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.
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Answer key: Number worksheet — GCSE Higher
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- (d) 9 — List all the factors of 36 in pairs that multiply to give 36: 1 × 36, 2 × 18, 3 × 12, 4 × 9, and 6 × 6. This gives the factors 1, 2, 3, 4, 6, 9, 12, 18 and 36 — nine factors in total, with 6 counted only once even though it appears in a pair with itself. Forgetting that 36 is itself a factor of 36 and leaving it off the list gives 8. Counting the number of factor pairs, five of them, rather than the number of individual factors gives 5. Treating the repeated pair 6 × 6 as two separate factors, 6 and 6 again, gives 10 instead of 9. So 36 has 9 factors.
- (b) 20 — Each length has its own error interval: 11.5 ≤ a < 12.5 and 6.5 ≤ b < 7.5. The upper bound of a sum is found by adding the upper bounds of both quantities: 12.5 + 7.5 = 20. Bounding only one of the two lengths and adding the other quantity's given value unbounded, 12.5 + 7 = 19.5, misses that both measurements carry their own uncertainty. Adding the lower bounds instead of the upper bounds, 11.5 + 6.5 = 18, gives the lower bound of the sum, not the upper one. Using a whole centimetre of error either side instead of half a centimetre, (12 + 1) + (7 + 1) = 21, comes from forgetting the error is half the rounding unit.
- (c) 1/4 — A negative index means the reciprocal of the positive power, so 2⁻² = 1 ÷ 2² = 1/4. Treating the negative sign as making the answer negative instead gives −(2²) = −4. Ignoring the negative sign altogether gives just 2² = 4. Finding the reciprocal correctly but then also applying a negative sign gives −1/4.
- (b) 3/50 — Method: write the decimal over the power of ten that matches the number of digits after the point, counting every digit including a zero, then divide the numerator and the denominator by their highest common factor. Working: 0.06 has two digits after the point, so it is 6 hundredths and can be written as 6/100; the highest common factor of 6 and 100 is 2, and 6 ÷ 2 = 3 with 100 ÷ 2 = 50. Answer: 3/50. The distractors: 3/5 comes from ignoring the zero straight after the point and converting 0.6 instead, giving 6/10, which cancels to 3/5; 3/500 comes from counting three decimal places instead of two and writing 6/1000, which cancels to 3/500; 1/6 comes from putting 1 over the digits after the point, as though 0.06 meant one sixth.
- (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.
- (d) 12 — Method: if the bracelets are identical and no beads are left over, the number of bracelets must divide exactly into both totals, so it is the highest common factor of 24 and 36. Working: 24 = 2³ × 3 and 36 = 2² × 3²; taking the lower index of each shared prime gives 2² × 3 = 4 × 3 = 12. Each bracelet then has 2 red beads and 3 blue beads. Answer: 12. The distractors: 6 comes from taking each shared prime once rather than at its lower index, giving 2 × 3, which is a common factor but not the highest; 72 is the lowest common multiple of 24 and 36, from taking the higher index of each prime instead of the lower; 60 comes from adding the two bead totals instead of looking for a common factor.
- (d) 9 — Method: split into two cases — the soups with no restriction, and the mushroom soup on its own — then add the totals. Working: the 2 soups other than mushroom can be paired with any of the 4 sandwiches: 2 × 4 = 8. The mushroom soup can only be paired with the cheese sandwich: 1 combination. Total = 8 + 1 = 9. Answer: 9. 12 comes from working out 3 × 4 = 12 without applying the restriction at all. 8 comes from correctly finding the 2 unrestricted soups' 8 combinations, but forgetting to add back the 1 allowed mushroom-and-cheese combination. 11 comes from taking the unrestricted total of 12 and removing only 1 mushroom combination instead of all 3 disallowed ones.
- (c) 25/36 — Method: square a fraction by squaring its numerator and its denominator separately, then add the two results over a common denominator. Working: (2/3)² = 4/9 and (1/2)² = 1/4; the lowest common denominator of 9 and 4 is 36, so 4/9 = 16/36 and 1/4 = 9/36, and 16 + 9 = 25. Answer: 25/36. The distractors: 49/36 comes from adding the two fractions first and squaring the total, giving (7/6)²; 5/13 comes from squaring correctly but then adding the numerators and the denominators, as (4 + 1)/(9 + 4); 7/3 comes from doubling each fraction instead of squaring it, giving 4/3 + 1.
- (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.
- (a) 3 — Method: the fourth root of a number is the positive value that gives that number when it is multiplied by itself four times. Working: 2 × 2 × 2 × 2 = 16, which is too small, and 3 × 3 × 3 × 3 = 9 × 9 = 81. Answer: 3. The distractors: 9 comes from taking the square root of 81 instead of its fourth root; 4.5 comes from taking the square root and then halving it, as though a fourth root were half a square root; 20.25 comes from dividing 81 by 4, treating the root's index as a divisor.
- (d) 60 mph — Average speed = distance ÷ time, with the time measured in hours. 45 minutes is 45/60 of an hour, which is 0.75 of an hour, so the journey takes 1.75 hours. Speed = 105 ÷ 1.75 = 60 mph. 52.5 mph rounds the time up to 2 hours, 72.4 mph writes 1 hour 45 minutes as 1.45 hours, and 183.75 mph multiplies the distance by the time instead of dividing.
- (a) 1/4 — First find the fraction of the people who are children: the pool is 3 + 5 = 8 equal shares and the children take 5, so the children are 5/8 of the people. The boys are 2/5 of that, and 'of' means multiply: 2/5 × 5/8 = 10/40 = 1/4. Answering 2/5 gives the boys as a fraction of the children only, 3/20 multiplies by the adults' fraction 3/8 instead of the children's, and 3/5 is the fraction of the children who are girls.
- (a) 14 — Multiply both numbers by 10 to clear the decimals: 8.4 becomes 84 and 0.6 becomes 6. Then divide: 84 ÷ 6 = 14, so 14 complete pieces can be cut. Scaling only the divisor by 10 and leaving the dividend as 8.4 gives 8.4 ÷ 6 = 1.4, which rounds down to 1 complete piece — the dividend was never converted. Scaling only the dividend by 10 and leaving the divisor as 0.6 gives 84 ÷ 0.6 = 140. Rounding the divisor from 0.6 to 0.7 before dividing, trading accuracy for a rounder number, gives 8.4 ÷ 0.7 = 12. So 14 complete pieces of ribbon can be cut.
- (a) 2/5 — The ratio 2:3 has 2 + 3 = 5 parts in total, and the larger plot is 3 of those parts. Since the larger plot is 18 m², each part is 18 ÷ 3 = 6 m², so the total area is 5 × 6 = 30 m² and the smaller plot is 2 × 6 = 12 m². The fraction of the total area taken up by the smaller plot is 12/30, which simplifies to 2/5. Giving the fraction for the larger plot instead of the smaller one gives 3/5. Comparing the smaller plot to the larger plot instead of to the total area gives 2/3. Assuming the two plots split the area evenly, ignoring the given ratio altogether, gives 1/2.
- (b) 28 — Method: count the ordered choices with the product rule and then correct for the double counting, because the two captains have equal standing and so a pair is the same pair whichever captain is named first. Working: there are 8 players who could be named first and 7 who could be named second, giving 8 × 7 = 56 ordered choices; each pair has been counted twice, once in each order, so the number of pairs is 56 ÷ 2 = 28. Answer: 28. The distractors: 56 comes from stopping at 8 × 7 and never halving, which counts each pair of captains twice; 64 comes from working out 8 × 8, which allows the same player to be chosen as both captains; 16 comes from multiplying the 8 players by the 2 captaincies instead of pairing the players with one another.
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