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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- (b) 1 : 2 — 2/5 of 50 is 20, so there are 20 red counters and 50 − 20 = 30 blue counters. Taking 5 red counters out leaves 15 red and 30 blue, so red : blue = 15 : 30. Dividing both parts by 15 gives 1 : 2. 2 : 3 is the ratio before any counters are removed, 2 : 1 has the two parts the wrong way round, and 4 : 5 comes from taking the 5 counters out of the blue instead of the red.
- (d) 3/10 — Total parts = 7 + 3 = 10. Gold beads are 3 of those 10 parts, so the fraction is 3/10. 7/10 comes from finding the fraction of red beads instead of gold beads. 7/3 comes from writing the ratio itself as a fraction, without adding the parts to find the total. 3/7 comes from comparing gold beads with red beads instead of with the total number of beads.
- (d) x⁽¹⁄²⁾ — Method: dividing two powers of the same letter subtracts the index of the divisor from the index of the term being divided, and fractional indices are subtracted like any other fractions. Working: 3/4 − 1/4 = 2/4, which simplifies to 1/2, so the result is x⁽¹⁄²⁾. Answer: x⁽¹⁄²⁾. The distractors: x comes from adding the indices, 3/4 + 1/4 = 1, as though the powers were being multiplied; x³ comes from dividing the indices, so that 3/4 divided by 1/4 gives 3; x⁽³⁄¹⁶⁾ comes from multiplying the indices, 3/4 × 1/4.
- (a) 3.84 × 10⁵ — 384,000 = 3.84 × 100,000 = 3.84 × 10⁵, with the decimal point moved five places and the coefficient kept between 1 and 10. Moving the point six places instead of five gives 3.84 × 10⁶, ten times too large. Leaving the coefficient as 38.4 gives 38.4 × 10⁴, which is not between 1 and 10. Using a negative exponent instead of a positive one gives 3.84 × 10⁻⁵, a number far smaller than 1.
- (c) 0.001 — Method: a negative index means 'one over' the positive power, so $10^{-3}$ means one over $10^{3}$. Working: ten cubed is 1000, and one over 1000 is 0.001. 1000 comes from ignoring the negative sign and working out ten cubed instead of its reciprocal. −1000 comes from ignoring what the negative index does to the size, while still writing a negative sign on the large value. −0.001 comes from correctly finding the size, 0.001, but wrongly keeping a negative sign because the index was negative. Answer: 0.001.
- (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.
- (a) 300 N/m² — Pressure is force ÷ area, and the unit N/m² says so: newtons divided by square metres. The calculation is 240 ÷ 0.8. Multiplying both numbers by 10 clears the decimal: 2400 ÷ 8 = 300 N/m². 192 N/m² multiplies the force by the area instead of dividing, 30 N/m² divides by 8 and so loses the decimal point, and 3000 N/m² multiplies only the force by 10 when clearing the decimal.
- (b) 16π — Use the circumference formula C = 2πr to find the radius: 8π = 2πr, so r = 8π ÷ 2π = 4 cm. Then use the area formula A = πr²: A = π × 4² = 16π cm². Using C = πr instead of C = 2πr gives r = 8, and squaring that gives 64π. Finding r = 4 correctly but then substituting it back into the circumference formula instead of the area formula gives 2π × 4 = 8π. Finding r = 4 correctly but forgetting to square it in the area formula, using A = πr instead of A = πr², gives 4π.
- (b) 14 — Method: the greatest number of identical rows is the highest common factor of the two bulb totals, found by taking every prime factor the two totals share. Working: 42 = 2 × 3 × 7 and 56 = 2 × 2 × 2 × 7, so the prime factors common to both are 2 and 7, giving a highest common factor of 2 × 7 = 14. 2 comes from taking only the common factor 2 and forgetting the common factor 7. 7 comes from taking only the common factor 7 and forgetting the common factor 2. 168 is the lowest common multiple of 42 and 56, not their highest common factor. Answer: 14.
- (a) 8 — To undo 'multiply by 4, then add 8', reverse the operations in reverse order: subtract 8 first, then divide by 4. 40 − 8 = 32, and 32 ÷ 4 = 8, so the number is 8. A candidate who added 8 again instead of subtracting worked out 40 + 8 = 48, then 48 ÷ 4 = 12. A candidate who divided before subtracting, doing the inverse operations in the wrong order, worked out 40 ÷ 4 = 10, then 10 − 8 = 2. A candidate who multiplied instead of dividing at the last step worked out (40 − 8) × 4 = 32 × 4 = 128.
- (d) −1.7 — Since the numbers have different signs, find the difference between their sizes: 4.5 − 2.8 = 1.7, then keep the sign of the number further from zero. So −4.5 + 2.8 = −1.7. A candidate who drops the negative sign gets 1.7. A candidate who adds the magnitudes instead of finding the difference gets −(4.5 + 2.8) = −7.3. A candidate who takes the smaller digit from the larger in the tenths column, doing 8 − 5 = 3 instead of borrowing to make 15 − 8 = 7, gets 2.3 and so −2.3.
- (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.
- (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.
- (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.
- (d) 636 g — Let x = 0.636363... . Since two digits repeat, multiply by 100: 100x = 63.636363... . Subtracting removes the recurring part exactly: 100x − x = 63.636363... − 0.636363... = 63, so 99x = 63, giving x = 63/99 = 7/11 kg. Converting to grams: 7/11 × 1000 = 7000/11 = 636.3636... g, which rounds to 636 g. Treating the decimal as if it terminated, writing 0.63 as 63/100 kg, gives 630 g when multiplied by 1000 — this drops the recurring part entirely. Subtracting 10x instead of x, using 100x − 10x = 90x = 63, is the wrong power of ten for a two-digit block, giving x = 63/90 = 7/10 kg, which is 700 g. A numerator slip in the subtraction, 63 − 1 = 62 instead of 63, gives x = 62/99 kg, which is 62000/99 = 626.26... g, rounding to 626 g.
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