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) 8,350 — Rounding to the nearest 100 means the true number can be up to half of 100, which is 50, below the recorded figure before it would round down to a lower hundred. The smallest possible number is therefore 8,400 − 50 = 8,350. Adding 50 instead of subtracting it gives 8,450, which is the upper end of the interval rather than the smallest value, and 8,450 is not itself possible because it would round up to 8,500. Subtracting a whole 100 instead of half of it gives 8,300, going too far below the recorded value. Subtracting 10 instead of half of the rounding unit gives 8,390, treating the rounding unit as 100 but the tolerance as only 10.
- (d) 24 km — Method: a fraction acts as an operator, so finding 8/12 of a distance means dividing by the denominator and multiplying by the numerator. Working: 36 ÷ 12 = 3, so one twelfth of the walk is 3 km, and eight twelfths is 3 × 8 = 24 km. Answer: 24 km. The distractors: 12 km comes from working out the part of the walk still left, the other four twelfths, instead of the part already completed; 288 km comes from multiplying by the numerator without dividing by the denominator, giving 36 × 8 = 288; 4.5 km comes from dividing by the numerator instead of multiplying by it, giving 36 ÷ 8 = 4.5.
- (b) No, the true mass could be as high as 852.5 kg — Method: find the upper bound of the true mass and compare it with the weight limit. Working: the display is correct to the nearest 5 kg, so half of 5 kg is 2.5 kg, and the true mass, m kg, satisfies 847.5 ≤ m < 852.5. Part of that interval lies above 850 kg, so the parcels are not definitely within the limit. Answer: the true mass could be as high as 852.5 kg, which is above the limit. ("Yes, the display reads 850 kg, which is not above the limit" compares the limit with the displayed value instead of with the largest value the true mass could take. "Yes, the true mass is at least 847.5 kg and at most 850 kg" uses the correct half unit below but caps the interval at the limit instead of at 852.5 kg. "No, 850 kg on the display rounds up to 855 kg" wrongly treats the displayed value as if it rounds again.)
- (c) 3.2 litres — The blue paint is 3 of the 8 equal shares in the mix, that is 3/8 of the total. Three shares are 1.2 litres, so one share is 1.2 ÷ 3 = 0.4 litres. The whole mix is 8 shares: 8 × 0.4 = 3.2 litres. 2 litres is the volume of white paint, 1.92 litres divides the blue paint by the white paint's 5 shares instead of its own 3, and 9.6 litres treats the 1.2 litres as a single share.
- (a) 2/5 — Method: find the number of vegetable plots in each allotment, add them, then divide by the total number of plots in both allotments. Working: Allotment A: 3/5 × 20 = 12 vegetable plots. Allotment B: 1/5 × 20 = 4 vegetable plots. Total vegetable plots = 12 + 4 = 16. Total plots = 20 + 20 = 40. Fraction = 16/40 = 2/5. Answer: 2/5. 3/10 comes from using only Allotment A's 12 vegetable plots over the combined total of 40 plots, forgetting to add Allotment B's vegetable plots. 1/5 comes from using only Allotment B's ratio (1:4) as the fraction of vegetables, ignoring Allotment A altogether. 3/5 comes from working out the fraction of the combined plots that grow flowers instead of vegetables.
- (b) (60 + 4.5π) cm² — Method: find the area of the rectangle and the area of the semicircle separately, then add them. Working: the rectangle has area 10 × 6 = 60 cm². The semicircle has radius 3 cm, so its area is half of π × 3² = half of 9π = 4.5π cm². Total area = (60 + 4.5π) cm². Answer: (60 + 4.5π) cm². (60 + 18π) cm² comes from using the diameter (6 cm) as the radius in the semicircle area formula: half of π × 6² = 18π. (60 + 9π) cm² comes from forgetting to halve the full circle's area: π × 3² = 9π. (60 + 3π) cm² comes from finding the semicircle's arc length instead of its area: half of 2 × π × 3 = 3π.
- (b) 364 — Divide in stages using multiples of 12. 12 × 300 = 3600, leaving a remainder of 4368 − 3600 = 768. Then 12 × 64 = 768, so 4368 ÷ 12 = 300 + 64 = 364. Placing the decimal point as though dividing 436.8 by 12 gives 36.4. Transposing the last two digits of 364 gives 346. Working out 768 ÷ 12 as 4 instead of 64, losing the tens digit, and adding 300 + 4 gives 304. So 4368 ÷ 12 = 364.
- (a) 7 — Convert the mixed number to an improper fraction: 5 1/4 = 21/4. Dividing by 3/4 means multiplying by its reciprocal, 4/3: 21/4 × 4/3 gives 84/12, which simplifies to 7. So exactly 7 complete pieces of 3/4 m can be cut. Ignoring the 1/4 m and dividing only the whole number, 5 ÷ 3/4, gives 20/3, which is 6 complete pieces with some wood left over. Multiplying by 3/4 instead of its reciprocal, 21/4 × 3/4, gives 63/16, which is 3 complete pieces. Misreading 5 1/4 as the fraction 5/4, then dividing by 3/4, gives 5/3, which is only 1 complete piece. So 7 complete pieces can be cut from the plank.
- (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) 6 + 2√3 — Multiply √3 by each term in the bracket separately. First term: √3 × 2 = 2√3. Second term: √3 × √12 = √(3 × 12) = √36 = 6. Adding the two results in the order they were found, and writing the whole-number term first, gives 6 + 2√3. Adding the numbers under the root for the second term instead of multiplying them (3 + 12 = 15) gives √15 in place of 6, leading to √15 + 2√3. Multiplying √3 by the 2 but never distributing to the √12 term at all leaves just 2√3. Treating √3 × 2 as if the 3 were multiplied by the 2 inside the root, √3 × 2 → √6, while still getting the second term correct, gives 6 + √6.
- (a) 30 cm — The tile's side length must be a common factor of 90 and 120. The factors of 90 include 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90; the factors of 120 include 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120. The highest number common to both lists is 30, so the largest square tile has a side length of 30 cm. Picking 15 cm, a common factor but not the largest, gives tiles that are smaller than necessary. Picking 10 cm, also a common factor but smaller still, wastes even more of the possible tile size. Working out the lowest common multiple instead of the highest common factor gives 360 cm, a length far bigger than either side of the patio. So the largest square tile Ben can use has a side length of 30 cm.
- (a) 5 — Method: each index is worked out first, and subtracting a negative number is the same as adding the positive. Working: (−2)³ = (−2) × (−2) × (−2) = −8 and (−3)² = (−3) × (−3) = 9, while − (−4) becomes + 4, so the calculation becomes −8 + 9 + 4 = 5. Answer: 5. The distractors: −13 comes from taking (−3)² as −9, giving −8 − 9 + 4 = −13; −3 comes from reading − (−4) as − 4, giving −8 + 9 − 4 = −3; 21 comes from treating every power of a negative number as positive, so that (−2)³ is taken as 8 and the calculation becomes 8 + 9 + 4 = 21.
- (c) 23.375 — The error intervals are 7.5 ≤ base < 8.5 and 4.5 ≤ height < 5.5. The upper bound of the area uses the upper bound of both the base and the height, then halves the product: 8.5 × 5.5 ÷ 2 = 23.375 cm². Using the lower bound of both dimensions instead, 7.5 × 4.5 ÷ 2 = 16.875, gives the lower bound of the area rather than the upper one. Multiplying the two upper bounds together but forgetting to halve for the triangle formula, 8.5 × 5.5 = 46.750, treats the triangle as if it were a rectangle. Using the given values directly without applying any bound at all, 8 × 5 ÷ 2 = 20.000, ignores that each rounded measurement has its own range of possible values.
- (b) £45 — Since 2/3 of the amount is £30, one third is £30 ÷ 2 = £15, and the whole amount is three thirds: £15 × 3 = £45. Applying the fraction forwards to £30 instead of reversing it, £30 × 2/3 = £20, treats the given amount as the whole rather than as two thirds of it. Finding one third correctly as £15 but forgetting to multiply by 3 to get the whole amount leaves £15 as the final answer. Reading £30 as one third of the amount rather than as two thirds, and so multiplying straight by 3, gives £30 × 3 = £90.
- (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.
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