Warm-up
Three quick ones from earlier work. Try each in your head or on paper before you open the answer.
- From Year 7: round 3.476 to 1 decimal place.
Answer
3.5. The next digit, 7, is 5 or more, so the 4 rounds up. - From Year 9: write 4.2 × 10³ as an ordinary number.
Answer
4200. Multiplying by 10³ moves every digit 3 places to the left. - Work out 30 ÷ 0.5.
Answer
60. Dividing by a half doubles: there are 60 halves in 30.
Words used in this lesson
- significant figure
- a digit that tells you something about the size or accuracy of a number, counted from the first digit that is not zero. For example: 0.0305 has 3.
- round
- replace a number by a nearby simpler one with fewer significant figures. For example: 4718 to 2 s.f. is 4700.
- estimate
- a quick approximate answer from rounded numbers. For example: 21.3 × 3.9 is about 20 × 4 = 80.
- halfway point
- the value exactly between the two numbers you could round to. For example: 4750 between 4700 and 4800.
- accuracy
- how close a value is to the true value, shown by how many significant figures it has. For example: 2.40 m, given to 3 significant figures.
- leading zero
- a zero before the first digit that is not zero; it only shows place value. For example: the zeros in 0.007.
One crowd, two headlines
A city marathon in Addis Ababa had 38 461 runners. One headline says “38 461 runners”; another says “about 38 000 runners”. Both are true, but they say different things about how carefully the runners were counted.
Predict first: if a reporter keeps 3 significant figures, does 38 461 become 38 400 or 38 500? And how many significant figures does 0.0070 have: 2, 4 or 5? Then use the figure to check.
38 461 to 2 significant figures: it lies between 38 000 and 39 000, below the halfway point 38 500, so it rounds to 38 000.
Show me the steps
- 38 461 to 2 significant figures: it lies between 38 000 and 39 000. It is below the halfway point 38 500, so it rounds to 38 000. The zeros keep its size.
- Slide to 3 significant figures: it lies between 38 400 and 38 500, above the halfway point 38 450, so it rounds to 38 500.
- Press 0.0070496 and choose 2 significant figures. The zeros before the 7 are not significant. It lies between 0.0070 and 0.0071, below the halfway point 0.00705, so it rounds to 0.0070.
- Choose 1 significant figure: 0.007. It has the same value as 0.0070, but the final zero in 0.0070 says the number is known to the nearest ten-thousandth.
The first significant figure is the first digit that is not zero. Count from there. Zeros between other digits count (405 has 3 significant figures); zeros that only place a number do not (0.007 has 1); a zero at the end of a decimal is written on purpose and counts (0.0070 has 2).
To round to n significant figures, find the two numbers with n significant figures either side, and go to the nearer one. Halfway rounds up. Fill the gap with zeros so the number keeps its size: 38 461 is about 38 000, never 38.
To estimate, round every number to 1 significant figure and work it out in your head. Give a final answer to the same accuracy as the data, usually 3 significant figures.
Two common mistakes. Rounding to decimal places when the question says significant figures (0.0070496 to 3 decimal places is 0.007, but to 3 significant figures it is 0.00705). And chopping digits off instead of rounding: 38 461 to 3 significant figures is 38 500, not 38 400, because the next digit is 6.
Worked examples
1. Round 0.050371 to 2 and to 3 significant figures.
- The first digit that is not zero is the 5. The 0 after it is between digits, so it counts.
- 2 significant figures: keep 5 and 0. The next digit is 3, so round down: 0.050.
- 3 significant figures: keep 5, 0 and 3. The next digit is 7, so round up: 0.0504.
2. Estimate (6.12 × 29.4) ÷ 0.198.
- Round each number to 1 significant figure: 6, 30 and 0.2.
- 6 × 30 = 180, and 180 ÷ 0.2 = 900, because dividing by 0.2 is the same as multiplying by 5.
- The estimate is 900. The exact value is 908.7…, so the estimate is close.
Your go. The first step is given. Round 7096 to 2 significant figures.
- The first two significant figures are 7 and 0, and the next digit is 9.
- Now decide whether to round up or down, and fill the gap with zeros.
Check the answer
9 is 5 or more, so 70 rounds up to 71: 7100. Writing 71 would lose the size of the number.
With a GDC (optional)
A GDC can round its display for you. In the mode or set-up menu, choose Sci (scientific) and set it to show 3 significant figures (some models ask for the number of decimal places instead, which is 2 here): 38461 is then shown as 3.85E4. That is calculator notation for 3.85 × 10⁴ = 38 500, which is 38461 to 3 significant figures; in a written answer, use the × 10⁴ form. The Fix setting is different: it rounds to a number of decimal places, so Fix 3 shows 0.0070496 as 0.007, which is only 1 significant figure.
Use the GDC to check, not to think. Estimate first, then type the whole calculation in one line with brackets, such as (6.12 × 29.4) ÷ 0.198, and compare the result with your estimate. Set the mode back to Normal or Float when you finish.
Check before you go on
1. How many significant figures does 0.04050 have?
Your turn
Start at Level 1 and go as far as you can. Type negative numbers with a minus sign, for example -7. A wrong answer gets a reason, not just a cross: read it, then try again.
Level 1 · Getting started
2. Round 63 852 to 2 significant figures.
3. Round 0.0068271 to 3 significant figures.
Level 2 · Building
4. Estimate (48.7 × 3.92) ÷ 0.512 by rounding each number to 1 significant figure.
5. A rectangular tea field in Kericho, Kenya, measures 132.6 m by 87.4 m. Find its area in m², correct to 3 significant figures.
6. A 1.5 litre bottle of mango juice is shared equally between 7 cups. A calculator gives 1.5 ÷ 7 = 0.214285714. Which is the most sensible answer for the juice in each cup?
Level 3 · Stretch
7. The crowd at a football match in Buenos Aires is reported as 47 000, correct to 2 significant figures. What is the largest whole number of people there could have been?
8. A chemist estimates (0.0391 × 612) ÷ 19.8 by rounding each number to 1 significant figure. What estimate does she get?
Something to think about
Rounding to 1 significant figure always changes a number a little. What is the largest change it can make, as a fraction of the number? Try 149, 150 and 951.
One way to think about it
149 rounds to 100, a change of 49, which is about a third of 149. 150 rounds to 200, a change of 50, exactly a third of 150. 951 rounds to 1000, a change of only about 5%. The worst case is a number starting 1.5, which changes by a third; numbers starting with a 9 hardly change at all. So a 1 significant figure estimate can be off by up to about a third, and that is why it checks the size of an answer, not its digits.
Exit ticket
9. Round 0.00995 to 2 significant figures.