Estimating the Result of a Calculation

Finding a rough answer quickly by rounding each number to 1 significant figure or to a convenient number, and telling whether the estimate is too high or too low.

Before you start

To round a number to 1 significant figure, keep its first non-zero digit and look at the digit after it. If that digit is 5 or more, add 1 to the digit you keep. Digits dropped before the decimal point are replaced by zeros to hold the place value; digits dropped after it are left out.

So $5870 \approx 6000$ , $23.92 \approx 20$ , $4.801 \approx 5$ , $0.31 \approx 0.3$ and $0.0782 \approx 0.08$ , each correct to 1 significant figure. The sign $\approx$ is read "is approximately equal to".

The order of operations is: brackets first, then powers and roots, then multiplication and division from left to right, then addition and subtraction from left to right. So $60 - 5 \times 20 = 60 - 100 = -40$ .

What an Estimate Is, and Why We Estimate

An estimate of a calculation is a rough value for its result, found quickly by replacing each number with a simpler number close to it. The simpler numbers are chosen so that the working can be done in your head.

There are three reasons to estimate.

Sometimes a rough answer is all that is needed.
An estimate made before a long calculation tells you what size of answer to expect.
An estimate shows up a wrong answer. A calculator works out whatever is keyed into it, and one wrong key gives a wrong answer that looks just as neat as a right one. If a calculator shows $1148.3992$ for $23.92 \times 4.801$ , the estimate $20 \times 5 = 100$ tells you at once that something was keyed wrongly: the display is about ten times too large. Judging an answer in this way, and finding the slip, is the subject of the lesson on checking a calculator answer by estimating.

How to estimate a calculation

Round each number to 1 significant figure, unless the question tells you to round in another way.
Work out the calculation with the rounded numbers, in the usual order of operations.
Write $\approx$ , not $=$ , at the step where the numbers are rounded, because the two sides are no longer exactly equal.

An estimate is not the exact answer. In this lesson every estimate is set beside the exact value, or the value a calculator gives, so that you can see how close it comes.

Example 1: Sums and differences

Estimate each of the following by rounding each number to 1 significant figure.
(a) $347 - 482 + 659$
(b) $58.72 + 9.36 - 31.8$

Solution

(a) $347 \approx 300$ , $482 \approx 500$ and $659 \approx 700$ .

$347 - 482 + 659 \approx 300 - 500 + 700 = 500$

Exact value: $347 - 482 + 659 = 524$ . The estimate is 24 below it.

(b) $58.72 \approx 60$ , $9.36 \approx 9$ and $31.8 \approx 30$ .

$58.72 + 9.36 - 31.8 \approx 60 + 9 - 30 = 39$

Exact value: $58.72 + 9.36 - 31.8 = 36.28$ . The estimate is 2.72 above it.

(a) $347 - 482 + 659 \approx 500$
(b) $58.72 + 9.36 - 31.8 \approx 39$

Estimating Products and Quotients

Rounding to 1 significant figure turns a long multiplication into a times-table fact and some zeros. Multiply the non-zero digits, then put the place value back: $20 \times 5 = 100$ , and $0.08 \times 600 = 8 \times 6 = 48$ , because $0.08 \times 100 = 8$ .

Example 2: Products

Estimate each product by rounding each number to 1 significant figure.
(a) $23.92 \times 4.801$
(b) $0.0782 \times 613$

Solution

(a) $23.92 \approx 20$ and $4.801 \approx 5$ .

$23.92 \times 4.801 \approx 20 \times 5 = 100$

Exact value: $23.92 \times 4.801 = 114.83992$ . The estimate has the right size, though it is about 15 too low. A later section shows why.

(b) $0.0782 \approx 0.08$ and $613 \approx 600$ .

$0.0782 \times 613 \approx 0.08 \times 600 = 48$

Exact value: $0.0782 \times 613 = 47.9366$ .

(a) $23.92 \times 4.801 \approx 100$
(b) $0.0782 \times 613 \approx 48$

To divide by a decimal in your head, first multiply both numbers by 10 (or by 100, or by 1000) so that the number you divide by becomes a whole number. The quotient does not change: $60 \div 0.3 = 600 \div 3 = 200$ . Notice that dividing a positive number by a number between 0 and 1 gives an answer larger than the number you started with.

Example 3: Quotients

Estimate each of the following by rounding each number to 1 significant figure.
(a) $5870 \div 28.6$
(b) $59.8 \div 0.31$
(c) $\dfrac{59.7 \times 4.02}{19.8}$

Solution

(a) $5870 \approx 6000$ and $28.6 \approx 30$ .

$5870 \div 28.6 \approx 6000 \div 30 = 600 \div 3 = 200$

Exact value: $5870 \div 28.6 = 205.2\ldots$

(b) $59.8 \approx 60$ and $0.31 \approx 0.3$ .

$59.8 \div 0.31 \approx 60 \div 0.3 = 600 \div 3 = 200$

Exact value: $59.8 \div 0.31 = 192.9\ldots$

(c) The fraction bar means that the whole numerator is divided by the denominator. $59.7 \approx 60$ , $4.02 \approx 4$ and $19.8 \approx 20$ .

$\dfrac{59.7 \times 4.02}{19.8} \approx \dfrac{60 \times 4}{20} = \dfrac{240}{20} = 12$

Exact value: $\dfrac{59.7 \times 4.02}{19.8} = \dfrac{239.994}{19.8} = 12.12\ldots$

(a) $5870 \div 28.6 \approx 200$
(b) $59.8 \div 0.31 \approx 200$
(c) $\dfrac{59.7 \times 4.02}{19.8} \approx 12$

Convenient Numbers

Rounding to 1 significant figure does not always give an easy calculation, and sometimes it moves a number a long way: 147 becomes 100. When a question says only "estimate", you may replace each number with a convenient number instead: a number close to it that makes the working easy. Convenient numbers are also called compatible numbers.

Choosing convenient numbers

For a division, choose a pair of numbers that divide exactly: $4370 \div 6.2 \approx 4200 \div 6$ , because $42 \div 6 = 7$ .
For a multiplication, look for a number that is easy to multiply by, such as 25 or 50: $36.2 \times 24.6 \approx 36 \times 25$ .
Under a root sign, choose the nearest perfect square or perfect cube: $\sqrt{61} \approx \sqrt{64}$ .
For a sum or a difference of numbers of very different sizes, round them all to the same place value: $3458 - 86 \approx 3500 - 100$ .

If a question says "by rounding each number to 1 significant figure", do exactly that. Whichever way you round, write down the rounded numbers so that your estimate can be followed.

Example 4: Convenient numbers

Estimate each of the following.
(a) $4370 \div 6.2$
(b) $147 \div 0.52$
(c) $36.2 \times 24.6$
(d) $2764 - 93$

Solution

(a) Rounding to 1 significant figure gives $4000 \div 6$ , which does not divide exactly. 4370 is nearly 44 hundreds, and the multiple of 6 nearest to 44 is $42 = 6 \times 7$ . So take 4200 for 4370 and 6 for 6.2.

$4370 \div 6.2 \approx 4200 \div 6 = 700$

Exact value: $4370 \div 6.2 = 704.8\ldots$

(b) To 1 significant figure, 147 is 100, which is a long way from 147. The number 150 is much closer and still easy to work with. Take 0.5 for 0.52.

$147 \div 0.52 \approx 150 \div 0.5 = 1500 \div 5 = 300$

Exact value: $147 \div 0.52 = 282.69\ldots$ Rounding to 1 significant figure would give $100 \div 0.5 = 200$ , which is much further from it.

$36.2 \times 24.6 \approx 36 \times 25 = 9 \times 4 \times 25 = 9 \times 100 = 900$

Exact value: $36.2 \times 24.6 = 890.52$ . Rounding to 1 significant figure would give $40 \times 20 = 800$ , which is further from it.

(d) The two numbers are of very different sizes. To 1 significant figure 2764 is 3000, which moves it by 236, more than the whole of the 93 that is taken away. Round both numbers to the same place value instead, the nearest hundred: $2764 \approx 2800$ and $93 \approx 100$ .

$2764 - 93 \approx 2800 - 100 = 2700$

Exact value: $2764 - 93 = 2671$ . Rounding to 1 significant figure would give $3000 - 90 = 2910$ , which is much further from it.

(a) $4370 \div 6.2 \approx 700$
(b) $147 \div 0.52 \approx 300$
(c) $36.2 \times 24.6 \approx 900$
(d) $2764 - 93 \approx 2700$

Squares and Roots

To estimate a square, round the number first and then square the rounded number. To estimate a square root or a cube root, do not round the number under the root sign to 1 significant figure: $\sqrt{60}$ is no easier to find than $\sqrt{61}$ . Replace the number with the nearest perfect square or perfect cube, whose root you know.

Numbers to look for

Perfect squares: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225

Perfect cubes: 1, 8, 27, 64, 125, 216

Example 5: Squares and roots

Estimate each of the following.
(a) $6.87^{2}$
(b) $\sqrt{61} \times \sqrt[3]{128}$
(c) $\sqrt{15.8} \times 2.94^{2}$

Solution

(a) $6.87 \approx 7$ , so $6.87^{2} \approx 7^{2} = 49$ .

Exact value: $6.87^{2} = 47.1969$ .

(b) 61 is close to $64 = 8^{2}$ , so $\sqrt{61} \approx \sqrt{64} = 8$ . 128 is close to $125 = 5^{3}$ , so $\sqrt[3]{128} \approx \sqrt[3]{125} = 5$ .

$\sqrt{61} \times \sqrt[3]{128} \approx 8 \times 5 = 40$

By calculator: $\sqrt{61} \times \sqrt[3]{128} = 39.361\ldots$ , which is 39.4 correct to 3 significant figures.

(c) 15.8 is close to $16 = 4^{2}$ , so $\sqrt{15.8} \approx \sqrt{16} = 4$ . Also $2.94 \approx 3$ , so $2.94^{2} \approx 3^{2} = 9$ .

$\sqrt{15.8} \times 2.94^{2} \approx 4 \times 9 = 36$

By calculator: $\sqrt{15.8} \times 2.94^{2} = 34.357\ldots$ , which is 34.4 correct to 3 significant figures.

(a) $6.87^{2} \approx 49$
(b) $\sqrt{61} \times \sqrt[3]{128} \approx 40$
(c) $\sqrt{15.8} \times 2.94^{2} \approx 36$

Expressions with Several Operations

Round every number first. Then work out the rounded expression in the correct order: brackets, powers and roots, multiplication and division, addition and subtraction. A fraction bar acts as brackets round the numerator and round the denominator, so work out each of them before you divide.

Example 6: Several operations

Estimate each of the following by rounding each number to 1 significant figure.
(a) $125 + 3.91 \times 27.48$
(b) $\dfrac{412 + 187}{0.52 \times 5.9}$

Solution

(a) $125 \approx 100$ , $3.91 \approx 4$ and $27.48 \approx 30$ .

Multiply before adding: $125 + 3.91 \times 27.48 \approx 100 + 4 \times 30 = 100 + 120 = 220$

Exact value: $125 + 3.91 \times 27.48 = 125 + 107.4468 = 232.4468$ .

(b) $412 \approx 400$ , $187 \approx 200$ , $0.52 \approx 0.5$ and $5.9 \approx 6$ .

Numerator: $400 + 200 = 600$ . Denominator: $0.5 \times 6 = 3$ .

$\dfrac{412 + 187}{0.52 \times 5.9} \approx \dfrac{600}{3} = 200$

Exact value: $\dfrac{412 + 187}{0.52 \times 5.9} = \dfrac{599}{3.068} = 195.2\ldots$

(a) $125 + 3.91 \times 27.48 \approx 220$
(b) $\dfrac{412 + 187}{0.52 \times 5.9} \approx 200$

Is the Estimate Too High or Too Low?

Rounding a number up or down pushes the result of the calculation up or down as well. Which way it is pushed depends on what the number does in the calculation. With positive numbers, a larger number added or multiplied makes the result larger, but a larger number subtracted or divided by makes the result smaller. An estimate that is too high is called an overestimate, and one that is too low is an underestimate.

For positive numbers $a$ and $b$ :

Calculation	The estimate is too high if	The estimate is too low if
A sum, $a + b$	both numbers are rounded up	both numbers are rounded down
A product, $a \times b$	both numbers are rounded up	both numbers are rounded down
A difference, $a - b$	$a$ is rounded up and $b$ is rounded down	$a$ is rounded down and $b$ is rounded up
A quotient, $a \div b$	$a$ is rounded up and $b$ is rounded down	$a$ is rounded down and $b$ is rounded up

Rule: too high or too low

For positive numbers, if every rounding pushes the result the same way, the estimate is too high (every push is upwards) or too low (every push is downwards). A number that is not changed pushes neither way. If some roundings push the result up and others push it down, the directions alone cannot tell you which it is.

Example 7: Too high or too low

Each estimate below was made by rounding each number to 1 significant figure. Without working out the exact value, say whether the estimate is too high or too low, if the directions of rounding can tell you.
(a) $6.8 \times 49 \approx 7 \times 50 = 350$
(b) $912 - 288 \approx 900 - 300 = 600$
(c) $7.7 \div 0.24 \approx 8 \div 0.2 = 40$
(d) $23.92 \times 4.801 \approx 20 \times 5 = 100$

Solution

(a) Both numbers were rounded up and they are multiplied, so both push the product up. The estimate is too high. Check: $6.8 \times 49 = 333.2$ , which is less than 350.

(b) 912 was rounded down, which pushes the difference down. 288, the number subtracted, was rounded up, and taking away more also pushes the difference down. The estimate is too low. Check: $912 - 288 = 624$ , which is more than 600.

(c) 7.7 was rounded up, which pushes the quotient up. 0.24, the number divided by, was rounded down, and dividing by less also pushes the quotient up. The estimate is too high. Check: $7.7 \div 0.24 = 32.08\ldots$ , which is less than 40.

(d) 23.92 was rounded down and 4.801 was rounded up. One pushes the product down and the other pushes it up, so the directions alone cannot tell.

The exact value, $114.83992$ , shows that this estimate is too low. The sizes of the two changes explain it: 23.92 lost nearly 4, about a sixth of its value, while 4.801 gained only about 0.2, about a twenty-fifth of its value.

(a) Too high
(b) Too low
(c) Too high
(d) The directions cannot tell. The exact value shows that it is too low.

Several Numbers Close to One Value

Sometimes the numbers to be added are all close to one value. Then each of them can be replaced by that value, and a long addition becomes one multiplication. The value is a convenient number chosen for the whole list.

Rule: numbers close to one value

If several numbers are all close to the same value, their sum is about (how many numbers there are) $\times$ (that value). Four prices that are all near $20 come to about $4 \times 20 = 80$ dollars.

"About" is not "enough"

An estimate of "about $80" may be a little more than the exact total or a little less. So it cannot answer the question "is $80 enough?". To answer that, find the exact total, or round every price up, so that the estimate is certainly not too low. How to round so as to be sure of having enough, or of reaching a target, is the subject of the lesson on rounding to suit the situation.

Example 8: Numbers close to one value

(a) Four storybooks cost $19.90, $20.45, $18.95 and $21.30. Estimate their total cost.
(b) The numbers of visitors to a school exhibition on six days were 487, 512, 496, 523, 478 and 509. Estimate the total number of visitors.
(c) Mrs Lee has $80. Does the estimate in (a) show that she can pay for the four storybooks?

Solution

(a) Every price is close to $20, so the total is about $4 \times 20 = 80$ dollars.

Exact value: $19.90 + 20.45 + 18.95 + 21.30 = 80.60$ , so the storybooks cost $80.60.

(b) Every number is close to 500, so the total is about $6 \times 500 = 3000$ visitors.

Exact value: $487 + 512 + 496 + 523 + 478 + 509 = 3005$ .

(c) No. Two of the prices were rounded up to $20 and two were rounded down, so the directions of rounding cannot tell whether $80 is too high or too low. The exact total, $80.60, is 60 cents more than Mrs Lee has.

Rounding every price up to the next dollar gives $20 + 21 + 19 + 22 = 82$ dollars. This estimate is certainly not too low, so $82 would be enough.

(a) About $80
(b) About 3000 visitors
(c) No. The estimate says only "about $80". The exact total is $80.60, so $80 is not enough.

Common Mistakes

Rounding a small number to 0. To the nearest whole number, 0.31 is 0, and then $59.8 \div 0.31$ cannot be estimated at all. Round to 1 significant figure instead: $0.31 \approx 0.3$ .
Dropping the zeros that hold the place value. $5870 \approx 6000$ , not 6. The estimate $6 \div 30$ would be a thousand times too small.
Forgetting the order of operations in the estimate. $100 + 4 \times 30$ is $100 + 120 = 220$ . It is not $104 \times 30 = 3120$ .
Thinking that dividing always makes a number smaller. $60 \div 0.3 = 200$ . Dividing a positive number by a number between 0 and 1 gives a larger answer, and multiplying it by such a number gives a smaller one.
Rounding the number under a root sign to 1 significant figure. $\sqrt{60}$ is no easier than $\sqrt{61}$ . Use the nearest perfect square: $\sqrt{61} \approx \sqrt{64} = 8$ .
Writing $=$ where $\approx$ is meant. $23.92 \times 4.801$ is not equal to 100. Write $23.92 \times 4.801 \approx 20 \times 5 = 100$ .
Taking "about" to mean "enough". Four prices near $20 come to about $80, but the exact total may be a little more. To be sure that the money is enough, add the prices exactly or round every one of them up.

Practice

Work each question on paper before you open its solution. Make every estimate without a calculator. Where a question says only "estimate" or "by choosing convenient numbers", your numbers may differ from the ones in the solution. Your estimate is acceptable if each number you used is close to the number it replaces and your working is shown.

Question 1

Estimate each of the following by rounding each number to 1 significant figure.
(a) $6182 + 2940$
(b) $78.3 - 21.6$
(c) $0.482 + 0.317 - 0.094$

ShowHide solution

(a) $6182 + 2940 \approx 6000 + 3000 = 9000$ . Exact value: $9122$ .

(b) $78.3 - 21.6 \approx 80 - 20 = 60$ . Exact value: $56.7$ .

(a) About 9000
(b) About 60
(c) About 0.71

Question 2

Estimate each product by rounding each number to 1 significant figure.
(a) $68.3 \times 4.12$
(b) $0.0612 \times 487$
(c) $13.24 \times 4.83 \times 6.09$

ShowHide solution

(a) $68.3 \times 4.12 \approx 70 \times 4 = 280$ . Exact value: $281.396$ .

(b) $0.0612 \times 487 \approx 0.06 \times 500 = 6 \times 5 = 30$ . Exact value: $29.8044$ .

In (c) the estimate is well below the exact value, because 13.24 lost about a quarter of its value when it was rounded to 10. The convenient number 13 gives a much closer estimate: $13 \times 5 \times 6 = 13 \times 30 = 390$ .

(a) About 280
(b) About 30
(c) About 300

Question 3

Estimate each of the following by rounding each number to 1 significant figure.
(a) $8190 \div 38.7$
(b) $41.7 \div 0.193$
(c) $0.627 \div 0.029$
(d) $\dfrac{7.94 \times 30.6}{0.39}$

ShowHide solution

(a) $8190 \div 38.7 \approx 8000 \div 40 = 800 \div 4 = 200$ . Exact value: $211.6\ldots$

(b) $41.7 \div 0.193 \approx 40 \div 0.2 = 400 \div 2 = 200$ . Exact value: $216.06\ldots$

(d) $\dfrac{7.94 \times 30.6}{0.39} \approx \dfrac{8 \times 30}{0.4} = \dfrac{240}{0.4} = \dfrac{2400}{4} = 600$ . Exact value: $\dfrac{242.964}{0.39} = 622.98\ldots$

(a) About 200
(b) About 200
(c) About 20
(d) About 600

Question 4

Estimate each of the following by choosing convenient numbers.
(a) $3470 \div 6.8$
(b) $258 \div 0.49$
(c) $24.8 \times 8.1$

ShowHide solution

(a) 6.8 is close to 7, and $35 = 7 \times 5$ , so take 3500 for 3470: $3470 \div 6.8 \approx 3500 \div 7 = 500$ . Exact value: $510.29\ldots$

(b) 0.49 is close to 0.5, and 250 is close to 258 and easy to divide by 0.5: $258 \div 0.49 \approx 250 \div 0.5 = 2500 \div 5 = 500$ . Exact value: $526.5\ldots$

(a) About 500
(b) About 500
(c) About 200

Question 5

Estimate each of the following.
(a) $19.6^{2}$
(b) $\sqrt{47.6} + \sqrt[3]{28.3}$
(c) $\sqrt{8.9^{2} + 12.2^{2}}$

ShowHide solution

(a) $19.6 \approx 20$ , so $19.6^{2} \approx 20^{2} = 400$ . Exact value: $384.16$ .

(b) 47.6 is close to $49 = 7^{2}$ and 28.3 is close to $27 = 3^{3}$ , so $\sqrt{47.6} + \sqrt[3]{28.3} \approx 7 + 3 = 10$ . By calculator: $9.946\ldots$ , which is 9.95 correct to 3 significant figures.

(c) $8.9 \approx 9$ and $12.2 \approx 12$ , so $8.9^{2} + 12.2^{2} \approx 81 + 144 = 225$ , and $225 = 15^{2}$ . So $\sqrt{8.9^{2} + 12.2^{2}} \approx \sqrt{225} = 15$ . By calculator: $15.10\ldots$ , which is 15.1 correct to 3 significant figures.

(a) About 400
(b) About 10
(c) About 15

Question 6

Estimate each of the following by rounding each number to 1 significant figure.
(a) $81.6 - 5.2 \times 9.7$
(b) $\dfrac{58.7 + 21.9}{0.41 \times 9.8}$
(c) $4.8 \times (61.3 - 18.9)$

ShowHide solution

(a) Multiply before subtracting: $81.6 - 5.2 \times 9.7 \approx 80 - 5 \times 10 = 80 - 50 = 30$ . Exact value: $81.6 - 50.44 = 31.16$ .

(b) Numerator: $60 + 20 = 80$ . Denominator: $0.4 \times 10 = 4$ . So $\dfrac{58.7 + 21.9}{0.41 \times 9.8} \approx \dfrac{80}{4} = 20$ . Exact value: $\dfrac{80.6}{4.018} = 20.059\ldots$

(c) Brackets first: $4.8 \times (61.3 - 18.9) \approx 5 \times (60 - 20) = 5 \times 40 = 200$ . Exact value: $4.8 \times 42.4 = 203.52$ .

(a) About 30
(b) About 20
(c) About 200

Question 7

Each estimate was made by rounding each number to 1 significant figure. Without working out the exact value, decide whether the estimate is too high or too low, or say that the directions of rounding cannot tell.
(a) $8.7 \times 59 \approx 9 \times 60 = 540$
(b) $6130 \div 28 \approx 6000 \div 30 = 200$
(c) $743 - 268 \approx 700 - 300 = 400$
(d) $43.2 \times 0.76 \approx 40 \times 0.8 = 32$

ShowHide solution

(a) Both numbers were rounded up and they are multiplied, so the estimate is too high. Check: $8.7 \times 59 = 513.3$ .

(b) 6130 was rounded down, which pushes the quotient down. 28, the number divided by, was rounded up, and dividing by more also pushes the quotient down. The estimate is too low. Check: $6130 \div 28 = 218.9\ldots$

(c) 743 was rounded down, which pushes the difference down. 268, the number subtracted, was rounded up, which also pushes it down. The estimate is too low. Check: $743 - 268 = 475$ .

(d) 43.2 was rounded down and 0.76 was rounded up. One pushes the product down and the other pushes it up, so the directions cannot tell. The exact value, $43.2 \times 0.76 = 32.832$ , shows that the estimate is a little too low.

(a) Too high
(b) Too low
(c) Too low
(d) The directions cannot tell.

Question 8

(a) The numbers of books borrowed from a school library on six days were 192, 197, 188, 195, 199 and 190. Estimate the total. Without adding, say whether your estimate is too high or too low.
(b) Farid's lunches on five school days cost $4.80, $5.20, $5.10, $4.70 and $5.30. Estimate the total cost.
(c) Farid has $25 for the five lunches. Does your estimate in (b) show that this is enough? Find the exact total to decide.

ShowHide solution

(a) Every number is close to 200, so the total is about $6 \times 200 = 1200$ books. Every number was rounded up and they are added, so the estimate is too high. Check: $192 + 197 + 188 + 195 + 199 + 190 = 1161$ .

(b) Every price is close to $5, so the total is about $5 \times 5 = 25$ dollars.

(c) The estimate says only "about $25". Three of the prices were rounded down to $5 and two were rounded up, so the directions of rounding cannot tell whether $25 is too high or too low.

Exact total: $4.80 + 5.20 + 5.10 + 4.70 + 5.30 = 25.10$ . The lunches cost $25.10, which is 10 cents more than Farid has.

(a) About 1200 books. The estimate is too high.
(b) About $25
(c) No. The exact total is $25.10, so $25 is not enough.

Question 9

Ravi walks 1890 m from his home to school. Each of his steps is 0.62 m long.
(a) Estimate the number of steps he takes by rounding each number to 1 significant figure.
(b) Make a second estimate with the convenient numbers 1800 and 0.6.
(c) A calculator gives $1890 \div 0.62 = 3048.3\ldots$ Which estimate is closer to it, and why?

ShowHide solution

(a) $1890 \approx 2000$ and $0.62 \approx 0.6$ , so the number of steps is about $2000 \div 0.6 = 20\,000 \div 6$ . This does not divide exactly: it is $3333.3\ldots$ , or about 3300 steps.

(b) $1890 \div 0.62 \approx 1800 \div 0.6 = 18\,000 \div 6 = 3000$ , so he takes about 3000 steps.

(c) The estimate in (b) is about 48 below the calculator value. The estimate in (a), 3300, is about 250 above it. So the convenient numbers give the closer estimate.

The reason is not that 1800 is much nearer to 1890 than 2000 is: 1890 moves 90 to become 1800 and 110 to become 2000. The reason is the way each rounding pushes the quotient.

In (a), 1890 was rounded up, which pushes the quotient up, and 0.62 was rounded down, and dividing by less also pushes the quotient up. Both pushes are upwards, so they add together.

In (b), 0.62 was rounded down as before, but 1890 was rounded down as well, which pushes the quotient down. The two pushes are in opposite directions, so most of one is cancelled by the other.

The numbers show it. Dividing by 0.6 in place of 0.62 raises the quotient by about 100, to $1890 \div 0.6 = 3150$ . Taking 2000 for 1890 then raises it further, to $3333.3\ldots$ Taking 1800 for 1890 lowers it by 150 instead, to 3000.

The convenient numbers also give the easier working, because 18 is a multiple of 6.

(a) About 3300 steps
(b) About 3000 steps
(c) The estimate in (b) is closer. In (a) both roundings push the quotient up. In (b) one pushes it up and the other pushes it down, so most of the change is cancelled.

Question 10

Joyce had $3458 in her bank account. She then made four payments, of $361, $86, $405 and $299.
(a) Estimate how much is left in the account by rounding each amount to the nearest $100.
(b) Find the exact amount left and compare it with your estimate.

ShowHide solution

(a) To the nearest $100: $3458 \approx 3500$ (the tens digit is 5, so round up), $361 \approx 400$ , $86 \approx 100$ , $405 \approx 400$ and $299 \approx 300$ .

Total of the payments: about $400 + 100 + 400 + 300 = 1200$ dollars.

Money left: about $3500 - 1200 = 2300$ dollars.

(b) Exact total of the payments: $361 + 86 + 405 + 299 = 1151$ . Exact amount left: $3458 - 1151 = 2307$ . The estimate is only $7 below it.

Rounding every amount to the same place value works well for a sum or a difference. Rounding each amount to 1 significant figure instead would give $3000 - (400 + 90 + 400 + 300) = 1810$ , which is nearly $500 out, because 3458 loses 458 when it becomes 3000.

(a) About $2300 is left.
(b) Exactly $2307 is left, which is $7 more than the estimate.

Question 11

A teacher buys tickets to an exhibition for 38 pupils. Each ticket costs $11.60.
(a) Estimate the total cost by rounding each number to 1 significant figure.
(b) Devi estimates the cost as $40 \times 12 = 480$ dollars. Without working out the exact cost, say whether her estimate is too high or too low.
(c) Can the directions of rounding tell whether your estimate in (a) is too high or too low?

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(a) $38 \approx 40$ and $11.60 \approx 10$ , so the cost is about $40 \times 10 = 400$ dollars.

(b) Devi rounded both numbers up, 38 to 40 and 11.60 to 12, and they are multiplied, so her estimate is too high.

Exact cost: $38 \times 11.6 = 440.8$ , so the tickets cost $440.80. This is indeed less than Devi's $480, and it shows that the estimate in (a), $400, is too low.

(a) About $400
(b) Too high
(c) No: one number was rounded up and the other was rounded down. The exact cost, $440.80, shows that $400 is too low.

Key Points

To estimate a calculation, round each number to 1 significant figure (or to a convenient number), then work it out in the usual order of operations. Write $\approx$ where the numbers are rounded.
To divide by a decimal, multiply both numbers by 10 or by 100 first: $60 \div 0.3 = 600 \div 3 = 200$ .
Convenient numbers are a pair that divides exactly, a number that is easy to multiply by, and the nearest perfect square or perfect cube under a root sign.
For positive numbers, rounding up a number that is added or multiplied pushes the result up; rounding up a number that is subtracted or divided by pushes the result down. If every push is the same way, you know whether the estimate is too high or too low.
When several numbers are all close to one value, their sum is about (how many numbers there are) $\times$ (that value).
An estimate is "about", not "enough": to be sure that an amount is enough, find the exact total or round every number up.