Estimation — sense-checking answers — Mathematics

Before this

This leads to

A calculator that is confidently wrong

You are working on a problem, and you need this:

\frac{38.7 \times 492}{19.4}

You type it into your calculator, and the screen says 9.81. It looks precise. It has two decimal places. Do you write it down?

Stop for five seconds and think about the size. 38.7 is close to 40. 492 is close to 500. And 19.4 is close to 20. So the answer should be close to

\frac{40 \times 500}{20} = \frac{20 000}{20} = 1000

The calculator said 9.81. The two numbers differ by a factor of about 100. Something went wrong, and it was not the calculator. Here, 492 was typed as 4.92. The true answer is about 981.5, and the five-second check caught the slip.

On a line marked in powers of ten, the estimate (1000) and the exact answer (about 981.5) sit almost on top of each other. The calculator's 9.81 is two powers of ten away. An estimate does not need to be close to the exact answer. It needs to be the right size.

A calculator does exactly what you type. It cannot tell you that you typed the wrong thing. Only your own sense of size can do that. In the last lecture, you rounded to keep the digits that matter. In this lecture, you round on purpose, much more roughly, to get a quick answer you can check against.

The method: round everything to 1 significant figure

To estimate a calculation:

Step 1. Round every number to 1 significant figure (1 s.f.).
Step 2. Do the easier calculation, in your head or on paper.
Step 3. Write the result with ≈, never =, because it is an estimate.

Rounding to 1 s.f. leaves one non-zero digit, with zeros to keep the size. For example, 38.7 becomes 40, 492 becomes 500, 0.0617 becomes 0.06, and 1430 becomes 1000. The rules are the ones from the last lecture, with $n = 1$ .

Why 1 significant figure? Because the goal is speed and size, not accuracy. With one non-zero digit in each number, the arithmetic becomes a single-digit calculation followed by counting zeros. $40 \times 500$ is $4 \times 5 = 20$ , followed by three zeros: 20 000.

Rounding to 1 s.f. can move a number a long way. 150 becomes 200, which is a third larger. 149 becomes 100, which is a third smaller. So an estimate can easily be 20% or 30% away from the exact answer. That does not matter. A mistyped decimal point or a missing zero makes an answer 10, 100 or 1000 times too big or too small. An estimate is far more accurate than that, so it catches those slips every time.

That is it. That is the whole idea. An estimate checks the size of an answer, not its digits.

Try it below. Choose a calculation, and compare estimates made with 1, 2 and 3 significant figures.

Explore — how good is a 1 s.f. estimate?

Choose a calculation. Then change how many significant figures each number is rounded to. Watch the estimate, the exact answer, and how far apart they are.

Round each number to: 1 s.f.

Four calculations, each estimated with every number rounded to 1 s.f.:

Calculation	Estimate	Exact	Gap
(38.7 × 492) ÷ 19.4	(40 × 500) ÷ 20 = 1000	≈ 981.5	+1.9%
(52 × 318) ÷ 63	(50 × 300) ÷ 60 = 250	≈ 262.5	−4.8%
0.347 × 1.29	0.3 × 1 = 0.3	≈ 0.4476	−33.0%
√(386 × 2.1)	√(400 × 2) = √800 ≈ 28.28	≈ 28.47	−0.7%

Even the worst estimate here, 33% too small, is the right size: about a half, not 5 or 0.05. With 2 s.f., three of the gaps become smaller (+0.6%, +1.6%, +0.5%), the first grows to +2.5%, and the arithmetic is much harder.

Did you notice two things? First, the 1 s.f. estimate for 0.347 × 1.29 is 33% too small, yet it still shows that the answer is about a half, not 5 and not 0.05. Second, more significant figures do not always mean a closer estimate. For the first calculation, 2 s.f. gives 1006, which is further from 981.5 than the 1 s.f. estimate of 1000. Rounding errors can cancel or add up, partly by luck.

Overestimate or underestimate?

Some questions ask whether your estimate is bigger or smaller than the exact answer. You can often tell without working out the exact answer, by looking at which way each number was rounded.

In a multiplication, rounding a number up makes the result bigger.
In a division, rounding the number you divide by up makes the result smaller. Dividing by a bigger number gives a smaller answer.

For $4.8 \times 2.6 \approx 5 \times 3 = 15$ , both numbers were rounded up, so 15 is an overestimate. (The exact answer is 12.48.) But many calculations pull in both directions. Test yourself on the opening calculation.

Predict — all rounded up, so too big?

In $(38.7 \times 492) \div 19.4 \approx (40 \times 500) \div 20 = 1000$ , all three numbers were rounded up. Which statement is right? Choose one, then read the explanation.

The three roundings pull in different directions:

38.7 → 40 and 492 → 500 are in the multiplication. Rounding them up makes the estimate bigger.
19.4 → 20 is the number we divide by. Rounding it up makes the estimate smaller.

So the directions alone cannot decide. Here the two upward changes (about 3.4% and 1.6%) outweigh the downward one (about 3.1%), and the estimate of 1000 is slightly too big: the exact answer is about 981.5.

Stop and think

Ana estimates $2.45 \times 3.47$ as $2 \times 3 = 6$ . The exact answer is about 8.50, so her estimate is about 29% too small. Was her estimate a failure?

Show the answer

No. It did its job. It shows that the answer is a single-digit number, not 85 or 0.85. If her calculator had said 85.0, the estimate would have caught it at once. If she needs a closer estimate, she can round to 2 s.f. instead: $2.5 \times 3.5 = 8.75$ . An estimate is judged by whether it catches mistakes, not by how close it is.

Worked examples

Worked Example 1 — A three-number calculation

Estimate $(52 \times 318) \div 63$ .

Round to 1 s.f.: 52 → 50, 318 → 300, 63 → 60.

$50 \times 300 = 15000$ , and $15000 \div 60 = 250$ . So the answer ≈ 250.

Check: the exact answer is about 262.5, so the estimate is about 4.8% too small. It is the right size.

Worked Example 2 — A square root

Estimate $386 \times 2.1$ .

Inside the root: $386 \times 2.1 \approx 400 \times 2 = 800$ .

Now find the square numbers on either side of 800: $2 8^{2} = 784$ and $2 9^{2} = 841$ .

800 lies between the square numbers 784 and 841, so √800 lies between 28 and 29, nearer to 28.

So $800$ is between 28 and 29, a little above 28. The answer ≈ 28.

Check: the exact answer is about 28.47. Knowing your square numbers up to about $3 0^{2} = 900$ makes these estimates fast.

Worked Example 3 — Painting a wall

A wall is 4.8 m wide and 2.6 m high. One tin of paint covers 12 m² and costs $18.99. Estimate the cost of painting the wall once.

Area ≈ $5 \times 3 = 15$ m². Tins ≈ $15 \div 10 = 1.5$ , so 2 tins, because you cannot buy half a tin. (Rounding 12 to 10 keeps the division easy.) Cost ≈ $2 \times $20 = $40$ .

Check exactly: the area is $4.8 \times 2.6 = 12.48$ m², which needs $12.48 \div 12 = 1.04$ tins. One tin is not quite enough, so 2 tins are needed, costing $2 \times $18.99 = $37.98$ .

Notice that the estimate rounded the area up, from 12.48 to 15. When you are buying materials, an overestimate is the safe direction. It is better to have a little paint left over than to stop with part of the wall unpainted.

Worked Example 4 — Catching a slip

Kofi works out $0.347 \times 1.29$ on his calculator and gets 44.8. Is that right?

Estimate: $0.3 \times 1 = 0.3$ . The answer should be a fraction of 1, not about 45.

His answer is about 100 times too big. He probably typed 34.7 instead of 0.347: $34.7 \times 1.29 \approx 44.76$ . The correct answer is $0.347 \times 1.29 \approx 0.448$ .

Check: 0.448 and the estimate 0.3 are the same size. 44.8 is not.

Worked Example 5 — Estimating in standard form

Estimate $(6.2 \times 1 0^{5}) \times (3.9 \times 1 0^{- 2})$ , and say whether the estimate is too big or too small.

Round the numbers in front: $6 \times 4 = 24$ . Add the powers: $1 0^{5 + (- 2)} = 1 0^{3}$ . So the answer ≈ $24 \times 1 0^{3} = 24000$ , or $2.4 \times 1 0^{4}$ .

6.2 was rounded down and 3.9 was rounded up, so the directions pull against each other. The exact answer is 24 180, so this estimate is slightly too small.

Three classic mistakes

Trap 1 — Rounding to 1 decimal place instead of 1 s.f.

0.0617 to 1 s.f. is 0.06, not 0.1. Rounding to 1 decimal place would change the number by more than half its size. Start counting at the first non-zero digit.

Trap 2 — Writing = for an estimate

$(40 \times 500) \div 20 = 1000$ is true. But the original calculation is not equal to 1000, so write "≈ 1000". In an exam, writing the rounded numbers you used often earns the method mark.

Trap 3 — Rounding a count the wrong way

1.04 tins of paint means 2 tins, not 1. When the answer counts real things, the context decides the rounding, not the usual rule. The same is true for buses, boxes and people.

Try it yourself

Work each one on paper first. Open a hint only when you are stuck.

Estimate: (a) $61.3 \times 4.87$ (b) $897 \div 31.2$ (c) $0.0583 \times 712$

Hint 1
Round each number to 1 s.f. first.

Hint 2
(a) $60 \times 5$ . (b) $900 \div 30$ . (c) $0.06 \times 700$ .

Full answer
(a) ≈ 300 (exact 298.531). (b) ≈ 30 (exact ≈ 28.75). (c) ≈ 42 (exact ≈ 41.51). Each estimate is within about 5% of the exact answer.
A student's calculator gives these results. Use estimates to decide which one is the wrong size. (a) $18.9 \times 0.52 = 98.28$ (b) $7210 \div 3.9 = 1848.7$ (c) $0.36 \times 250 = 9.49$

Hint 1
Estimate each one to 1 s.f., then compare sizes.

Hint 2
(a) $20 \times 0.5$ . (b) $7000 \div 4$ . (c) $0.4 \times 300 = 120$ .

Full answer
(a) ≈ 10, so 98.28 is 10 times too big: the correct answer is 9.828. (b) ≈ 1750, the same size as 1848.7, so it is plausible (and correct). (c) $120$ is between 10 and 11, the same size as 9.49, so it is plausible (and correct: $90 \approx 9.49$ ).
Estimate $\frac{4.12 \times 19.7}{0.48}$ . Is your estimate too big or too small? Explain using the direction of each rounding, then check with a calculator.

Hint 1
0.48 to 1 s.f. is 0.5. Dividing by 0.5 is the same as multiplying by 2.

Hint 2
4.12 → 4 (down), 19.7 → 20 (up), 0.48 → 0.5 (up, and it is the divisor).

Full answer
Estimate: $(4 \times 20) \div 0.5 = 80 \div 0.5 = 160$ . Directions: 4.12 → 4 pushes down, 19.7 → 20 pushes up, and rounding the divisor 0.48 up to 0.5 pushes down. Two of the three push down, and the 0.48 → 0.5 change is the largest (about 4%), so expect an underestimate. Exact: $4.12 \times 19.7 \div 0.48 \approx 169.1$ . The estimate, 160, is about 5% too small.
Estimate how many times a human heart beats in a lifetime of 75 years, at an average of about 70 beats per minute. Give your answer in standard form to 1 s.f. Then work it out on a calculator (use 365 days a year) and compare.

Hint 1
Beats = beats per minute × minutes per hour × hours per day × days per year × years.

Hint 2
$70 \times 60 \times 20 \times 400 \times 80$ . Multiply the leading digits, then count the zeros.

Full answer
Estimate: $7 \times 6 \times 2 \times 4 \times 8 = 2688$ , followed by six zeros (one each from 70, 60, 20 and 80, and two from 400), so about 2 688 000 000 ≈ $3 \times 1 0^{9}$ . Calculator: $70 \times 60 \times 24 \times 365 \times 75 = 2759400000 \approx 2.8 \times 1 0^{9}$ , which is also $3 \times 1 0^{9}$ to 1 s.f. The estimate is very close because rounding 24 down to 20 and 365 up to 400 partly cancel. About three billion heartbeats is the right size.

The bigger idea

Estimation is the everyday form of a skill called number sense: a feel for how big an answer should be. A height of 183 cm for a person is believable. 1830 cm is not, and neither is 18.3 cm. You can see this before doing any detailed arithmetic.

Scientists use the same skill on questions that have no exact data at all. The physicist Enrico Fermi was famous for it. At the first nuclear test in 1945, he dropped small pieces of paper as the blast wave passed and, from how far they moved, estimated the energy released as about 10 kilotons. Later measurements put it at about 25 kilotons. His estimate was not exact, but it was the right size, made in seconds with paper scraps. Rough calculations like this are now called Fermi problems.

This lecture ends the unit on number. Surds, standard form, significant figures and estimation all ask the same question in different ways: what do these numbers really tell me, and how much should I trust them?

Key concepts in this lesson

To estimate, round every number to 1 significant figure and do the simpler calculation, writing the result with ≈.
An estimate checks the size of an answer, so it catches slips that make an answer 10, 100 or 1000 times too big or too small.
Rounding a number up in a multiplication makes the estimate bigger, but rounding the divisor up makes the estimate smaller.
A square root can be estimated by finding the square numbers on either side of the number inside the root.
When the answer counts real things, such as tins of paint, the context decides whether to round up or down.