Printable · GCSE Foundation · ages 14-16
Rounding, significant figures and error intervals worksheet — GCSE Foundation
Fifteen questions on "rounding, significant figures and error intervals" — DfE statement N15. Print it, or print three versions so neighbours cannot copy by letter; the key gives the letter for each version.
Rounding, significant figures and error intervals worksheet — GCSE Foundation
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- 1.A charity trek covers 830 miles over roughly 19 days. By rounding each number to 1 significant figure, work out an estimate for the number of miles walked per day.
- 2.Round 0.006852 to 2 significant figures.
- 3.Four friends share a restaurant bill of £53.90 equally. A calculator gives each share as 13.475. Work out how much each friend should pay.
- 4.A train journey takes 45 minutes, correct to the nearest 5 minutes. Using t for the actual time of the journey in minutes, write down the error interval for t.
- 5.Round 4,685 to the nearest 100.
- 6.Round 6.283 to 1 significant figure.
- 7.Round 592.5 to the nearest 10.
- 8.A quantity surveyor calculates a length as 24.6851 m. Round this length to 2 decimal places.
- 9.A block of butter has a mass of 250 g, correct to the nearest 10 g. Using m for the mass of the block in grams, write down the error interval for m.
- 10.A number, x, is truncated (not rounded) to 1 decimal place and the result is 6.2. Write down the error interval for x.
- 11.Round 0.006482 to 2 significant figures.
- 12.Sam rounds 0.070268 to 3 significant figures and writes 0.0703. Which of these statements is correct?
- 13.A tin of paint has a mass of 5.672 kg. Round this mass to 1 decimal place.
- 14.A van has a mass of 2,000 kg, correct to 1 significant figure. Using m for the mass of the van in kilograms, write down the error interval for m.
- 15.To estimate the cost of buying 38.7 m of rope at £21.40 per metre, both numbers are first rounded to 1 significant figure. Work out the estimate.
Answer key
- (d) 40 miles — Method: round each number to 1 significant figure first, then divide to estimate the daily distance. Working: 830 rounds to 800, and 19 rounds to 20, and 800 ÷ 20 = 40, so the estimate is 40 miles per day. 41.5 miles comes from rounding only the number of days and working out 830 ÷ 20 = 41.5, without rounding the distance too. 830 miles is the total distance for the whole trek, given as the answer without dividing by the number of days at all. 4 miles comes from working out 80 ÷ 20 = 4, misplacing a digit in the rounded distance. Answer: 40 miles.
- (a) 0.0069 — Leading zeros are not significant, so the significant figures in 0.006852 start at 6: 6, 8, 5, 2. Rounding to 2 significant figures means keeping 6 and 8, and looking at the next digit, 5, to decide whether to round up. Since 5 rounds up, the second significant figure increases from 8 to 9: 0.006852 rounds to 0.0069. A candidate who rounded to 1 significant figure instead of 2 wrote 0.007. A candidate who rounded to 3 significant figures instead of 2 wrote 0.00685. A candidate who did not round up despite the next digit being 5 wrote 0.0068.
- (b) £13.48 — Method: an amount of money is written to the nearest penny, which is 2 decimal places, so the calculator display has to be rounded to 2 decimal places. Working: 53.90 ÷ 4 = 13.475, and the digit in the third decimal place is 5, so the penny digit goes up from 7 to 8. Answer: £13.48. The distractors: £13.47 comes from chopping the third decimal place off instead of rounding with it; £13.50 comes from rounding to the nearest 10p rather than to the nearest penny; £13.40 comes from cutting the display short at 1 decimal place, which is both the wrong degree of accuracy and a truncation rather than a rounding.
- (a) 42.5 ≤ t < 47.5 — Rounding to the nearest 5 minutes means the actual time can be up to half of 5 minutes, 2.5 minutes, below or above 45 before it would round to a different multiple of 5. The lower bound is 45 − 2.5 = 42.5 and the upper bound is 45 + 2.5 = 47.5. A time of exactly 47.5 minutes would round up to 50, not 45, so 47.5 is excluded while 42.5 does still round to 45. Writing 42.5 ≤ t ≤ 47.5 wrongly includes 47.5. Writing 40 ≤ t < 50 uses a whole rounding unit, 5, either side instead of half of it. Writing 44.5 ≤ t < 45.5 treats the rounding unit as 1 minute instead of 5 minutes.
- (b) 4,700 — To round to the nearest 100, look at the digit in the tens column, which decides whether the hundreds column rounds up or stays the same. In 4,685 that digit is 8, and since 8 is 5 or more, the 6 in the hundreds column rounds up to 7, giving 4,700. Simply changing the last two digits to zero without checking the tens digit gives 4,600, which rounds down when it should round up. Rounding to the nearest 10 instead of the nearest 100 gives 4,690. Rounding to the nearest 1,000 instead gives 5,000, one place value too coarse.
- (c) 6 — Method: the first significant figure is the first non-zero digit; round using the digit after it to decide whether to round up or down. Working: the first significant figure of 6.283 is the 6; the next digit is 2, which rounds down, so 6.283 rounds to 6. 6.3 comes from rounding to 2 significant figures instead of 1. 10 comes from rounding up to the nearest 10 instead of finding 1 significant figure of the number itself. 0.6 comes from misplacing the decimal point after rounding. Answer: 6.
- (c) 590 — To round to the nearest 10, decide which multiple of 10 the number is nearer to. 592.5 lies between 590 and 600. It is 592.5 − 590 = 2.5 above 590, but 600 − 592.5 = 7.5 below 600, so it is much nearer to 590. Equivalently, the units digit is 2, and 2 is less than 5, so round down: 592.5 rounds to 590. A candidate who wrote 600 rounded up because of the 5 in the tenths place, but that digit decides rounding to the nearest whole number, not to the nearest 10 — the units digit is what matters here. A candidate who wrote 595 rounded to the nearest 5 instead of the nearest 10. A candidate who wrote 500 cut the number down to its hundreds digit instead of rounding to the nearest 10.
- (d) 24.69 — To round to 2 decimal places, look only at the third decimal digit to decide whether the second decimal digit rounds up. In 24.6851 the third decimal digit is 5, and since 5 is 5 or more, the second decimal digit rounds up from 8 to 9, giving 24.69. Rounding to 1 decimal place instead of 2 gives 24.7, one place value too coarse. Keeping the third decimal digit rather than dropping it gives 24.685, which is 3 decimal places. Looking at the fourth decimal digit, 1, instead of the third one, and wrongly deciding that no rounding is needed, leaves the length unrounded at 24.68.
- (b) 245 ≤ m < 255 — Method: a mass given to the nearest 10 g lies within half of 10 g, that is 5 g, of the figure written down. The lower limit is included because it rounds up to that figure, and the upper limit is excluded because it rounds up to the next multiple of 10. Working: 250 − 5 = 245 and 250 + 5 = 255, and a mass of 245 g rounds to 250 g while a mass of 255 g rounds to 260 g. Answer: 245 ≤ m < 255. The distractors: 240 ≤ m < 260 goes a whole 10 g either side instead of half of it; 245 < m ≤ 255 has the two limits the wrong way round; 240 ≤ m < 250 treats the stated 250 g as the largest mass possible, as though rounding were always upwards.
- (b) 6.2 ≤ x < 6.3 — Truncating simply cuts off the digits after the required decimal place instead of rounding them, so every value from 6.2 up to (but not reaching) 6.3 truncates to 6.2. This gives the error interval 6.2 ≤ x < 6.3, with no allowance made on the lower side because truncation never rounds a smaller value up into this interval. Using 6.15 ≤ x < 6.25 applies the rounding rule of going half a unit either side, which does not apply to truncation. Writing 6.1 < x ≤ 6.2 puts the interval below 6.2 instead of above it. Writing 6.2 ≤ x ≤ 6.3 wrongly includes 6.3, which truncates down to itself, not to 6.2.
- (c) 0.0065 — Leading zeros are never significant, so counting from the first non-zero digit, the first two significant figures of 0.006482 are 6 and 4. Look at the next digit along, 8, to decide whether the second figure rounds up: since 8 is 5 or more, the 4 rounds up to 5, giving 0.0065. Rounding to 2 decimal places instead of 2 significant figures gives 0.01, which answers a different question. Wrongly counting one of the leading zeros as a significant figure and stopping one figure short gives 0.006. Keeping an extra digit, as in 0.00648, gives 3 significant figures rather than 2.
- (c) Sam is correct — The leading zeros in 0.070268 are not significant, so the first three significant figures are 7, 0 and 2. The next digit along is 6, and since 6 is 5 or more, the third significant figure rounds up from 2 to 3, giving 0.0703. This means Sam's answer is correct. Writing 0.070 keeps only 2 significant figures, one short of what was asked. Writing 0.0702 ignores the digit 6 that follows and leaves the third figure unrounded. Writing 0.0704 rounds the third figure up twice, as if a later digit had also pushed it up.
- (c) 5.7 kg — Method: to round to 1 decimal place, keep one digit after the decimal point and let the digit in the second decimal place decide whether that digit stays as it is or goes up. Working: 5.672 has 6 in the first decimal place and 7 in the second decimal place; 7 is 5 or more, so the 6 goes up to 7 and the digits beyond the first decimal place are dropped. Answer: 5.7 kg. The distractors: 5.6 kg comes from chopping the digits after the first decimal place off instead of rounding them, which is truncation rather than rounding; 6.0 kg comes from rounding to the nearest whole kilogram instead of to 1 decimal place; 5.0 kg comes from chopping everything after the decimal point off, so the mass is both truncated and given to the wrong degree of accuracy.
- (a) 1,500 ≤ m < 2,500 — Method: a four-digit figure written to 1 significant figure has been rounded to the nearest 1,000, so the mass lies within half of 1,000, that is 500, of the figure given. Working: 2,000 − 500 = 1,500 and 2,000 + 500 = 2,500. The lower limit is included, because 1,500 kg rounds up to 2,000 kg to 1 significant figure, while 2,500 kg rounds up to 3,000 kg, so the upper limit is not. Answer: 1,500 ≤ m < 2,500. The distractors: 1,950 ≤ m < 2,050 comes from rounding to the nearest 100 instead of to 1 significant figure; 1,000 ≤ m < 3,000 goes a whole 1,000 either side instead of half of it; 1,500 < m ≤ 2,500 has the two limits the wrong way round.
- (d) £800 — Rounding 38.7 to 1 significant figure gives 40, and rounding 21.40 to 1 significant figure gives 20. Multiplying the rounded values gives an estimate of 40 × 20 = £800. Rounding 21.40 to the nearest whole number instead of to 1 significant figure gives 21, and 40 × 21 = £840, one place value too fine for the price. Adding the rounded values instead of multiplying them gives 40 + 20 = £60. Rounding both numbers to 2 significant figures instead of 1, giving 39 and 21, produces 39 × 21 = £819.
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