Maths · 1 Number

1.9 Estimation

Before you start

Place value in whole numbers and decimals: which digit is the tens, the tenths, the hundredths.
Multiplying and dividing by 10, 100 and 1000 in your head, e.g. $0.05 \times 300 = 15$ .
Dividing by a decimal by first multiplying both numbers by 10, e.g. $6 \div 0.2 = 60 \div 2 = 30$ .
The square numbers up to $15^{2} = 225$ , so that you can spot $\sqrt{36} = 6$ or $\sqrt{100} = 10$ .

Syllabus objectives covered (3)

Round values to a specified degree of accuracy.
Make estimates for calculations involving numbers, quantities and measurements.
Round answers to a reasonable degree of accuracy in the context of a given problem.

Part 1 of 6

Rounding to the nearest 10, 100 or 1000

Rounding replaces a number with a simpler one that is close to it. The question tells you how simple: to the nearest thousand, to 2 decimal places, to 3 significant figures. That instruction is the degree of accuracy.
Every kind of rounding uses the same method. Find the last digit you are allowed to keep. Then look at the deciding digit, the one straight after it. If the deciding digit is 5 or more, the kept digit goes up by 1. If it is 4 or less, the kept digit stays as it is.
Round $5764$ to the nearest thousand. The thousands digit is 5, and the deciding digit after it is 7. Seven is 5 or more, so the 5 goes up to 6. The answer is $6000$ .
$5764 \approx 6000 (nearest thousand)$
The digits you remove become zeros. They are not deleted. $5764$ to the nearest thousand is $6000$ , not $6$ . The zeros keep every digit in its place, so the rounded number is still about the same size as the original.
A deciding digit of exactly 5 rounds up. $39 650$ to the nearest hundred: the hundreds digit is 6, the deciding digit is 5, so the 6 becomes 7. The answer is $39 700$ .
Sometimes the kept digit is a 9, and going up makes it carry. $9970$ to the nearest hundred: the hundreds digit is 9, the deciding digit is 7, so the 9 goes up. That gives 10 hundreds, which carries into the thousands. The answer is $10 000$ .
Check your answer against the two numbers either side. $5764$ is between $5000$ and $6000$ , so the answer to the nearest thousand must be one of those two. If your answer is neither, you have lost a place value.

Part 2 of 6

Rounding to decimal places

Decimal places (dp) count the digits after the decimal point. $3.142$ has 3 decimal places. Rounding to 2 dp means you keep exactly two digits after the point.
The method is the same. $3.14159$ to 2 dp: keep $3.14$ , and the deciding digit is 1, so nothing changes. The answer is $3.14$ . To 3 dp, keep $3.141$ ; the deciding digit is 5, so the last 1 goes up and the answer is $3.142$ .
Write every decimal place the question asks for, even when it is a zero. $3.996$ to 2 dp: keep $3.99$ , and the deciding digit 6 makes the last 9 go up. That carries twice and gives $4.00$ . Write $4.00$ , not $4$ . The question asked for 2 decimal places, so two must be on the page.
Round once, from the original number. $2.447$ to 1 dp: keep $2.4$ , the deciding digit is 4, so the answer is $2.4$ . If you first round to $2.45$ and then round that to 1 dp, you get $2.5$ , which is wrong. The first rounding moved the number, and the second rounding used the moved number.

Part 3 of 6

Rounding to significant figures

Significant figures (sf) start at the first non-zero digit, wherever it is. In $48 350$ the first significant figure is the 4. In $0.004072$ it is the 4 as well. From there, every digit counts, zeros included.
Zeros at the front of a decimal are not significant. They only show how small the number is. $0.004072$ has 4 significant figures: 4, 0, 7 and 2.
Zeros between non-zero digits are significant. In $20 503$ the first three significant figures are 2, 0 and 5. To 3 sf the deciding digit is the next 0, so the answer is $20 500$ .
In a whole number, the digits you remove become zeros, exactly as before. $48 350$ to 2 sf: keep 4 and 8, the deciding digit is 3, so the answer is $48 000$ . To 3 sf: keep 4, 8 and 3, the deciding digit is 5, so the 3 goes up and the answer is $48 400$ .
In a decimal, you stop after the last kept digit. You do not add zeros after it. $0.004072$ to 2 sf: the kept digits are 4 and 0, the deciding digit is 7, so the 0 goes up to 1. The answer is $0.0041$ .
A zero at the end of a rounded decimal can be one of the significant figures. Then you must write it. $0.0602$ to 2 sf: the kept digits are 6 and 0, the deciding digit is 2, so the answer is $0.060$ . If you write $0.06$ , you have given only 1 sf.
To 1 sf you keep only the first non-zero digit. $351$ becomes $400$ , $0.0486$ becomes $0.05$ , and $9.79$ becomes $10$ . That last one is still 1 sf: the 9 went up and carried into the next place. You will use 1 sf for estimating in the next section.

Part 4 of 6

Estimating a calculation

An estimate is an approximate answer that you can work out without a calculator. The standard method: round every number to 1 significant figure, then do the calculation with the rounded numbers.
Write the rounded numbers down before you calculate. They are your method. An answer on its own does not show that you rounded first.
Estimate $\frac{62.8}{4.13 \times 0.487}$ . To 1 sf, $62.8$ is $60$ , $4.13$ is $4$ and $0.487$ is $0.5$ . The denominator is $4 \times 0.5 = 2$ , and $60 \div 2 = 30$ .
$\frac{62.8}{4.13 \times 0.487} \approx \frac{60}{4 \times 0.5} = \frac{60}{2} = 30$
Use $\approx$ , which means 'is approximately equal to', between the real calculation and the rounded one. After that, the arithmetic with the rounded numbers is exact, so you can use $=$ .
Dividing by a decimal is where estimates go wrong. For $60 \div 0.2$ , multiply both numbers by 10 to get $600 \div 2 = 300$ . Notice the answer is bigger than 60. That is correct: dividing a positive number by a number between 0 and 1 always makes the answer bigger.
Read the instruction. If the question says 'by writing each number correct to 1 significant figure', use exactly the 1 sf values, even if a nearby number would be easier. If it only says 'estimate', you may choose a convenient nearby number, such as $\sqrt{35.6} \approx \sqrt{36} = 6$ , as long as you show it.
The same method works for quantities and measurements. A room that measures $4.85$ m by $3.12$ m has a floor area of about $5 \times 3 = 15$ m². The exact area is $15.132$ m², so the estimate is close.

Part 5 of 6

Using an estimate to check, and whether it is too big or too small

A calculator does exactly what you type, so an estimate is your check that you typed the right thing. Do the estimate first, then compare it with the calculator answer.
Type $62.8 \div 4.13 \times 0.487$ without brackets, and the calculator divides by $4.13$ and then multiplies by $0.487$ . It shows $7.405 \dots$ . Your estimate of 30 tells you at once that this is wrong. Put brackets around the whole denominator, $62.8 \div (4.13 \times 0.487)$ , and it shows $31.22 \dots$ , which agrees with the estimate.
An estimate checks the size of the answer, not its exact digits. $31.2$ against an estimate of 30 is fine. $7.4$ or $312$ against an estimate of 30 means something went wrong.
To decide whether an estimate is too big or too small, look at which way each number moved. In $4.6 \times 2.8 \approx 5 \times 3 = 15$ , both numbers were rounded up, so the product is too big. The estimate is an overestimate. The exact value is $12.88$ .
In a division, the number on the bottom works the other way. Dividing by a smaller number gives a bigger answer. In $8.7 \div 2.2 \approx 9 \div 2 = 4.5$ , the 8.7 went up and the 2.2 went down. Both changes make the answer bigger, so $4.5$ is an overestimate. The exact value is $3.95 \dots$
If the roundings pull in opposite directions, you cannot tell from the directions alone. In $3.68 \times 52.4 \approx 4 \times 50 = 200$ , one number went up and the other went down. Say that you cannot be sure without more work. Do not guess.

Part 6 of 6

Rounding your answer to suit the question

If the question states the accuracy ('give your answer correct to 2 decimal places'), use exactly that.
If the question does not state it, the rule for your paper is: give a non-exact answer to 3 significant figures, and an angle in degrees to 1 decimal place. A calculator answer of $55.769230 \dots$ is written as $55.8$ .
If the calculator answer is exact and has 4 or 5 significant figures, do not round it. $12.4 \times 7.85 = 97.34$ exactly, so the answer is $97.34$ .
Write the longer calculator value first, then the rounded answer: $55.769 \dots = 55.8$ . This shows where your answer came from, and it lets you check the rounding.
Round only at the end. If an answer feeds into a later step, carry the full value forward (use the calculator's memory or Ans key). $300 \div 7 = 42.857 \dots = 42.9$ to 3 sf. But if you round $\frac{1}{7}$ to $0.14$ first and then multiply by 300, you get $42$ , which is wrong to 3 sf.
Money is written to the nearest cent, which is 2 decimal places of a dollar. Twelve pens for $5.39 cost $5.39 \div 12 = 0.449166 \dots$ dollars each, which is $0.45 per pen.
If the answer counts whole objects, it must be a whole number. The context decides the direction. Round up when everything must be covered: 265 people in coaches that seat 52 gives $265 \div 52 = 5.09 \dots$ , so 6 coaches are needed, because 5 coaches leave people behind.
Round down when only complete items count. Pieces $0.35$ m long cut from $2.5$ m of ribbon: $2.5 \div 0.35 = 7.14 \dots$ , so you get 7 pieces. The $0.14 \dots$ is only part of a piece (just $2.5 - 7 \times 0.35 = 0.05$ m of ribbon is left), so it does not count.
A reasonable degree of accuracy also means you do not claim more accuracy than the data has. Lengths measured to the nearest tenth of a metre do not give an area that is correct to six decimal places. Three significant figures is enough.

Worked examples

Example 1. Write (a) $0.0050738$ correct to 2 significant figures, (b) $0.0050738$ correct to 3 decimal places, (c) $0.0050738$ correct to 3 significant figures, (d) $7.9963$ correct to 2 decimal places, (e) $29 951$ correct to the nearest hundred.

Show the working

(a) The zeros at the front are not significant. The first significant figure is the 5, the second is the 0 after it. The deciding digit is 7, so the 0 goes up to 1: $0.0051$ .
(b) Decimal places count from the point, zeros included. The first three decimal places are 0, 0 and 5. The deciding digit is 0, so nothing changes: $0.005$ . Notice that (a) and (b) give different answers. Significant figures and decimal places are different instructions.
(c) The three significant figures are 5, 0 and 7. The deciding digit is 3, so nothing changes: $0.00507$ .
(d) Keep $7.99$ . The deciding digit is 6, so the last 9 goes up. It carries through the other 9 into the units, giving $8.00$ . Write both zeros, because the question asked for 2 decimal places.
(e) The hundreds digit is 9 and the deciding digit is 5, so the 9 goes up and carries: 299 hundreds become 300 hundreds. The answer is $30 000$ . Check: $29 951$ is between $29 900$ and $30 000$ , and it is closer to $30 000$ .

Answer: (a) $0.0051$ (b) $0.005$ (c) $0.00507$ (d) $8.00$ (e) $30 000$

Example 2. By writing each number correct to 1 significant figure, work out an estimate for $\frac{\sqrt{83.1 + 17.6}}{0.198}$ . Show your working.

Show the working

Round every number to 1 sf and write each one down: $83.1 \to 80$ , $17.6 \to 20$ , $0.198 \to 0.2$ . The $0.198$ has 1 as its first significant figure and 9 as the deciding digit, so it rounds up to $0.2$ .
Work out the top: $80 + 20 = 100$ , and $\sqrt{100} = 10$ . Rounding each number before adding is what the instruction asks for, and here it happens to give a square number.
$\frac{\sqrt{83.1 + 17.6}}{0.198} \approx \frac{\sqrt{80 + 20}}{0.2} = \frac{10}{0.2}$
Divide by the decimal: multiply both numbers by 10, so $10 \div 0.2 = 100 \div 2 = 50$ . The answer is bigger than 10, which it should be, because you divided by a number less than 1.
Check with a calculator if you have one: the exact value is $50.68 \dots$ , so the estimate is close.

Answer: $\frac{\sqrt{80 + 20}}{0.2} = 50$

Example 3. Leo works out $\frac{738}{18.2 \times 0.39}$ on his calculator and gets $15.81 \dots$ (a) By writing each number correct to 1 significant figure, work out an estimate. (b) Use your estimate to explain why Leo's answer is wrong. (c) Find the correct answer, correct to 3 significant figures.

Show the working

(a) To 1 sf: $738 \to 700$ , $18.2 \to 20$ , $0.39 \to 0.4$ . The denominator is $20 \times 0.4 = 8$ , and $700 \div 8 = 87.5$ .
$\frac{738}{18.2 \times 0.39} \approx \frac{700}{20 \times 0.4} = \frac{700}{8} = 87.5$
(b) The estimate is about 90. Leo's answer is about 16, more than five times too small, so it cannot be right. An answer that far from the estimate means the calculation was entered wrongly.
Find the slip. $738 \div 18.2 \times 0.39$ typed without brackets gives $15.81 \dots$ , because the calculator divides by 18.2 and then multiplies by 0.39. The 0.39 must be part of the denominator.
(c) Type it with brackets: $738 \div (18.2 \times 0.39) = 738 \div 7.098 = 103.972 \dots$ Round to 3 sf: the deciding digit after $103$ is 9, so the 3 goes up to give $104$ .
Compare: $104$ is close to the estimate of $87.5$ . It is a bit bigger because $738$ was rounded down and both numbers on the bottom were rounded up. All three changes make the estimate smaller than the exact value.

Answer: (a) $87.5$ (b) $15.8$ is far too small compared with the estimate of about $90$ , so the calculation was entered wrongly (no brackets on the denominator). (c) $104$

Example 4. A school trip has 247 students and 18 teachers. (a) Each coach seats 52 people. Find the number of coaches needed. (b) The coaches cost $1385 in total, shared equally by the 247 students. Calculate the cost for each student. (c) A drinks container holds 20 litres. Each cup holds $0.35$ litres. Find the number of full cups the container fills.

Show the working

(a) Total people: $247 + 18 = 265$ . Then $265 \div 52 = 5.096 \dots$ Five coaches seat only $5 \times 52 = 260$ people, which leaves 5 people with no seat. Everyone must travel, so round up: 6 coaches.
(b) $1385 \div 247 = 5.607287 \dots$ This is money, so give it to the nearest cent. The deciding digit after $5.60$ is 7, so the answer is $5.61.
(c) $20 \div 0.35 = 57.14 \dots$ Only full cups count, so round down: 57 cups. A 58th cup would need $58 \times 0.35 = 20.3$ litres, more than the container holds.
Notice that (a) went up and (c) went down, though both started as decimals. The direction comes from the question, not from the deciding digit.

Answer: (a) 6 coaches (b) $5.61 (c) 57 cups

Where people go wrong

Mistake: Writing $5764$ correct to the nearest thousand as $6$ .

Instead: Replace the removed digits with zeros: $6000$ .

Rounding must give a number of about the same size as the original. $6$ is nowhere near $5764$ ; the zeros hold the 6 in the thousands column.

Mistake: Giving $0.00483$ correct to 2 significant figures as $0.00$ .

Instead: Count from the first non-zero digit, the 4. The answer is $0.0048$ .

$0.00$ is 2 decimal places, a different instruction. Zeros at the front of a decimal only place the digits; they are not significant figures.

Mistake: Rounding in two stages: $2.447 \to 2.45 \to 2.5$ to 1 decimal place.

Instead: Round once, from the original number. The deciding digit after $2.4$ is 4, so the answer is $2.4$ .

The first rounding moves the number. The second rounding then works on the moved number and can push it over the halfway point, when the original was below it.

Mistake: Working out the exact answer on a calculator and rounding it, when the question asks for an estimate.

Instead: Round each number to 1 significant figure first, write the rounded numbers down, and calculate with them.

The estimate is a different number from the rounded exact answer. For $\frac{62.8}{4.13 \times 0.487}$ the estimate is $30$ but the exact answer rounds to $31.2$ . The question tests the rounding method, so the rounded numbers must be visible.

Mistake: Rounding a count to the nearest whole number: 265 people at 52 per coach gives $5.09 \dots$ , so '5 coaches'.

Instead: Ask what the context needs. People who cannot be left behind need 6 coaches. Only complete items (full cups, full pieces) round down.

The deciding-digit rule tells you which whole number is nearer. It does not tell you which whole number makes sense in the situation.

Check yourself

Write $68 449$ correct to the nearest thousand.
Answer
$68 000$ (the deciding digit is 4)
Write $0.07096$ correct to 3 significant figures.
Answer
$0.0710$ (the final zero is the third significant figure, so it must be written)
Write $14.9752$ correct to 2 decimal places.
Answer
$14.98$
Write $405 620$ correct to 3 significant figures.
Answer
$406 000$
By writing each number correct to 1 significant figure, work out an estimate for $\frac{7.82 \times 31.4}{0.48}$ .
Answer
$\frac{8 \times 30}{0.5} = \frac{240}{0.5} = 480$ (exact value $511.5 \dots$ )
Work out an estimate for $\sqrt{35.6} \times 19.7$ . Show the numbers you use.
Answer
$\sqrt{36} \times 20 = 6 \times 20 = 120$ (exact value $117.5 \dots$ )
Floor tiles are sold in boxes of 8. A floor needs 107 tiles. Find the number of boxes needed.
Answer
$107 \div 8 = 13.375$ , so 14 boxes (13 boxes hold only 104 tiles)
Ana estimates $4.6 \times 2.8$ as $5 \times 3 = 15$ . Explain whether her estimate is bigger or smaller than the exact value.
Answer
Bigger. Both numbers were rounded up, so the product of the rounded numbers is bigger than the exact product ( $12.88$ ).

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1.

(a) Write $48 627$ correct to the nearest hundred. 1 mark No calculator

Mark scheme

B1 48 600 cao

(b) Write $0.030549$ correct to 3 significant figures. 1 mark No calculator

Mark scheme

B1 0.0305 cao

(c) Write $6.9971$ correct to 2 decimal places. 1 mark No calculator

Mark scheme

B1 7.00 cao

Question 2. You are not allowed to use a calculator in this question.

(a) By writing each number correct to 1 significant figure, work out an estimate for $\frac{287 \times 0.0496}{6.13 - 3.82}$ . You must show your working. 2 marks No calculator

Mark scheme

M1 300, 0.05, 6 and 4 all seen, or $\frac{300 \times 0.05}{6 - 4}$ seen, or $\frac{15}{2}$ seen
A1 7.5 cao

(b) Without working out the exact value, explain whether your estimate in part (a) is bigger or smaller than the exact value. 2 marks No calculator

Mark scheme

B1 Numerator: 287 and 0.0496 were both rounded up, so $300 \times 0.05$ is bigger than $287 \times 0.0496$
B1 Denominator: 6.13 was rounded down and 3.82 was rounded up, so $6 - 4$ is smaller than $6.13 - 3.82$ ; dividing by a smaller number gives a bigger result, so the estimate is bigger than the exact value

Question 3. A rectangular field measures $186.4$ m by $93.7$ m.

(a) By writing each measurement correct to 1 significant figure, work out an estimate of the area of the field. 2 marks

Mark scheme

M1 $200 \times 90$ seen
A1 18 000 cao

(b) Calculate the area of the field. Give your answer correct to 3 significant figures. 2 marks

Mark scheme

M1 $186.4 \times 93.7$ (17 465.68)
A1 17 500 cao

(c) Fence panels are $2.4$ m long. A panel can be cut to fit. Calculate the least number of panels needed to go all the way round the edge of the field. 3 marks

Mark scheme

M1 $2 \times (186.4 + 93.7)$ oe, or 560.2 seen
M1 their perimeter $\div 2.4$ (233.4…), or $2 \times \frac{186.4}{2.4} + 2 \times \frac{93.7}{2.4}$ oe
A1 234 cao

Question 4. Mei buys a $3.75$ kg bag of rice for $8.95.

(a) Calculate the cost of 1 kg of the rice. Give your answer to a reasonable degree of accuracy. 2 marks

Mark scheme

M1 $8.95 \div 3.75$ (2.3866…)
A1 2.39 cao

(b) Each portion of rice uses $0.35$ kg. Work out the greatest number of full portions Mei can make from the bag. 2 marks

Mark scheme

M1 $3.75 \div 0.35$ (10.71…)
A1 10 cao

(c) Explain why the answer to part (b) is not 11. 1 mark

Mark scheme

B1 11 portions need $11 \times 0.35 = 3.85$ kg, which is more than 3.75 kg oe (e.g. the 0.71… is not a full portion)

How this is examined

Estimating a calculation by rounding every number to 1 significant figure: A non-calculator question that starts 'By writing each number correct to 1 significant figure…', with a fraction, a product or a square root to estimate, worth 2 marks M1 for the rounded values written down (e.g. $\frac{60}{4 \times 0.5}$ ) or the rounded calculation seen. A1 for the answer worked out from those values (e.g. $30$ ). The exact answer rounded off is not the estimate, and it does not show the rounding method.
Choosing the whole-number answer that the context needs: A division inside a word problem (coaches, boxes, pieces of a length, full containers) where the calculator answer is a decimal, worth 2 or 3 marks The division seen (M1), then a whole number in the right direction: up when everything must be covered, down when only complete items count. If the command is Explain, a sentence such as '5 coaches seat only 260 people' or '11 portions would need 3.85 kg' earns the mark.
Saying whether an estimate is too big or too small, or using it to check a calculator answer: A 1- or 2-mark Explain straight after an estimate part A statement about the direction of each rounding, linked to its effect: 'both numbers on top were rounded up, so the numerator is too big' and 'the denominator was rounded down, and dividing by a smaller number gives a bigger answer', ending with the conclusion 'so the estimate is bigger than the exact value'.
Rounding a given number to decimal places, significant figures, or the nearest 10, 100 or 1000: A short Write or Write down part, 1 mark each, sometimes several in a row on the same number The exact rounded value in the form asked for: placeholder zeros in whole numbers ( $6000$ , not $6$ ), the required decimal places shown even when they are zero ( $8.00$ , not $8$ ), and a significant final zero kept ( $0.0710$ ).
Giving a final calculator answer to the right accuracy: The last mark of any calculation question, whether or not the accuracy is stated The stated accuracy if there is one. If not: 3 significant figures for a non-exact answer, 1 decimal place for an angle in degrees, and an exact answer of 4 or 5 significant figures left as it is. For money, the nearest cent (2 decimal places) is the reasonable choice. Write the longer value first, and round only the final answer.

Words to know

Degree of accuracy: How precisely a value is given, e.g. to the nearest hundred, to 2 decimal places, or to 3 significant figures.
Deciding digit: The digit straight after the last one you keep. If it is 5 or more, the kept digit goes up by 1; if it is 4 or less, the kept digit stays.
Decimal places (dp): The number of digits after the decimal point. $3.142$ has 3 decimal places.
Significant figures (sf): The digits of a number counted from the first non-zero digit. Zeros at the front of a decimal are not significant; zeros between non-zero digits are.
Placeholder zero: A zero that holds the other digits in their place value, such as the zeros in $6000$ when $5764$ is rounded to the nearest thousand.
Estimate: An approximate answer found by rounding each number (normally to 1 significant figure) and calculating with the rounded values.
$\approx$: The symbol for 'is approximately equal to'.
Overestimate: An estimate that is bigger than the exact value.
Underestimate: An estimate that is smaller than the exact value.
Exact answer: An answer that needs no rounding, such as $12.4 \times 7.85 = 97.34$ . A non-exact answer, such as $55.769 \dots$ , has to be rounded.

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