Printable · GCSE Higher · ages 14-16
Number worksheet — GCSE Higher
Fifteen questions across the number statements at Higher tier. Choose the non-calculator filter to rehearse Paper 1, which counts for a third of the marks.
Answer key: Number worksheet — GCSE Higher
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- (d) 6,000,000 — Method: round each number to 1 significant figure, then subtract. Working: 8,340,000 rounds to 8,000,000 (1 s.f.); 1,950,000 rounds to 2,000,000 (1 s.f.); 8,000,000 − 2,000,000 = 6,000,000. Answer: 6,000,000. 6,390,000 is the exact difference, found without rounding the numbers first. 8,000,000 comes from rounding the population correctly but forgetting to subtract the capital's population at all. 6,300,000 comes from rounding 8,340,000 to the nearest hundred thousand, 8,300,000, instead of to 1 significant figure, then subtracting the correctly rounded 2,000,000.
- (b) 5 — Method: find the square root first, then divide. Working: √225 = 15, and 15 ÷ 3 = 5. Answer: 5. (75 comes from dividing 225 by 3 first and forgetting to take the square root at all. 8.7 comes from dividing 225 by 3 inside the root, √(225 ÷ 3) ≈ 8.7, instead of taking the root first. 45 comes from misreading the divisor as 5 instead of 3, working out 225 ÷ 5 = 45.)
- (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.
- (d) 28 — Method: round each number to the nearest whole number, then multiply the rounded numbers to get an estimate that can be compared with the assistant's answer. Working: 7.2 rounds to 7, and 3.9 rounds to 4, so the estimate is 7 × 4 = 28. Since 28 is much smaller than 56.16, the assistant's answer cannot be correct. 56 comes from rounding the assistant's answer to the nearest whole number, instead of rounding the two numbers being multiplied and then multiplying them. 35 comes from rounding both numbers correctly but then slipping in the seven times table, writing 7 × 5 = 35 in place of 7 × 4 = 28. 21 comes from rounding 3.9 down to 3 instead of 4, giving 7 × 3 = 21. Answer: 28.
- (c) 18 — The error intervals are 35 ≤ distance < 45 and 2.5 ≤ time < 3.5. Average speed is distance ÷ time, and to make a quotient as large as possible you divide the largest possible numerator by the SMALLEST possible denominator: 45 ÷ 2.5 = 18 km/h. Using the upper bound of time as well as the upper bound of distance, 45 ÷ 3.5, gives roughly 12.9 km/h — dividing by a bigger number produces a smaller result, so this actually finds a value smaller than the true maximum. Using the lower bound of distance with the lower bound of time, 35 ÷ 2.5 = 14, mixes up which bound belongs to a maximum calculation. Using the lower bound of distance with the upper bound of time, 35 ÷ 3.5 = 10, is in fact the correct method for the MINIMUM speed, not the maximum.
- (d) 1 7/12 — Method: convert both mixed numbers to improper fractions with a common denominator, then subtract. Working: 3 1/4 = 39/12 and 1 2/3 = 20/12, so 39/12 − 20/12 = 19/12 = 1 7/12. Answer: 1 7/12. Chloe's method, subtracting whole numbers (3−1=2) and fraction parts (2/3−1/4=5/12) separately without exchanging, gives 2 5/12. 1 3/4 comes from converting 2/3 to twelfths incorrectly as 6/12 instead of 8/12, then subtracting. 8 comes from converting both mixed numbers to improper fractions correctly (13/4 and 5/3) but then subtracting numerators and denominators separately: (13−5)/(4−3) = 8/1.
- (c) 0.0479 — Method: round each option to 2 significant figures and check which one gives 0.048. Working: for 0.0479, the first two significant figures are 4 and 7; the next digit is 9, so 7 rounds up to 8, giving 0.048. For 0.0485, the first two significant figures are 4 and 8; the next digit is 5, so 8 rounds up to 9, giving 0.049, not 0.048. 0.052 already has exactly 2 significant figures, 5 and 2, so it stays as 0.052 and does not round to 0.048 at all. 0.04 has only 1 significant figure, so it is already less precise than the 2 significant figures asked for. Answer: 0.0479.
- (a) 71 — 2.9⁴ = (2.9²)². First, 2.9² = 8.41. Then square that: 8.41² = 70.7281, since 841² = 707281 and the decimal point moves four places. To 2 significant figures this rounds to 71, because the figure after the first two significant figures (7 and 0) is a 7, which rounds the 0 up to 1. Rounding the working down instead of up — taking 70.7281 to 70 — ignores that the next figure is 5 or more. Rounding 2.9 up to 3 before doing any working at all, then computing 3⁴ = 81, uses a much cruder approximation and overshoots the true value. Rounding 8.41 all the way down to 8 before squaring, 8² = 64, rounds far too aggressively and loses the accuracy needed for 2 significant figures.
- (d) 5 — Method: the greatest number of identical bunches is the highest common factor of the two flower totals; then divide the red roses by that number of bunches. Working: 60 = 2² × 3 × 5 and 84 = 2² × 3 × 7, so their highest common factor is 2² × 3 = 12. That means 12 bunches, and 60 ÷ 12 = 5 red roses in each. 7 is the number of white roses in each bunch, since 84 ÷ 12 = 7, not red roses. 12 is the number of bunches itself, not the number of red roses in one bunch. 20 comes from working out 60 ÷ 3 = 20, dividing by only part of the highest common factor. Answer: 5.
- (d) 31, which is prime — Method: work out the value, remembering that multiplication comes before addition, then test it for primality by dividing by each prime up to its square root. Working: 2 × 3 × 5 = 30, so the value is 30 + 1 = 31. Since 6² = 36 is larger than 31, only 2, 3 and 5 need testing: 31 is odd, 31 ÷ 3 leaves a remainder of 1, and 31 does not end in 0 or 5. It therefore has exactly two factors, 1 and itself. Answer: 31, which is prime. The distractors: 30, which is not prime comes from working out 2 × 3 × 5 and forgetting to add the 1; the claim that 31 = 1 × 31 makes it non-prime comes from treating any factor pair as proof, forgetting that a prime is allowed the pair 1 and itself; the claim that 31 is a multiple of 3 comes from assuming that a number containing the digit 3 divides by 3, when in fact 31 ÷ 3 leaves a remainder.
- (b) £45.90 — The cost is 18 × £3.45 = £62.10. Profit = £108 − £62.10 = £45.90. A candidate who does not borrow in the tenths column, doing 1 − 0 = 1 instead of borrowing to make 10 − 1 = 9 and so leaving the units as 8 − 2 = 6, gets £46.10. A candidate who adds the cost to the selling price instead of subtracting gets £108 + £62.10 = £170.10. A candidate who gives the cost instead of the profit gets £62.10.
- (c) 3/11 — Method: let a letter stand for the recurring decimal, multiply by the power of ten that shifts exactly one repeating block past the point, subtract the original equation so that the recurring tail cancels, then solve and cancel. Working: let x = 0.272727...; the repeating block is two digits long, so multiply by 100 to give 100x = 27.272727...; subtracting gives 99x = 27, so x = 27/99; the highest common factor of 27 and 99 is 9, and 27 ÷ 9 = 3 with 99 ÷ 9 = 11. Answer: 3/11. The distractors: 27/100 comes from writing the repeating block over 100 instead of over 99, forgetting that subtracting x leaves 99x rather than 100x; 3/10 comes from rounding the decimal to one place and converting 0.3; 2/9 comes from treating only the 2 as recurring and converting 0.222... instead.
- (b) 2/5 — The statement compares girls with boys, so girls : boys = 2 : 3. The club is therefore 2 + 3 = 5 equal shares and the girls take 2 of them, so 2/5 of the members are girls. 2/3 repeats the part-to-part fraction the question gives, 3/5 is the fraction that are boys, and 1/3 comes from treating 2/3 as a fraction of the whole club and taking what is left.
- (d) 10 — Method: list the pairs systematically, taking each sweet in turn and pairing it only with the sweets that come after it, so that no pair is written down twice. Working: numbering the sweets 1 to 5, the first sweet pairs with 4 others, the second pairs with 3 sweets that come after it, the third with 2 and the fourth with 1, so the total is 4 + 3 + 2 + 1 = 10. Answer: 10. The distractors: 20 comes from working out 5 × 4 and never halving, which counts each pair twice, once in each order; 25 comes from working out 5 × 5, which allows the same sweet to be chosen twice; 9 comes from adding the 5 choices and the 4 remaining choices instead of combining them as a selection of two.
- (d) 1.2 × 10⁵ — 4 × 3 = 12, and 3 + 1 = 4, giving 12 × 10⁴ — but 12 is not between 1 and 10, so this must be rewritten as 1.2 × 10⁵. Stopping at 12 × 10⁴ without rewriting it leaves the coefficient out of range. Rewriting 12 as 1.2 but leaving the exponent at 4 instead of increasing it to 5 gives 1.2 × 10⁴, which is ten times too small. Adding the coefficients instead of multiplying them gives 4 + 3 = 7, so 7 × 10⁴.
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