Place value and estimating

Last Lesson and Before

Answer each one, then check.

1

Round 3.846 to 1 decimal place.

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\(3.8\)
2

Work out \(4500 \div 100\).

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\(45\)
3

Work out \(0.3 \times 0.2\).

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\(0.06\)
4

Work out \(7 \div 0.1\).

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\(70\) - dividing by 0.1 is the same as multiplying by 10.
5

Last lesson: list the meals from 2 starters (A, B) and 2 mains (X, Y).

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AX, AY, BX, BY

Learning Objectives

1Use a known calculation to work out related calculations.
2Round numbers to decimal places and significant figures.
3Estimate the answer to a calculation by rounding to 1 significant figure.
4Estimate square roots of numbers that are not square numbers.
5Decide whether an estimate is too big or too small.

Place Value

The value of a digit depends on its position.

Multiplying by 10
Every digit moves one place to the left: \(3.7 \times 10 = 37\).
Dividing by 10
Every digit moves one place to the right: \(3.7 \div 10 = 0.37\).
Multiplying by 0.1
The same as dividing by 10: \(45 \times 0.1 = 4.5\).
Dividing by 0.1
The same as multiplying by 10: \(45 \div 0.1 = 450\).

One Fact, Many Answers

Start from \(38 \times 142 = 5396\) and use place value.

\(38 \times 142\)
Answer: \(5396\). Why: The fact we are given
\(3.8 \times 142\)
Answer: \(539.6\). Why: One number is 10 times smaller, so the answer is 10 times smaller
\(3.8 \times 1.42\)
Answer: \(5.396\). Why: 10 times smaller and 100 times smaller: 1000 times smaller
\(0.38 \times 14.2\)
Answer: \(5.396\). Why: 100 times smaller and 10 times smaller: 1000 times smaller
\(5396 \div 142\)
Answer: \(38\). Why: Division is the inverse of multiplication
\(5396 \div 14.2\)
Answer: \(380\). Why: Dividing by a number 10 times smaller gives an answer 10 times bigger
\(53.96 \div 3.8\)
Answer: \(14.2\). Why: \(5396 \div 38 = 142\), and both numbers are 100 times smaller

Using a Known Fact

Given that \(23 \times 47 = 1081\), work out (a) \(2.3 \times 0.47\) and (b) \(108.1 \div 4.7\).

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1 (a) Compare with the fact: 2.3 is \(23 \div 10\) and 0.47 is \(47 \div 100\) \(\div 10\) and \(\div 100\) make \(\div 1000\)
2 So divide the answer by 1000 \(1081 \div 1000 = 1.081\)
3 (b) From the fact, \(1081 \div 47 = 23\) division undoes multiplication
4 108.1 is \(1081 \div 10\) and 4.7 is \(47 \div 10\) both divided by 10
5 Dividing both numbers by 10 does not change the answer \(108.1 \div 4.7 = 23\)

Answer(a) \(1.081\) (b) \(23\)

Significant Figures

The first significant figure is the first digit that is not zero.

Counting
4 is the first significant figure of 45 678 and of 0.004 12.
Zeros in the middle count
7.0496 has 5 significant figures; the 0 counts.
Zeros at the start do not
0.003 082 has 4 significant figures: 3, 0, 8, 2.
Keep the place value
45 678 to 2 significant figures is 46 000, not 46.

Rounding to Significant Figures

45 678
1 s.f.: 50 000. 2 s.f.: 46 000. 3 s.f.: 45 700
0.003 082
1 s.f.: 0.003. 2 s.f.: 0.0031. 3 s.f.: 0.003 08
7.0496
1 s.f.: 7. 2 s.f.: 7.0. 3 s.f.: 7.05
299 512
1 s.f.: 300 000. 2 s.f.: 300 000. 3 s.f.: 300 000
0.098 76
1 s.f.: 0.1. 2 s.f.: 0.099. 3 s.f.: 0.0988

How to Estimate

To estimate, round every number to 1 significant figure, then work out the calculation.

Round first
Round each number to 1 s.f. before doing anything else.
Never round to 0
0.048 rounds to 0.05, not 0.
Dividing by a decimal
\(150 \div 0.5 = 300\): dividing by 0.5 is the same as doubling.
Show it
Write the rounded calculation down: the method marks are for the rounded numbers, not just the answer. Use \(\approx\) for "is approximately equal to".

Estimating a Calculation

Work out an estimate for \(\dfrac{48.7 \times 3.14}{0.52}\)

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1 Round each number to 1 significant figure \(48.7 \approx 50\), \(3.14 \approx 3\), \(0.52 \approx 0.5\)
2 Write the rounded calculation \(\dfrac{50 \times 3}{0.5}\)
3 Work out the top \(50 \times 3 = 150\)
4 Divide by 0.5 (the same as doubling) \(150 \div 0.5 = 300\)
5 Compare with the calculator The exact answer is \(294.07\ldots\), close to 300

Answer\(\approx 300\)

Too Big or Too Small?

Look at which way each number was rounded.

Multiplying
Rounding the numbers UP gives an overestimate; rounding them DOWN gives an underestimate.
Dividing by a number rounded DOWN
The answer gets bigger: an overestimate.
Dividing by a number rounded UP
The answer gets smaller: an underestimate.
Mixed rounding
If some numbers went up and some went down, you cannot always tell without working it out.

Estimating a Square Root

40 is not a square number, so \(\sqrt{40}\) lies between two whole numbers.

1 \(\sqrt{36} = 6\)
The square number just below 40
2 \(\sqrt{40} \approx 6.3\)
40 is 4 of the 13 steps from 36 to 49, so about 0.3 of the way
3 \(\sqrt{42.25} = 6.5\)
Halfway between 6 and 7 - 40 is below this
4 \(\sqrt{49} = 7\)
The square number just above 40

Estimating a Square Root

Estimate \(\sqrt{70}\) to 1 decimal place without a calculator.

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1 Find the square numbers either side of 70 \(64 = 8^2\) and \(81 = 9^2\)
2 So \(\sqrt{70}\) is between 8 and 9 \(8 < \sqrt{70} < 9\)
3 How far is 70 from 64, out of the gap from 64 to 81? 6 out of 17
4 6 out of 17 is a bit more than a third about \(0.35\)
5 Check by squaring \(8.4^2 = 70.56\), which is close to 70

Answer\(\sqrt{70} \approx 8.4\) (the exact value is \(8.366\ldots\))

Estimating a Crowd

Nobody counts a crowd one person at a time.

Count a sample
Count the people in one block of seats: about 400.
Count the blocks
The stadium has about 150 blocks.
Multiply
\(400 \times 150 = 60\,000\).
Report it sensibly
"About 60 000" - rounded to 1 significant figure, because the inputs were only estimates.

Case study

Enrico Fermi and "Fermi Problems"

The Italian-American physicist Enrico Fermi, who won the 1938 Nobel Prize in Physics, was famous for making quick, surprisingly accurate estimates from almost no information. He liked to ask his students questions such as "How many piano tuners are there in Chicago?" and expected them to reason it out: the population of the city, how many homes have a piano, how often a piano is tuned, and how many pianos one tuner can tune in a year. Each number is rounded, but the errors tend to cancel out, and the answer is usually the right size. Questions like this are now called Fermi problems.

1938 Fermi wins the Nobel Prize in Physics

Estimation Station

Estimate each calculation, then use a calculator to find the exact answer and say whether your estimate was over or under. (a) \(6.8 \times 41.2\) (b) \(897 \div 2.9\) (c) \(\dfrac{19.6 \times 0.48}{0.21}\) (d) \(\sqrt{98.6 + 23.1}\) (e) \(3.14 \times 7.9^2\)

1. Estimate first, writing the rounded calculation.

2. Then use a calculator.

3. Say whether the estimate was over or under, and why.

A good answer shows: (a) \(7 \times 40 = 280\) (exact 280.16). (b) \(900 \div 3 = 300\) (exact 309.3). (c) \(\dfrac{20 \times 0.5}{0.2} = 50\) (exact 44.8). (d) \(\sqrt{120} \approx 11\) (exact 11.03). (e) \(3 \times 8^2 = 192\) (exact 195.97).

Can I...?

1Use a known fact to work out related calculations.
2Multiply and divide by 0.1 and 0.01.
3Round to a given number of decimal places.
4Round to a given number of significant figures.
5Estimate a calculation by rounding to 1 s.f.
6Divide by a decimal such as 0.5 or 0.2.
7Estimate a square root.
8Say whether an estimate is over or under.

Summary & Exam Focus

A known fact plus place value gives many related answers.
Significant figures start at the first non-zero digit; keep the place value when rounding.
Estimate by rounding each number to 1 significant figure.
Estimate square roots using the square numbers either side.
Over or under: look at which way each number was rounded.

Exam focus

Work out an estimate for \(\dfrac{3.78 \times 48.2}{0.52}\) (3 marks) (3 marks)

Write the rounded numbers down before you calculate. An answer with no rounded calculation shown can score zero, even if it is right.

Key terms

The vocabulary this lesson expects you to use. Each one is linked from the first place it appears above.

Place value: The value of a digit, which depends on its position in the number.
Decimal place (d.p.): A digit after the decimal point.
Significant figure (s.f.): Any digit of a number starting from the first non-zero digit.
Estimate: An approximate answer found by rounding the numbers first.
Approximately equal to: The symbol \(\approx\), used for an estimate.
Overestimate: An estimate bigger than the exact answer.
Underestimate: An estimate smaller than the exact answer.

Practice questions

Have a go at each one before you open its answer.

Question 1 Non-calculator 2 marks

Using the information that \(38 \times 142 = 5396\), write down the value of (a) \(3.8 \times 14.2\) (b) \(5396 \div 1.42\)
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Model answer

(a) \(53.96\) (b) \(3800\)

Mark scheme
- (a) \(53.96\) — B1
- (b) \(3800\) — B1
Question 2 Non-calculator 1 mark

Round 0.070 49 to 3 significant figures.
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Model answer

\(0.0705\)

Mark scheme
- \(0.0705\) — B1
Question 3 Non-calculator 3 marks

Work out an estimate for \(\dfrac{3.78 \times 48.2}{0.52}\)
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Model answer

\(\dfrac{4 \times 50}{0.5} = \dfrac{200}{0.5} = 400\)

Mark scheme
- At least two of the numbers rounded correctly to 1 s.f. (4, 50, 0.5) — M1
- \(200 \div 0.5\) or \(4 \times 100\) or equivalent — M1
- \(400\) — A1
Question 4 Non-calculator 1 mark

Is your answer to the last question an overestimate or an underestimate? Give a reason for your answer.
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Model answer

An overestimate: 3.78 and 48.2 were both rounded up, making the top bigger, and 0.52 was rounded down, and dividing by a smaller number gives a bigger answer.

Mark scheme
- Overestimate, with a correct reason referring to the rounding — C1
Question 5 Non-calculator 2 marks

Estimate the value of \(\sqrt{70}\). Give your answer to 1 decimal place.
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Model answer

\(64 = 8^2\) and \(81 = 9^2\), so \(\sqrt{70}\) is between 8 and 9. 70 is 6 of the 17 steps from 64 to 81, so \(\sqrt{70} \approx 8.4\).

Mark scheme
- \(8^2 = 64\) and \(9^2 = 81\) seen, or \(\sqrt{70}\) stated to be between 8 and 9 — M1
- Answer in the range 8.3 to 8.4 — A1
Question 6 Calculator 2 marks

(a) Use your calculator to work out \(\dfrac{(6.2 + 3.84)^2}{\sqrt{0.7}}\). Write down all the figures on your calculator display. (b) Write your answer to part (a) correct to 4 significant figures.
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Model answer

(a) \(120.4809562\) (b) \(120.5\)

Mark scheme
- (a) \(120.48095\ldots\) (at least 6 figures) — B1
- (b) \(120.5\), follow through from (a) — B1

Quick check

What is 45 678 rounded to 2 significant figures?
1. A45
2. B46
3. C46 000
4. D45 700
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C: 46 000

Keep the place value: 45 678 rounds to 46 000.
Given that \(12 \times 34 = 408\), what is \(1.2 \times 0.34\)?
1. A\(0.408\)
2. B\(4.08\)
3. C\(40.8\)
4. D\(0.0408\)
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A: \(0.408\)

1.2 is 10 times smaller and 0.34 is 100 times smaller, so the answer is 1000 times smaller.
Between which two whole numbers does \(\sqrt{55}\) lie?
1. A5 and 6
2. B6 and 7
3. C8 and 9
4. D7 and 8
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D: 7 and 8

\(49 = 7^2\) and \(64 = 8^2\), and 55 is between them.

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