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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- (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.
- (b) 750,000,000 — Method: multiplying by 10⁸ moves the decimal point eight places to the right, and every empty place is filled with a zero. Working: 10⁸ = 100,000,000, and moving the decimal point in 7.5 eight places to the right gives 7.5 × 100,000,000. Answer: 750,000,000. The distractors: 7,500,000,000 comes from removing the decimal point first to make 75 and then writing eight zeros after it, which carries the digits one place too far; 600 comes from reading 10⁸ as 10 × 8 = 80 and working out 7.5 × 80; 0.000000075 comes from moving the decimal point eight places to the left, as though the index were negative.
- (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.
- (c) 90 — Multiply the number of choices for each course: 5 × 6 × 3 = 90. 14 comes from adding the three numbers instead of multiplying them. 30 comes from multiplying only the starters and mains, 5 × 6, and forgetting the dessert. 18 comes from multiplying only the mains and desserts, 6 × 3, and forgetting the starter.
- (a) £18 — Method: the percentage acts as an operator on the bill, and 15% can be built from 10% and 5%, where 5% is half of 10%. Working: 10% of £120 is £120 ÷ 10 = £12, and 5% is half of that, £6, so the charge is £12 + £6 = £18. Answer: £18. The distractors: £8 comes from reading 15% as one fifteenth and working out £120 ÷ 15 = £8; £12 comes from finding 10% of the bill and stopping there; £138 comes from finding the charge correctly and then giving the new total, £120 + £18, rather than the charge the question asks for.
- (c) 7 × 10⁻⁸ — '100 times smaller' means dividing by 100 = 10². Dividing 7 × 10⁻⁶ by 10² means subtracting 2 from the exponent: −6 − 2 = −8, giving 7 × 10⁻⁸. A candidate who multiplied by 100 instead of dividing added 2 to the exponent, getting 7 × 10⁻⁴. A candidate who divided by 10 instead of 100 subtracted only 1 from the exponent, getting 7 × 10⁻⁵. A candidate who did not apply the scale factor at all left the diameter as 7 × 10⁻⁶, the same as the red blood cell.
- (a) 33.2 — The radius was rounded to 1 decimal place, so its error interval is 3.15 ≤ r < 3.25. The upper bound for the area uses the upper bound of the radius, squared: area = π × 3.25² ≈ 33.183, which rounds to 33.2 m² (3 s.f.). Using the given value of the radius directly instead of its upper bound, π × 3.2² ≈ 32.2, ignores that the radius itself has a range of possible values. Bounding the radius correctly but forgetting to square it, using area = π × 3.25 ≈ 10.2 instead of π × 3.25², drops the whole squaring step from the area formula. Using the LOWER bound of the radius instead of the upper one, π × 3.15² ≈ 31.2, finds the lower bound of the area, not the upper one.
- (c) 150 — Since 90 students represent 3 of the 5 equal parts, one part is 90 ÷ 3 = 30, and the whole year group is five parts: 30 × 5 = 150. Applying the fraction forwards to 90 instead of reversing it, 90 × 3/5 = 54, treats the given number as the whole rather than as three fifths of it. Finding one part correctly as 30 but forgetting to scale up to the whole year group leaves 30 as the final answer. Treating 90 as the whole year group and adding on 2/5 of 90 for the students who do not walk, 90 + (90 × 2/5) = 126, applies the missing fraction to the wrong base amount.
- (c) 12 — Method: division and multiplication have equal priority, so they are carried out in the order they are written, from left to right; a negative divided by a negative is positive. Working: (−36) ÷ (−6) = 6, and then 6 × 2 = 12. Answer: 12. The distractors: 3 comes from carrying out the multiplication first, (−6) × 2 = −12 followed by (−36) ÷ (−12) = 3; −12 comes from treating a negative divided by a negative as negative, giving −6 and then −6 × 2 = −12; 6 comes from stopping at the division and never carrying out the multiplication by 2.
- (b) −18 — Method: the multiplication is carried out before the addition, and a negative multiplied by a positive is negative. Working: (−4) × 3 = −12, so the calculation becomes (−6) + (−12) = −18. Answer: −18. The distractors: −30 comes from adding first and multiplying afterwards, giving (−6 + −4) × 3 = −10 × 3 = −30; 6 comes from treating (−4) × 3 as +12 on the grounds that a minus sign makes a product positive, giving −6 + 12 = 6; 18 comes from ignoring both minus signs and working out 6 + 4 × 3 = 18.
- (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.
- (b) 93.9975 — Each measurement was rounded to 1 decimal place, so the error is half of 0.1: length is 12.35 ≤ L < 12.45, and width is 7.45 ≤ W < 7.55. The upper bound for the area comes from multiplying the upper bounds of both dimensions: 12.45 × 7.55 = 93.9975 m². Using the lower bound of both dimensions instead, 12.35 × 7.45 = 92.0075 m², gives the lower bound of the area rather than the upper one. Multiplying the two given rounded values directly, 12.4 × 7.5 = 93, forgets that a rounded measurement is not exact and needs its own error interval. Bounding only the length and leaving the width at its given value, 12.45 × 7.5 = 93.375, misses that the width also has an upper bound of its own.
- (c) 3 × 10⁶ — Daily revenue = (4 × 10³) × 2.5 = 1 × 10⁴ (£10,000). Multiplying by the number of days, (3 × 10²), gives annual revenue = (1 × 10⁴) × (3 × 10²) = 3 × 10⁶ (£3,000,000). A candidate who forgot to multiply by the price and just multiplied the number of items by the number of days worked out (4 × 10³) × (3 × 10²) = 1.2 × 10⁶. A candidate who added the number of days to the daily revenue instead of multiplying worked out 1 × 10⁴ + 3 × 10² = 1.03 × 10⁴. A candidate who misread £2.50 as £25 worked out a daily revenue of (4 × 10³) × 25 = 1 × 10⁵, giving an annual total of (1 × 10⁵) × (3 × 10²) = 3 × 10⁷.
- (a) 2 × 3² × 5 — Method: divide repeatedly by the smallest prime number until only prime factors remain. Working: 90 ÷ 2 = 45, 45 ÷ 3 = 15, 15 ÷ 3 = 5, and 5 is prime, so 90 = 2 × 3 × 3 × 5, written as 2 × 3² × 5. 2 × 3 × 15 stops before the 15 is broken down into 3 × 5, so it is not fully factorised. 3 × 3 × 10 stops before the 10 is broken down into 2 × 5. 2 × 45 stops after only one division. Answer: 2 × 3² × 5.
- (c) 150 m — Method: a fraction acts as an operator, so finding 3/7 of a length means dividing by the denominator and multiplying by the numerator. Working: 350 ÷ 7 = 50, so one seventh of the route is 50 m, and three sevenths is 50 × 3 = 150 m. Answer: 150 m. The distractors: 50 m comes from finding one seventh and stopping there instead of multiplying by 3; 1050 m comes from multiplying by the numerator without dividing by the denominator, giving 350 × 3 = 1050; 200 m comes from working out the stretch of the route the footpath does not run alongside, which is 4/7 of 350 m, instead of the stretch it does.
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