Round everything first
Before working out 3.8 × 52, it helps to know roughly how big the answer should be. Round each number to one significant figure.
The first figure of 3.8 is the 3, and the digit after it, 8, is 5 or more, so 3.8 rounds up to 4. The first figure of 52 is the 5, and the 2 after it is less than 5, so 52 rounds down to 50. So 3.8 × 52 is about 4 × 50 = 200.
The exact answer is close
The exact answer is 3.8 × 52 = 197.6, very close to the estimate of 200.
The two roundings pulled in opposite directions. Rounding 3.8 up to 4 made the product larger, and rounding 52 down to 50 made it smaller, so the two changes nearly cancel.
On a line from 0 to 1000, the exact 197.6 sits right beside 200.
only one factor moved, so the error is that strip alone: 38 × 52 − 1976 = +0, +0%
Round both factors to the nearest ten and get the error under 2%
The dashed rectangle is 38 by 52, and its area, 38 × 52, is ten times 3.8 × 52. Drag the sides of the gold rectangle to 40 and 50: rounding 38 up adds a strip, rounding 52 down takes one away, and 40 × 50 = 2000 is only 24 more than 38 × 52 = 1976.
Decimals round the same way
To estimate 0.52 × 88, find the first significant figure of 0.52. It is the 5, in the tenths, and the 2 after it is less than 5, so 0.52 rounds to 0.5. 88 rounds up to 90.
So 0.52 × 88 is about 0.5 × 90, which is half of 90, or 45. The exact answer is 45.76.
0.5 × 90 is half of 90, which is 45.
Checking an answer
An estimate is a check on a calculation. If a calculator, or a classmate, says that 3.8 × 52 = 40, the estimate says the answer should be about 200, so 40 is far too small and something has gone wrong.
An estimate cannot give the exact answer, but it catches the mistakes that change the size of an answer, such as a decimal point in the wrong place or a digit left out.
40 is nowhere near the estimate of 200, so it cannot be the answer to 3.8 × 52.
The usual mistakes
Keep the first significant figure in its own place. To one significant figure 0.0285 is 0.03: the first nonzero digit, the 2, stays in the hundredths, and the 8 after it rounds it up to 3. It is not 0, and it is not 0.3.
Round each number the right way. 3.8 rounds to 4, not 3, because the 8 after the first figure is 5 or more.
Worked example: A Calculator Answer Checked Against an Estimate
Question A factory makes 7840 bolts, each with a mass of 0.0285 kg. The bolts are packed into boxes that each hold 0.62 kg of bolts, so the number of boxes needed is 7840 × 0.02850.62. (a) Estimate the number of boxes by rounding each number to 1 significant figure. (b) A clerk works this out on a calculator and reads 36.04. Use your estimate to decide whether the clerk's answer is of the right size.
1.Round each number to 1 significant figure: 7840 becomes 8000, 0.0285 becomes 0.03 and 0.62 becomes 0.6. The calculation is now 8000 × 0.030.6.
To 1 significant figure, 7840 becomes 8000, 0.0285 becomes 0.03 and 0.62 becomes 0.6. 2.Work out the top first: 8000 × 0.03 = 80 × 3 = 240.
Work out the top first: 8000 × 0.03 = 240. 3.To divide by 0.6, multiply the top and the bottom by 10: 2400.6 = 24006 = 400. The answer is larger than 240, because dividing by a number less than 1 makes a number larger. (a) About 400 boxes are needed.
(a) 2400.6 = 24006 = 400 boxes. Dividing by a number less than 1 makes 240 larger. 4.(b) The clerk's 36.04 is about 10 times too small, so it is not of the right size. A correct answer should be a few hundred. The digits suggest a decimal point in the wrong place, and the calculation does give about 360 boxes.
(b) 36.04 is about 10 times too small, so it is not of the right size.
Answer: (a) about 400 boxes; (b) no: 36.04 is about 10 times too small, and the calculation gives about 360 boxes
Common mistakes
- Working out 240 ÷ 0.6 as 240 × 0.6 = 144, or expecting the answer to be smaller than 240. There are more than 240 lots of 0.6 in 240: 240 ÷ 0.6 = 2400 ÷ 6 = 400.
- Rounding 0.0285 to 0 or to 0.3. To 1 significant figure the first non-zero digit, the 2, is kept in its place and rounded up by the 8 after it, which gives 0.03.
More significant figures and estimation problems, worked step by step →
Worked example: A Bill Estimated by Rounding Each Price to One Significant Figure
Question Priya buys four items that cost $28.50, $16.40, $4.34 and $0.76. Before paying, she estimates the bill by rounding each price to 1 significant figure. (a) Find her estimate. (b) Find the exact bill and the percentage error of her estimate.
1.Round each price to 1 significant figure. $28.50 becomes $30 because the second digit is 8. $16.40 becomes $20 because the second digit is 6. $4.34 becomes $4 because the second digit is 3. In $0.76 the first significant figure is the 7 and the next digit is 6, so it becomes $0.80.
To 1 significant figure the prices become $30, $20, $4 and $0.80. 2.(a) Her estimate is 30 + 20 + 4 + 0.80 = $54.80.
(a) Her estimate is 30 + 20 + 4 + 0.80 = $54.80. 3.The exact bill is 28.50 + 16.40 + 4.34 + 0.76. Add in pairs: 28.50 + 16.40 = 44.90 and 4.34 + 0.76 = 5.10, so the bill is 44.90 + 5.10 = $50.
The exact bill is 44.90 + 5.10 = $50. 4.The error of the estimate is 54.80 − 50 = $4.80.
The error of the estimate is 54.80 − 50 = $4.80. 5.(b) The percentage error is 4.8050 × 100% = 9.6%. The estimate is too high because the two largest prices were both rounded up. Check: 9.6% of $50 is $4.80.
(b) The percentage error is 4.8050 × 100% = 9.6%.
Answer: (a) $54.80; (b) the exact bill is $50 and the percentage error is 9.6%
Common mistakes
- Rounding $0.76 to $1. That is rounding to the nearest dollar. To 1 significant figure the first non-zero digit, the 7, is kept, and 0.76 becomes 0.8.
- Dividing the error by the estimate, 4.8054.80. The percentage error is measured against the exact value, which is the bill of $50.
More significant figures and estimation problems, worked step by step →