c) Devi multiplies her answer to (a) by 500. What does she get, and why is it not the answer to (b)?

ShowHide solution

(a) $25 \div 11 = 2.272727\ldots$ The nearest 0.01 s is 2 decimal places. The last digit kept is the hundredths digit, 7. The next digit is 2, so round down: the 7 stays.

(b) Use the time before it was rounded: $500 \times 25 \div 11 = 1136.3636\ldots$ The nearest second is the nearest whole number. The last digit kept is the ones digit, 6. The next digit is 3, so round down: the 6 stays.

(c) $2.27 \times 500 = 1135$ . Devi calculated with a value that was already rounded. The $0.002727\ldots$ s dropped from the time for one bottle has been dropped 500 times, and that comes to more than 1 s.

(a) One bottle takes 2.27 s, correct to the nearest 0.01 s.
(b) 500 bottles take 1136 s, correct to the nearest second.
(c) She gets 1135 s. It is not the answer because she calculated with a rounded value: the rounding is done last.

Question 8

Twelve identical marbles have a total mass of 0.065 kg. Find the mass of one marble
(a) in kilograms, correct to 3 decimal places
(b) in grams, correct to 1 decimal place.
(c) Ben changes his answer to (a) into grams. What does he get, and how accurate is it?

ShowHide solution

$0.065 \div 12 = 0.0054166\ldots$ , so one marble has a mass of $0.0054166\ldots$ kg.

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

(b) Change the unit first. 1 kg $= 1000$ g, so the mass is $0.0054166\ldots \times 1000 = 5.4166\ldots$ g. The last digit kept is the tenths digit, 4. The next digit is 1, so round down: the 4 stays.

(c) 0.005 kg $= 5$ g. Three decimal places of a mass in kilograms is the nearest gram, so 5 g is correct only to the nearest gram. It is not the mass correct to 1 decimal place in grams, which is 5.4 g.

(a) One marble has a mass of 0.005 kg, correct to 3 decimal places.
(b) One marble has a mass of 5.4 g, correct to 1 decimal place.
(c) He gets 5 g, which is correct only to the nearest gram.

Question 9

A swimming pool is emptied at a steady rate. In 20 hours the water level changes by $-0.91$ m. Find the change in the water level in one hour
(a) in metres, correct to 2 decimal places
(b) in centimetres, correct to 1 decimal place
(c) in millimetres, correct to the nearest millimetre.

ShowHide solution

$-0.91 \div 20 = -0.0455$ , so the water level changes by $-0.0455$ m in one hour. The change is negative, so round its numerical value each time and put the sign back.

(a) The numerical value is $0.0455$ . The last digit kept is the hundredths digit, 4. The next digit is 5, so round up: $0.0455 \approx 0.05$ . Put the sign back: $-0.05$ .

(b) Change the unit first: $-0.0455 \times 100 = -4.55$ , so the change is $-4.55$ cm. The numerical value is $4.55$ . The last digit kept is the tenths digit, 5. The next digit is 5, so round up: $4.55 \approx 4.6$ . Put the sign back: $-4.6$ . (The change is exactly halfway between $-4.6$ cm and $-4.5$ cm, and a negative number that is exactly halfway goes away from zero.)

(c) $-0.0455 \times 1000 = -45.5$ , so the change is $-45.5$ mm. The numerical value is $45.5$ . The last digit kept is the ones digit, 5. The next digit is 5, so round up: $45.5 \approx 46$ . Put the sign back: $-46$ .

The answers to (b) and (c) are the same change, because 1 decimal place of a length in centimetres is the nearest millimetre. The answer to (a) is $-5$ cm: it is correct only to the nearest centimetre.

(a) The water level changes by $-0.05$ m in one hour, correct to 2 decimal places.
(b) The water level changes by $-4.6$ cm in one hour, correct to 1 decimal place.
(c) The water level changes by $-46$ mm in one hour, correct to the nearest millimetre.

Question 10

A pack of 6 cans of a drink costs $4.75. A pack of 10 cans of the same drink costs $7.85.
(a) Find the cost of one can in each pack, correct to the nearest cent.
(b) Which pack is the better buy? Explain how you know.

ShowHide solution

(a) Pack of 6: $4.75 \div 6 = 0.791666\ldots$ The nearest cent is 2 decimal places. The last digit kept is the hundredths digit, 9. The next digit is 1, so round down: $0.79.

Pack of 10: $7.85 \div 10 = 0.785$ . The last digit kept is the hundredths digit, 8. The next digit is 5, so round up: $0.79.

(b) The two rounded costs are the same, so they cannot decide. Compare the costs before rounding: $0.785$ is less than $0.791666\ldots$ , so a can in the pack of 10 costs less.

(a) A can costs $0.79 in each pack, correct to the nearest cent.
(b) The pack of 10 is the better buy. Before rounding, a can costs $0.785 in the pack of 10 and $ $0.791666\ldots$ in the pack of 6.

Key Points

For money in dollars, the nearest cent is 2 decimal places, the nearest 10 cents is 1 decimal place and the nearest dollar is the nearest whole number.
An amount in dollars and cents is written with two decimal places: $4.50, not $4.5.
The nearest 5 cents is the nearer of the two amounts ending in 0 or 5 cents on either side. An amount exactly halfway between them goes up.
A measurement keeps its unit. An accuracy named in a smaller unit is changed to the unit of the measurement: for a length in metres, the nearest centimetre is 2 decimal places.
The nearest 0.1 kg is 1 decimal place of a mass in kilograms, and the nearest 0.01 s is 2 decimal places of a time in seconds.
To give an answer in another unit, change the unit first and round last.
In a word problem, calculate first and round once, at the end. Use the values before rounding in every later part.