Printable · GCSE Foundation · ages 14-16
Estimation and checking worksheet — GCSE Foundation
Fifteen questions on "estimation and checking" — DfE statement N14. Print it, or print three versions so neighbours cannot copy by letter; the key gives the letter for each version.
Answer key: Estimation and checking worksheet — GCSE Foundation
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- (b) 7 — Method: trap the number between the two square numbers on either side of it, then decide which of them it is nearer to. Working: 6² = 36 and 7² = 49, so √45 lies between 6 and 7; 49 − 45 = 4 while 45 − 36 = 9, so 45 is nearer to 49. Answer: 7. The distractors: 6 comes from taking the square number below 45 and stopping there, without checking which of 36 and 49 is nearer; 22.5 comes from halving 45 instead of looking for the number that multiplies by itself to give 45; 2,025 comes from squaring 45 instead of taking its square root.
- (a) 300 — Method: round each number to 1 significant figure, then divide the rounded values. Working: 8,900 rounds to 9,000 and 29 rounds to 30; cancelling a zero from each gives 900 ÷ 3. Answer: 300. The distractors: 450 comes from rounding 29 down to 20 instead of to the nearest ten, giving 9,000 ÷ 20; 3,000 comes from rounding 29 to 3 rather than to 30, a place-value slip that divides by a number ten times too small; 307 is the exact quotient rounded to the nearest whole number, worked out in full when the question asks for an estimate.
- (a) £600 — Method: round the number of cakes and the price of each cake to 1 significant figure, then multiply the rounded values. Working: 187 rounds to 200, and £2.95 rounds to £3 (the digit after the first, 9, rounds the 2 up to 3), so the estimate is 200 × £3 = £600. £400 comes from rounding £2.95 down to £2 instead of up to £3, giving 200 × £2 = £400. £561 comes from rounding only the price and using the exact number of cakes, 187 × £3 = £561. £570 comes from rounding 187 to the nearest 10 as 190 instead of to 1 significant figure as 200, giving 190 × £3 = £570. Answer: £600.
- (c) 280 — Method: round each number to 1 significant figure, then multiply the rounded numbers. Working: 6.8 rounds to 7 because the next digit is 8, and 41 rounds to 40 because its next digit is 1, so the estimate is 7 × 40 = 280. Answer: 280. The distractors: 240 comes from cutting 6.8 down to 6 instead of rounding it up to 7, giving 6 × 40 = 240; 350 comes from rounding 41 up to 50 when the digit after its first significant figure is less than 5, giving 7 × 50 = 350; 28 comes from multiplying the leading digits only and losing the place value of the 40, which makes the estimate ten times too small.
- (d) 2 hours — Method: the time for a journey is the distance divided by the speed, so round the distance first and then divide by the speed. Working: 95 km rounds to 100 km, and 100 ÷ 50 = 2; the speed is in kilometres per hour, so the answer is a number of hours. Answer: 2 hours. The distractors: 1 hour comes from rounding the distance down to 50 km to match the speed, so that the journey looks like a single hour of driving; 30 minutes comes from dividing the speed by the distance, 50 ÷ 100, instead of the distance by the speed; 1 hour 54 minutes is the exact time, 95 ÷ 50 = 1.9 hours, worked out in full when the question asks for an estimate.
- (d) 60 — Method: round each number to 1 significant figure and multiply; the estimate then shows whether the calculator answer is sensible. Working: 3.1 rounds to 3 and 19.6 rounds to 20, so the estimate is 3 × 20 = 60. Answer: 60. Hannah's 6.076 is about ten times too small, which is what happens when 19.6 is keyed in as 1.96. The distractors: 62 comes from rounding 19.6 only and leaving 3.1 as it stands, giving 3.1 × 20 = 62; 6 comes from trusting the calculator display rather than checking it against an estimate; 600 comes from rounding 19.6 to 200 instead of to 20, a place-value slip, giving 3 × 200 = 600.
- (c) 1,000 m² — Method: round each length to 1 significant figure, then use area of a rectangle = length × width on the rounded lengths. Working: 19.6 m rounds to 20 m and 48.3 m rounds to 50 m, so the estimate is 20 × 50 = 1,000 and the area is about 1,000 m². Answer: 1,000 m². The distractors: 800 m² comes from rounding 48.3 down to 40 when the digit after its first significant figure is 8 and sends it up to 50, giving 20 × 40 = 800; 140 m² is the perimeter of the rounded rectangle, 2 × 20 + 2 × 50 = 140, not its area; 70 m² comes from adding the rounded lengths, 20 + 50 = 70, instead of multiplying them.
- (d) 1,000 — Method: round each number to 1 significant figure, then multiply the rounded values. Working: 49 rounds to 50 and 21 rounds to 20, and 50 × 20 = 1,000 because 5 × 2 = 10 and the two rounded numbers carry one zero each. Answer: 1,000. The distractors: 800 comes from rounding 49 down to 40 instead of to the nearest ten; 1,500 comes from rounding 21 up to 30 rather than down to 20; 1,029 is the exact product 49 × 21, worked out in full when the question asks for an estimate.
- (a) 20 — Method: round each number to 1 significant figure, then divide the rounded values. Working: 588 rounds to 600 (1 s.f.) and 31 rounds to 30 (1 s.f.). 600 ÷ 30 = 20. Answer: 20. 17 comes from cutting 588 down to 500, keeping the leading digit as it stands instead of rounding it up to 1 significant figure, 600, then dividing by the correctly rounded 30. 200 comes from misreading the rounded divisor 30 as 3, giving 600 ÷ 3 instead of 600 ÷ 30. 19 is the exact value of 588 ÷ 31 rounded to the nearest whole number, found without rounding the numbers first.
- (b) 20 — Method: check a calculator answer by replacing the percentage with a simple fraction — 25% is one quarter, so the calculation becomes a division by 4. Working: 80 ÷ 4 = 20, and the calculator answer of 320 is 80 × 4, which is what happens when the amount is multiplied by 4 instead of divided by it; a quarter of an amount must be smaller than the amount. Answer: 20. The distractors: 25 comes from writing the percentage itself down as the answer; 16 comes from dividing by 5 instead of by 4, which finds 20% rather than 25%; 3.2 comes from working out 80 ÷ 25 instead of a quarter of 80.
- (b) £1,000, so £900 is not enough — Method: round each number to 1 significant figure, multiply to estimate the total cost, then compare the estimate with the money available. Working: 187 rounds to 200 and £4.85 rounds to £5, so the estimate is 200 × 5 = 1,000, and £1,000 is more than the £900 the school has. Answer: £1,000, so £900 is not enough. The distractors: £800 comes from cutting £4.85 down to £4 instead of rounding it up to £5, giving 200 × 4 = 800, and that estimate wrongly suggests the money stretches; £935 comes from rounding the price only and keeping 187 lunches, giving 187 × 5 = 935; £950 comes from rounding 187 to the nearest 10 rather than to 1 significant figure, giving 190 × 5 = 950.
- (a) 3200 — Method: round each number to 1 significant figure, then multiply the rounded numbers. Working: 37 rounds to 40 (the digit after the first, 7, rounds the 3 up to 4), and 84 rounds to 80, so the estimate is 40 × 80 = 3200. 2400 comes from rounding 37 down to 30, keeping the first digit as it is instead of letting the 7 round it up, giving 30 × 80 = 2400. 3108 comes from multiplying the exact numbers, 37 × 84, without rounding either of them first. 120 comes from adding the rounded numbers, 40 + 80 = 120, instead of multiplying them. Answer: 3200.
- (a) 49 — Method: 7² means 7 multiplied by itself. Working: 7 × 7 = 49. Answer: 49. 14 comes from working out 7 × 2, treating the power 2 as a number to multiply by rather than an instruction to multiply 7 by itself. 77 comes from writing the digit 7 twice side by side, treating the power as an instruction to repeat the digit rather than to multiply. 9 comes from working out 7 + 2, adding the base and the power instead of multiplying the base by itself.
- (d) 1 × 10⁷ seconds — Method: the time for a journey is the distance divided by the speed, so round each number to 1 significant figure and then divide; dividing numbers in standard form means dividing the coefficients and subtracting the indices. Working: 3.1 × 10¹⁵ rounds to 3 × 10¹⁵ and 2.998 × 10⁸ rounds to 3 × 10⁸; 3 ÷ 3 = 1 for the coefficients, and 15 − 8 = 7 for the indices. Answer: about 1 × 10⁷ seconds. The distractors: 3 × 10⁷ seconds comes from subtracting the indices correctly but leaving the coefficient as 3 instead of dividing 3 by 3; 1 × 10⁻⁷ seconds comes from dividing the speed by the distance instead of the distance by the speed; 9 × 10²³ seconds comes from multiplying the two quantities instead of dividing them, since 3 × 3 = 9 and 15 + 8 = 23.
- (b) £220 — Method: round each ticket price to the nearest £5, multiply each rounded price by the number of tickets, then add the two totals. Working: the adult price £38.50 rounds to £40, and 4 × £40 = £160; the child price £19.75 rounds to £20, and 3 × £20 = £60; £160 + £60 = £220. Answer: £220. £213.25 is the exact total cost, found without rounding the prices first, so it is not an estimate. £160 comes from including the cost of the adult tickets only and forgetting the three children's tickets. £200 comes from swapping the two ticket quantities, using 3 adults and 4 children instead of 4 adults and 3 children.
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