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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Number worksheet — GCSE Higher
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- 1.A circle has a circumference of 8π cm. Work out the exact area of the circle in terms of π.
- 2.Write 200 as a product of its prime factors, using index notation.
- 3.A semicircle has a diameter of 8 cm. Work out the exact area of the semicircle, in terms of π.
- 4.Write 0.875 as a fraction in its simplest form.
- 5.At a swimming pool the ratio of adults to children is 3 : 5. Of the children, 2/5 are boys. What fraction of the people at the pool are boys?
- 6.A restaurant adds a service charge of 15% to a bill of £120. Work out the service charge.
- 7.A quarter-circle has a radius of 6 cm. Work out the exact perimeter of the quarter-circle, giving your answer in terms of π.
- 8.The number 36 can be written as 2² × 3², and the number 84 can be written as 2² × 3 × 7. Work out the highest common factor of 36 and 84.
- 9.Find the missing number: ▢ ÷ 15 = 24
- 10.Work out (5 + 2) × 3²
- 11.A carton of orange juice holds 1.35 litres. Ruby pours the juice equally into 4 identical glasses. Work out how much juice is in each glass, giving your answer as a fraction of a litre in its simplest form.
- 12.Work out (4 × 10⁻³) × (2 × 10⁵). Give your answer in standard form.
- 13.By considering fourth powers, work out which two consecutive integers ⁴√200 lies between.
- 14.Work out 2 × 3 × 5 + 1 and decide whether the result is a prime number.
- 15.Write these decimals in order, starting with the smallest: 0.6, 0.55, 0.601, 0.5, 0.56
Answer key
- (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.
- (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.
- (b) 8π cm² — A diameter of 8 cm gives a radius of 4 cm. The area of a full circle would be π × r² = π × 4² = 16π cm², and a semicircle is exactly half of this, giving 16π ÷ 2 = 8π cm². Forgetting to halve the area for the semicircle gives 16π cm², the area of the whole circle. Halving the diameter twice, using a radius of 2 instead of 4, gives π × 2² = 4π cm². Using the diameter itself as the radius, so π × 8² = 64π, and then halving that for the semicircle gives 32π cm².
- (a) 7/8 — 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: 0.875 has three digits after the point, so it is 875 thousandths and can be written as 875/1000; the highest common factor of 875 and 1000 is 125, and 875 ÷ 125 = 7 with 1000 ÷ 125 = 8. Answer: 7/8. The distractors: 8/7 comes from cancelling correctly but writing the two parts the wrong way round; 9/10 comes from rounding 0.875 to one decimal place as 0.9 before converting; 7/80 comes from counting four decimal places instead of three and using a denominator of 10000, giving 875/10000.
- (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) £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) (12 + 3π) cm — The perimeter of a quarter-circle is made up of two straight radii plus a quarter of the circumference. The two radii give 2 × 6 = 12 cm, and a quarter of the circumference is (1/4) × 2 × π × 6 = 3π cm, so the total perimeter is (12 + 3π) cm. Giving only the curved part, 3π cm, forgets the two straight edges entirely. Using the full circumference, 2 × π × 6 = 12π, instead of a quarter of it gives (12 + 12π) cm. Including only one radius instead of two gives (6 + 3π) cm.
- (a) 12 — Compare the powers of each prime that appears in both factorisations. In 2² × 3² and 2² × 3 × 7, the prime 2 appears with power 2 in both, and the prime 3 appears with power 2 in one and only power 1 in the other — take the lower power, 3¹. Multiplying the shared primes at their lower powers, 2² × 3, gives 12. Using power 1 for both primes instead of comparing the powers properly, 2 × 3, gives 6, which misses that 2 is common at power 2, not power 1. Multiplying the primes at their higher powers and including 7, which only appears in 84, gives 2² × 3² × 7, which comes to 252 — this is the lowest common multiple, not the highest common factor. Only spotting that 3 is a common prime and overlooking that 2 is common as well gives 3. So the highest common factor of 36 and 84 is 12.
- (a) 360 — The inverse of ÷ 15 is × 15, so the missing number is 24 × 15 = 360. Subtracting instead of multiplying gives 24 − 15 = 9. Dividing by 15 again instead of multiplying gives 24 ÷ 15 = 1.6. Adding instead of multiplying gives 24 + 15 = 39.
- (b) 63 — 5 + 2 = 7, then 3² = 9, then 7 × 9 = 63. Ignoring the brackets and applying BIDMAS as if the expression were unbracketed gives 3² = 9, then 2 × 9 = 18, then 5 + 18 = 23. Squaring the bracket instead of the 3 gives 7² = 49, then 49 × 3 = 147 — the power belongs to the 3 alone. Multiplying by 3 before squaring the whole product gives 7 × 3 = 21, then 21² = 441.
- (d) 27/80 — Method: write the total as a fraction of a litre, then divide by the number of glasses. Working: 1.35 = 27/20, so each glass holds 27/20 ÷ 4 = 27/80 of a litre. Answer: 27/80. 27/20 comes from converting the total correctly to a fraction but forgetting to divide by the number of glasses. 27/5 comes from multiplying the total by 4 instead of dividing. 17/50 comes from rounding 1.35 ÷ 4 to 0.34 before converting to a fraction.
- (a) 8 × 10² — Method: to multiply numbers written in standard form, multiply the coefficients and add the indices. Working: 4 × 2 = 8 for the coefficients, and −3 + 5 = 2 for the indices; 8 already lies between 1 and 10, so no adjustment is needed. Answer: 8 × 10². The distractors: 6 × 10² comes from adding the coefficients, 4 + 2, instead of multiplying them; 8 × 10⁸ comes from ignoring the minus sign and adding 3 + 5; 8 × 10⁻¹⁵ comes from multiplying the indices, −3 × 5, instead of adding them.
- (c) 3 and 4 — Find the two consecutive fourth powers either side of 200: 3⁴ = 81 and 4⁴ = 256. Since 81 < 200 < 256, ⁴√200 lies between 3 and 4. Answering 14 and 15 comes from taking a square root instead of a fourth root: √200 ≈ 14.14, which does lie between 14 and 15, but that is not the root the question asks for. Answering 5 and 6 comes from taking a cube root instead of a fourth root: ∛200 ≈ 5.85, which lies between 5 and 6. Answering 4 and 5 comes from working 4⁴ as though it were 4 × 4 × 4 = 64 and stopping a factor short, then concluding that 200 is already past 4⁴ and the root must be above 4.
- (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.
- (a) 0.5, 0.55, 0.56, 0.6, 0.601 — Compare the decimals by giving them all the same number of decimal places first: 0.600, 0.550, 0.601, 0.500, 0.560. Ordering these from smallest to largest gives 0.500, 0.550, 0.560, 0.600, 0.601, which is 0.5, 0.55, 0.56, 0.6, 0.601. Comparing the digits as though they were whole numbers, reading 0.601 as "601" and 0.5 as "5", without padding to the same number of decimal places, gives the wrong order 0.5, 0.6, 0.55, 0.56, 0.601, because it ignores the place value of each digit. Ordering largest to smallest instead of smallest to largest, as the question asks, gives 0.601, 0.6, 0.56, 0.55, 0.5. Misreading the close values 0.55 and 0.56 and swapping them gives 0.5, 0.56, 0.55, 0.6, 0.601. So the correct order, smallest to largest, is 0.5, 0.55, 0.56, 0.6, 0.601.
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