Printable · GCSE Foundation · ages 14-16
Number worksheet — GCSE Foundation
Fifteen questions across the number statements at Foundation 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 Foundation
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- (d) 7/20 — Method: write the decimal over the power of ten that matches the number of digits after the point, then divide the numerator and the denominator by their highest common factor. Working: 0.35 has two digits after the point, so it is 35 hundredths and can be written as 35/100; the highest common factor of 35 and 100 is 5, and 35 ÷ 5 = 7 with 100 ÷ 5 = 20. Answer: 7/20. The distractors: 3/10 comes from reading only the first digit after the point and converting 0.3; 7/25 comes from dividing the numerator by 5 but the denominator by 4, using a different factor on the top and on the bottom; 35/10 comes from counting one decimal place instead of two and writing the digits over 10.
- (b) 8/5 — The reciprocal of a fraction a/b is b/a — swap the numerator and denominator. So the reciprocal of 5/8 is 8/5. A candidate who wrote 5/8 gave the original fraction, not its reciprocal. A candidate who worked out 1 − 5/8 = 3/8 confused 'reciprocal' with subtracting from 1. A candidate who wrote −8/5 inverted the fraction correctly but introduced an unnecessary sign change.
- (b) (2 + 3) × 5 − 1 — 2 + 3 = 5, then 5 × 5 = 25, then 25 − 1 = 24, so the brackets belong around 2 + 3. Placing them around 5 − 1 instead gives 5 − 1 = 4, then 3 × 4 = 12, then 2 + 12 = 14. Leaving the multiplication bracketed instead changes nothing, because it already had priority: 3 × 5 = 15, then 2 + 15 = 17, then 17 − 1 = 16. Bracketing both 2 + 3 and 5 − 1 uses two pairs instead of the one asked for: 2 + 3 = 5, 5 − 1 = 4, then 5 × 4 = 20.
- (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) 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.
- (b) 30 — Method: the smallest matching total is the lowest common multiple of the two pack sizes. Working: multiples of 6 are 6, 12, 18, 24, 30 …; multiples of 10 are 10, 20, 30 …. The lowest common multiple is 30. 60 comes from working out 6 × 10 = 60, the product of the pack sizes rather than their lowest common multiple. 16 comes from working out 6 + 10 = 16, which is not a common multiple at all. 2 is the highest common factor of 6 and 10, not a total of tickets. Answer: 30.
- (a) 4,000 nanometres — Method: the number of nanometres is the diameter divided by the length of one nanometre, and dividing powers of ten means subtracting the indices. Working: −6 − (−9) = 3, so 10⁻⁶ ÷ 10⁻⁹ = 10³, and the diameter is 4 × 10³ nanometres. Answer: 4,000 nanometres. The distractors: 400 nanometres comes from taking the difference between the indices as 2 instead of 3; 4 nanometres comes from changing the name of the unit without converting, leaving the coefficient untouched; 0.004 nanometres comes from dividing by 10³ instead of multiplying by it, as though a nanometre were the larger of the two units.
- (b) 1 — Method: BIDMAS deals with the index before the subtraction, and a negative number multiplied by itself gives a positive result. Working: (−2)² = (−2) × (−2) = 4, so the calculation becomes 4 − 3 = 1. Answer: 1. The distractors: −7 comes from squaring only the 2 and leaving the minus sign outside the index, giving −(2²) − 3 = −4 − 3 = −7; −1 comes from subtracting the square from 3 instead of 3 from the square, giving 3 − 4 = −1; 25 comes from carrying out the subtraction before the index, giving (−2 − 3)² = (−5)² = 25.
- (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) 2 — Method: the cube root of a number is the value that multiplies by itself three times to give that number. Working: 2 × 2 × 2 = 8, so ∛8 = 2. 4 comes from working out 8 ÷ 2 = 4, halving the number instead of finding its cube root. 24 comes from working out 8 × 3 = 24, multiplying by 3 instead of cube-rooting it. 64 is 8², the square of 8, not its cube root. Answer: 2.
- (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.
- (b) 9 g/cm³ — Method: density = mass ÷ volume. Working: 540 ÷ 60 = 9. Answer: 9 g/cm³. (0.11 g/cm³ comes from dividing the volume by the mass instead of the mass by the volume. 480 g/cm³ comes from subtracting the volume from the mass instead of dividing. 32400 g/cm³ comes from multiplying the mass by the volume instead of dividing.)
- (d) 2³ × 5² — Method: divide repeatedly by the smallest prime number, then write any repeated prime using a power. Working: 200 ÷ 2 = 100, 100 ÷ 2 = 50, 50 ÷ 2 = 25, 25 ÷ 5 = 5, and 5 is prime, so 200 = 2 × 2 × 2 × 5 × 5, written as 2³ × 5². 2² × 5³ swaps the two powers, giving 4 × 125 = 500, not 200. 2³ × 5 leaves out one of the two 5s, giving 8 × 5 = 40, not 200. 2 × 5³ leaves out two of the three 2s, giving 2 × 125 = 250, not 200. Answer: 2³ × 5².
- (d) x⁴ — Method: dividing two powers of the same letter subtracts the index of the divisor from the index of the term being divided. Working: six factors of x on the top and two on the bottom cancel in pairs, leaving 6 − 2 = 4 factors of x. Answer: x⁴. The distractors: x³ comes from dividing the indices, 6 ÷ 2, instead of subtracting them; x⁸ comes from adding the indices, 6 + 2, as though the powers were being multiplied; x¹² comes from multiplying the indices, 6 × 2, as though a power were being raised to a power.
- (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.
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