Benchmarks and Estimating Measures

Estimating lengths, masses, volumes and times from measures you already know, giving an estimate to a sensible accuracy in a sensible unit, and measuring many things together to estimate one.

Before you start

The usual rule of rounding looks at the first digit that is dropped: 5 or more rounds up, less than 5 rounds down. To round to 1 significant figure, keep the first non-zero digit and round it by this rule: $0.47 \approx 0.5$ , $6.3 \approx 6$ and $384 \approx 400$ . The sign $\approx$ means "is approximately equal to".

To estimate the result of a calculation, round each number to 1 significant figure (or to a convenient number) and work it out in your head: $38.6 \times 5.2 \approx 40 \times 5 = 200$ . An estimate that is more than the exact value is an overestimate, and one that is less is an underestimate. Here 200 is a slight underestimate of the exact value, 200.72.

To change units: $1\text{ cm} = 10\text{ mm}$ , $1\text{ m} = 100\text{ cm}$ , $1\text{ km} = 1000\text{ m}$ , $1\text{ kg} = 1000\text{ g}$ and $1\text{ litre} = 1000\text{ ml}$ . There are 60 seconds in a minute and 60 minutes in an hour. A smaller unit gives a larger number: $3.4\text{ cm} = 34\text{ mm}$ and $1.8\text{ kg} = 1800\text{ g}$ .

When an answer has to fit a situation, the situation decides which way to round, not the usual rule: 4.2 buses for a school outing means 5 buses, because everyone needs a seat.

Benchmarks

How tall is your school block? You cannot hold a measuring tape against it, but you can still estimate its height: count the storeys, and use what you know about the height of one storey. A measure that you know well and use in this way is called a benchmark.

Benchmark

A benchmark is a measure you already know, used as a reference for estimating a measure you do not know.

Choose a benchmark of the same kind (a length for a length, a mass for a mass) and not too different in size.
Judge how many times the benchmark fits into what you are estimating.
Multiply, and give the answer with the word "about".

Benchmarks worth remembering

Kind of measure	Benchmark	About
Length	the width of a fingernail	1 cm
Length	a long ruler, or the long side of a sheet of A4 paper	30 cm
Length	the height of a door handle above the floor	1 m
Length	one long step of a grown-up	a little less than 1 m
Length	the height of a door	2 m
Length	the height of one storey of a building	3 m
Mass	a teaspoon of water	5 g
Mass	a 1-litre carton of milk, or 1 litre of water	1 kg
Mass	a 5 kg bag of rice	5 kg
Volume of liquid	a teaspoon	5 ml
Volume of liquid	a can of drink	a third of a litre (many cans say 320 ml or 330 ml)
Volume of liquid	a carton of milk of the 1-litre size	1 litre
Volume of liquid	a large bottle of water of the 1.5-litre size	1.5 litres
Time	saying "one thousand and one" aloud	1 second
Time	walking 1 km at an ordinary pace	12 to 15 minutes

Your own body gives benchmarks that are always with you: your hand span, the length of your step, and your arm span, which is about the same as your height. Measure yours once and remember them. Packets, bottles and cans give more: the mass or the volume is printed on the label.

Example 1: Estimating a height

Estimate the height of
(a) a school block with 4 storeys,
(b) a block of flats with 25 storeys.

Solution

The benchmark is the height of one storey, about 3 m. This agrees with another benchmark: a door is about 2 m high, and there is about 1 m of wall and floor above it before the next storey begins.

(a) Height of the school block $\approx 4 \times 3\text{ m} = 12\text{ m}$

(b) Height of the block of flats $\approx 25 \times 3\text{ m} = 75\text{ m}$

Both answers come from a rough benchmark, so they are rough too. Say "about 75 m". An answer of 75.0 m would claim an accuracy that this estimate does not have.

(a) The school block is about 12 m high.
(b) The block of flats is about 75 m high.

Example 2: Using your step as a benchmark

Mr Tan has measured one of his long steps: it is 0.87 m. He paces along two sides of a rectangular room and counts 9 steps for its length and 7 steps for its breadth. Estimate the perimeter of the room.

Solution

His step is the benchmark. Going once round the room takes $2 \times (9 + 7) = 2 \times 16 = 32$ steps.

Round the length of a step to 1 significant figure: $0.87 \approx 0.9$

Perimeter $\approx 32 \times 0.9\text{ m} = 28.8\text{ m}$ , which is about 29 m.

The step was rounded up, so this is a slight overestimate: a calculator gives $32 \times 0.87 = 27.84$ . Taking the step as 1 m would be quicker still, but its answer, 32 m, is more than 4 m too much.

The perimeter of the room is about 29 m.

Example 3: Estimating a time

Hafiz has timed himself: he walks 1 km in about 12 minutes. The path round a reservoir is 4.3 km long. About how long will he take to walk round it?

Solution

The benchmark is his time for 1 km, about 12 minutes.

Round the distance to 1 significant figure: $4.3 \approx 4$

Time $\approx 4 \times 12\text{ minutes} = 48\text{ minutes}$

The distance was rounded down, so this is a slight underestimate: a calculator gives $4.3 \times 12 = 51.6$ . Nobody keeps exactly the same pace all the way, so "51.6 minutes" would claim too much. To the nearest 10 minutes, 48 and 51.6 are both 50.

Hafiz will take about 50 minutes.

A Sensible Answer: Accuracy and Units

An estimate should look like an estimate. Two things make it sensible: the accuracy it is given to, and the unit it is given in.

Rule: a sensible answer

An estimate is only as accurate as the numbers it was made from. From "about 3 m for each storey" the answer is "about 75 m", not 75.0 m.
A measurement is given no more accurately than the instrument can measure: a ruler reads to the nearest millimetre, and a kitchen scale to the nearest gram.
An answer worked out from a reading on a ruler, a scale or a stopwatch deserves no more significant figures than the reading has. A length read as 5.2 cm has 2 significant figures, so an answer worked out from it is given to 2 significant figures at most.
Choose the unit that gives a number that is easy to picture: the thickness of a coin in millimetres, the height of a door in metres, the distance from Jurong to Changi in kilometres, the length of a bus ride in minutes.
Check the size of every answer against a benchmark. A door 20 m high or an egg with a mass of 6 kg must be a mistake.

Example 4: Choosing a sensible estimate

Choose the most sensible estimate, and give a reason.
(a) The height of a double-decker bus: 1.5 m, 4.5 m or 15 m
(b) The mass of a school bag full of books: 500 g, 5 kg or 50 kg
(c) The amount of tea in a full mug: 30 ml, 300 ml or 3 litres

Solution

(a) A door is about 2 m high. Each deck of the bus has to be about as high as a door, so the bus is a little more than two doors high: 4.5 m. A bus 1.5 m high would be lower than a door, and one 15 m high would be as tall as a block of 5 storeys.

(b) 500 g is half a carton of milk, which is too light for a bag full of books. 50 kg is ten 5 kg bags of rice, far too heavy to carry to school. 5 kg, one bag of rice, is sensible.

(c) A can of drink holds about a third of a litre, a little over 300 ml, and a mug holds about as much as a can: 300 ml. 30 ml is only 6 teaspoons, and 3 litres is three cartons of milk.

(a) 4.5 m
(b) 5 kg
(c) 300 ml

Example 5: A calculator answer that says too much

A ribbon 2 m long is to be cut into 7 equal pieces. A calculator shows $2 \div 7$ as 0.2857142857. How long should each piece be cut?

Solution

The calculator shows 0.2857142857 m. Nobody can measure ribbon as finely as that, so most of these digits are of no use.

Work in the unit you will measure in. A ruler is marked in centimetres and millimetres: $200\text{ cm} \div 7 = 28.571\ldots\text{ cm}$

A ruler reads to the nearest millimetre, which is 1 decimal place in centimetres. The length lies between 28.5 cm and 28.6 cm.

Let the situation decide between them. $7 \times 28.6 = 200.2$ , which is more ribbon than there is. $7 \times 28.5 = 199.5$ , which leaves 0.5 cm over. So round down, although the usual rule of rounding would give 28.6.

Each piece should be cut 28.5 cm long.

Measuring Many to Estimate One

Some things are too thin or too light to measure one at a time. A ruler cannot measure the thickness of one sheet of paper, and a kitchen scale cannot weigh one grain of rice. But a ruler can measure a whole pack of paper, and a scale can weigh a whole cup of rice.

Rule: measure many, then divide

To estimate the size of one small thing, measure a known number of identical things together and divide:

size of one $\approx$ size of them all $\div$ number of things

The other way round, when you know the size of one you can estimate how many there are: number of things $\approx$ size of them all $\div$ size of one.

The more things you measure together, the better the estimate of one, because the small error in the measurement is shared among them all.

Example 6: The thickness of a sheet of paper

A ream of printer paper has 500 sheets. Jia Hui measures the thickness of a ream with a ruler: 5.2 cm. Estimate the thickness of one sheet, in millimetres.

Solution

Change the unit first: $5.2\text{ cm} = 52\text{ mm}$

Thickness of one sheet $\approx 52\text{ mm} \div 500 = 0.104\text{ mm}$

The thickness was read as 5.2 cm, to 2 significant figures only, so the answer deserves no more than 2: the third figure of 0.104 cannot be trusted. Correct to 2 significant figures it is 0.10 mm, and a ream is never pressed perfectly flat, so "about 0.1 mm" is the sensible answer.

One sheet of paper is about 0.1 mm thick.

Example 7: The thickness of a coin

Sulin stacks 25 one-dollar coins. The stack is 6.0 cm high.
(a) Estimate the thickness of one coin, in millimetres.
(b) About how high would a stack of 100 one-dollar coins be?

Solution

(a) Change the unit first: $6.0\text{ cm} = 60\text{ mm}$

Thickness of one coin $\approx 60\text{ mm} \div 25 = 2.4\text{ mm}$

The height was read as 6.0 cm: the zero shows that it was read to the nearest millimetre, so the reading has 2 significant figures. The answer, 2.4 mm, has 2 significant figures too, which is as many as the reading allows.

Check the size: a coin a little more than 2 mm thick is sensible. (Dividing 6.0 by 25 without changing the unit gives 0.24, and that is in centimetres, not millimetres.)

(b) Height of 100 coins $\approx 100 \times 2.4\text{ mm} = 240\text{ mm} = 24\text{ cm}$

(a) One coin is about 2.4 mm thick.
(b) A stack of 100 coins would be about 24 cm high.

Example 8: Estimating how many

Daniel wants to know roughly how many grains of rice there are in a 5 kg bag. He counts out 500 grains and puts them on a kitchen scale: their mass is 10 g. Estimate the number of grains in the bag.

Solution

First estimate the mass of one grain: $10\text{ g} \div 500 = 0.02\text{ g}$

Change the unit of the bag: $5\text{ kg} = 5000\text{ g}$

Number of grains $\approx 5000 \div 0.02 = 250\,000$

Another way to see it: 5000 g is 500 lots of 10 g, and each lot is about 500 grains, so there are about $500 \times 500 = 250\,000$ grains.

The grains are not all the same size, and the scale reads only to the nearest gram, so this is a rough figure. Say "about $250\,000$ ", or "about a quarter of a million".

There are about $250\,000$ grains of rice in the bag.

Common Mistakes

Giving an estimate to too many figures. "About 75 m" is honest. "75.0 m", or "0.104 mm" from a reading of 5.2 cm, claims more than the estimate can give.
Forgetting the unit. 6.0 cm ÷ 25 is 0.24 cm, which is 2.4 mm, not 0.24 mm. Change the unit before you divide, and check the answer against a benchmark.
Taking a benchmark too roughly. A step of 0.87 m taken as 1 m makes a perimeter of 32 steps come to 32 m, which is more than 4 m too much. Round it to 1 significant figure, 0.9 m, not to the nearest metre.
Dividing the wrong way round. For the thickness of one sheet, divide the thickness of the ream by the number of sheets: $52 \div 500$ , not $500 \div 52$ . Check the size of the answer: a sheet of paper cannot be 9.6 mm thick.

Practice

Work each question on paper before you open its solution.

Question 1

Choose the most sensible estimate, and give a reason.
(a) The length of a new pencil: 1.8 cm, 18 cm or 1.8 m
(b) The mass of an apple: 15 g, 150 g or 1.5 kg
(c) The amount of water in a full pail: 100 ml, 10 litres or 1000 litres
(d) The height of a classroom from floor to ceiling: 30 cm, 3 m or 30 m

ShowHide solution

(a) A new pencil is a little more than half as long as a 30 cm ruler: 18 cm. 1.8 cm is less than two fingernails wide, and 1.8 m is nearly as high as a door.

(b) 15 g is only 3 teaspoons of water, which is far too light for an apple. 1.5 kg is the mass of a large 1.5-litre bottle of water, which is far too heavy for one. 150 g is sensible: six or seven apples of this size make 1 kg.

(c) About ten cartons of milk would fill a pail: 10 litres. 100 ml is less than half a can of drink, and 1000 litres is a thousand cartons.

(d) A classroom is one storey high: 3 m. 30 cm is the length of a ruler, and 30 m is the height of a block of 10 storeys.

(a) 18 cm
(b) 150 g
(c) 10 litres
(d) 3 m

Question 2

Each measurement is correct, but its unit makes it hard to picture. Write it in a more sensible unit.
(a) The thickness of a mobile phone: 0.008 m
(b) The length of a cycling path: $185\,000$ cm
(c) The mass of a paper clip: 0.0005 kg
(d) The amount of water in a tank: $250\,000$ ml
(e) The time for a bus journey: 4500 seconds

ShowHide solution

(a) A phone is thin, so use millimetres. $1\text{ m} = 100\text{ cm} = 1000\text{ mm}$ , so $0.008\text{ m} = 0.008 \times 1000\text{ mm} = 8\text{ mm}$

(b) A path is long, so use kilometres. $185\,000\text{ cm} = 1850\text{ m} = 1.85\text{ km}$

(d) A tank is large, so use litres. $250\,000\text{ ml} = 250\text{ litres}$

(e) A journey is timed in minutes and hours. $4500 \div 60 = 75$ , so it takes 75 minutes, which is 1 hour 15 minutes.

Other answers can be sensible too: 0.8 cm for (a) and 1850 m for (b) are also easy to picture. What matters is to choose a unit that gives a number of an ordinary size.

(a) 8 mm
(b) 1.85 km
(c) 0.5 g
(d) 250 litres
(e) 75 minutes, or 1 hour 15 minutes

Question 3

Priya's hand span is 17 cm.
(a) The teacher's table is 7 of her hand spans long. Estimate its length, and give your answer in metres.
(b) A lamp post looks about 4 times as high as the door of the shop beside it. Estimate the height of the lamp post.

ShowHide solution

(a) Length of the table $\approx 7 \times 17\text{ cm} = 119\text{ cm}$

Hand spans cannot be laid end to end exactly, so the last digit cannot be trusted: say about 120 cm, which is 1.2 m.

(b) The benchmark is the height of a door, about 2 m. Height of the lamp post $\approx 4 \times 2\text{ m} = 8\text{ m}$

(a) The table is about 1.2 m long.
(b) The lamp post is about 8 m high.

Question 4

Mr Chua fills an empty fish tank with a 1.5-litre bottle. The tank is full after he has poured in 23 bottles of water.
(a) Estimate the amount of water in the tank.
(b) 1 litre of water has a mass of 1 kg. Estimate the mass of the water in the tank.
(c) Mr Chua says that the tank holds exactly 34.5 litres. Why is "about 35 litres" a more sensible answer?

ShowHide solution

(a) The bottle is the benchmark. Amount of water $\approx 23 \times 1.5\text{ litres} = 34.5\text{ litres}$ , which is about 35 litres.

(b) Each litre of water has a mass of 1 kg, so about 35 litres of water has a mass of about 35 kg. That is as heavy as seven 5 kg bags of rice.

(c) The bottle is not filled to exactly 1.5 litres every time, and a little water is spilt or left behind in it. Over 23 bottles these small differences add up, so the tenths of a litre cannot be trusted.

(a) The tank holds about 35 litres of water.
(b) The water has a mass of about 35 kg.
(c) The estimate is made from 23 rough measurements, so it is not accurate to a tenth of a litre.

Question 5

Siti walks 10 steps along a corridor and measures how far she has gone: 5.8 m.
(a) Estimate the length of one of her steps, correct to 1 significant figure.
(b) Siti takes 47 steps to walk the length of the school's basketball court. Estimate the length of the court.
(c) Is your estimate in (b) likely to be too high or too low? Give a reason.

ShowHide solution

(a) Length of one step $\approx 5.8\text{ m} \div 10 = 0.58\text{ m}$ , which is 0.6 m correct to 1 significant figure.

(b) Length of the court $\approx 47 \times 0.6\text{ m} = 28.2\text{ m}$ , which is about 28 m.

(c) 0.58 was rounded up to 0.6, so each of the 47 steps was counted as a little longer than it is. The estimate is likely to be too high. (A calculator gives $47 \times 0.58 = 27.26$ .)

(a) One step is about 0.6 m long.
(b) The court is about 28 m long.
(c) The estimate is likely to be a little too high, because the length of a step was rounded up.

Question 6

Dinesh has timed himself on the school track: he jogs one lap of 400 m in about 3 minutes.
(a) Estimate the time he takes to jog 2.4 km at this pace.
(b) His friend says, "At that pace you could jog 10 km in half an hour." Use the benchmark to show that this is not sensible.

ShowHide solution

(a) Change the unit first: $2.4\text{ km} = 2400\text{ m}$

Number of laps $= 2400 \div 400 = 6$

Time $\approx 6 \times 3\text{ minutes} = 18\text{ minutes}$

(b) $10\text{ km} = 10\,000\text{ m}$ , which is $10\,000 \div 400 = 25$ laps.

Time $\approx 25 \times 3\text{ minutes} = 75\text{ minutes}$ , which is 1 hour 15 minutes. Half an hour is less than half of that.

(a) Dinesh takes about 18 minutes.
(b) At this pace 10 km takes about 75 minutes, so "half an hour" is not sensible.

Question 7

A ream of 500 sheets of A4 paper has a mass of 2.5 kg.
(a) Estimate the mass of one sheet, in grams.
(b) A pile of the same paper has a mass of 1.2 kg. About how many sheets are in the pile?

ShowHide solution

(a) Change the unit first: $2.5\text{ kg} = 2500\text{ g}$

Mass of one sheet $\approx 2500\text{ g} \div 500 = 5\text{ g}$

(b) $1.2\text{ kg} = 1200\text{ g}$

Number of sheets $\approx 1200 \div 5 = 240$

(a) One sheet has a mass of about 5 g.
(b) There are about 240 sheets in the pile.

Question 8

One swing of a playground swing, forward and back, is over too quickly to time well with a stopwatch. Kavita times 20 swings instead: they take 50 seconds.
(a) Estimate the time for one swing.
(b) About how many times does it swing in 5 minutes, if it is kept going at the same rate?
(c) Why does timing 20 swings give a better estimate than timing one?

ShowHide solution

(a) Time for one swing $\approx 50\text{ seconds} \div 20 = 2.5\text{ seconds}$

(b) Change the unit first: 5 minutes is $5 \times 60 = 300$ seconds.

Number of swings $\approx 300 \div 2.5 = 120$

(c) Starting or stopping the stopwatch a moment early or late changes the reading by a fraction of a second. For one swing of about 2.5 seconds that is a large part of the answer. Over 20 swings the same small error is shared among them all.

(a) One swing takes about 2.5 seconds.
(b) It swings about 120 times in 5 minutes.
(c) The small error in the timing is shared among 20 swings and does not all fall on one.

Question 9

Jun Wei has written four answers. Check the size of each against a benchmark. Say which answers cannot be right, and suggest what each of them should be.
(a) "One sheet of paper is 10.4 mm thick."
(b) "My school bag has a mass of 4.5 kg."
(c) "A full pail holds 12 ml of water."
(d) "It takes me 20 seconds to walk 1 km."

ShowHide solution

(a) Cannot be right. 10.4 mm is the width of a fingernail, and a ream of 500 sheets as thick as that would be $500 \times 10.4\text{ mm} = 5200\text{ mm}$ , which is 5.2 m. The decimal point is in the wrong place: a ream is about 5.2 cm thick, and $52\text{ mm} \div 500 = 0.104\text{ mm}$ , which is about 0.1 mm.

(b) Sensible. 4.5 kg is a little less than a 5 kg bag of rice, which is about right for a bag full of books.

(d) Cannot be right. Walking 1 km takes about 12 to 15 minutes, and 20 seconds is hardly time to cross a road. The unit is wrong: 20 minutes is sensible for a slow walk.

Answers (a), (c) and (d) cannot be right. They should be about 0.1 mm, 12 litres and 20 minutes. Answer (b) is sensible.

Question 10

A 1.5-litre bottle of juice is to be shared equally among 9 cups. A calculator shows $1500 \div 9$ as 166.6666667. The measuring jug is marked every 10 ml, so it measures to the nearest 10 ml. How much juice should be measured out for each cup?

ShowHide solution

The calculation is in millilitres: $1.5\text{ litres} = 1500\text{ ml}$ , so the calculator shows 166.6666667 ml. No jug can measure as finely as that.

The jug measures to the nearest 10 ml, so give the answer to the nearest 10 ml. The amount lies between 160 ml and 170 ml.

Let the situation decide between them. $9 \times 170 = 1530$ , which is more juice than there is. $9 \times 160 = 1440$ , which leaves 60 ml over. So round down, although the usual rule of rounding would give 170.

160 ml of juice should be measured out for each cup.

Question 11

A storage tray has 6 identical compartments in 2 rows of 3, as shown. Amir measures one compartment with a ruler: it is 9.8 cm long and 7.2 cm wide. (Ignore the thickness of the walls between the compartments.)
(a) Estimate the area of the base of the tray.
(b) Amir's calculator shows 423.36 for the area. Why does " $423.36\text{ cm}^{2}$ " claim too much, and how should the area be given?
(c) A sheet of A4 paper is about 30 cm long and 21 cm wide. Use it as a benchmark to check the size of your answer: about what fraction of the sheet would the tray cover?

ShowHide solution

(a) Round each measurement to 1 significant figure: $9.8 \approx 10$ and $7.2 \approx 7$

Length of the tray $\approx 3 \times 10\text{ cm} = 30\text{ cm}$

Breadth of the tray $\approx 2 \times 7\text{ cm} = 14\text{ cm}$

Area of the base $\approx 30 \times 14 = 420\text{ cm}^{2}$

(b) The calculator has worked with the measurements as they were read: $3 \times 9.8 = 29.4$ , $2 \times 7.2 = 14.4$ and $29.4 \times 14.4 = 423.36$ . Each measurement was read to 2 significant figures only, so the answer deserves no more than 2: the digits after 42 cannot be trusted. Correct to 2 significant figures, 423.36 is 420, which is the estimate from (a).

(c) The sheet is about 30 cm by 21 cm, and the tray is about 30 cm by 14 cm. They are about the same length, and the breadths are 14 cm against 21 cm: $\frac{14}{21} = \frac{2}{3}$ . So the tray would cover about two thirds of the sheet.

This agrees with the area: the sheet is about $30 \times 21 = 630\text{ cm}^{2}$ , and two thirds of 630 is 420. A slip that gave $4200\text{ cm}^{2}$ would be the area of more than six sheets of A4, which a tray of this size cannot have.

(a) The area of the base of the tray is about $420\text{ cm}^{2}$ .
(b) The measurements have only 2 significant figures, so the area is given as about $420\text{ cm}^{2}$ , not $423.36\text{ cm}^{2}$ .
(c) The tray would cover about two thirds of a sheet of A4, so $420\text{ cm}^{2}$ is a sensible size.

Key Points

A benchmark is a measure you know (a door is about 2 m high, a 1-litre carton of milk has a mass of about 1 kg, walking 1 km takes about 12 to 15 minutes), used to estimate a measure you do not know.
To use a benchmark, choose one of the same kind, judge how many times it fits, multiply, and say "about".
Give an estimate with the word "about", to no more figures than the numbers behind it deserve, and in a unit that suits its size.
An answer worked out from a reading has no more significant figures than the reading: from 5.2 cm, 2 significant figures at most.
Check the size of every answer against a benchmark. It shows up a wrong unit or a decimal point in the wrong place.
To estimate one small thing, measure many of them together and divide.
To estimate how many there are, divide the size of them all by the size of one.
Change everything to one unit before you multiply or divide.