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.
Number worksheet — GCSE Higher
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- 1.£60 is shared between Amir, Bo and Chen in the ratio 1:2:3. Work out the fraction of the £60 that Amir and Bo receive together.
- 2.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.
- 3.Write 5/6 as a decimal, showing clearly which digit recurs.
- 4.Priya has 24 red beads and 36 blue beads. She makes identical bracelets, using every bead and with none left over. Work out the greatest number of bracelets she can make.
- 5.The thickness of a sheet of card is 0.02384 cm. Write this thickness correct to 2 significant figures.
- 6.Work out (−2)² − 3
- 7.Nadia has £84. She spends 3/7 of it on a jacket and the rest on a bag. Work out the ratio of the amount spent on the jacket to the amount spent on the bag, in its simplest form.
- 8.Leah measures the length of her classroom with a tape measure marked in centimetres. She writes the length down as 7.3157 m. Give a reason why this is not an appropriate degree of accuracy.
- 9.Work out the value of .
- 10.A plank has length a = 12 cm and a second plank has length b = 7 cm, each correct to the nearest centimetre. Work out the upper bound of a + b.
- 11.p = 24 and q = 6, each correct to the nearest integer. Work out the greatest degree of accuracy to which p ÷ q can be guaranteed correct.
- 12.Sam rounds 0.070268 to 3 significant figures and writes 0.0703. Which of these statements is correct?
- 13.A pair of trainers is priced at £80 online. The shop takes 25% off the price and then adds £10 for next-day delivery. Work out the total cost.
- 14.Simplify 2³ × 2⁴, giving your answer as a single power of 2.
- 15.A bag of flour is labelled 1.5 kg, correct to the nearest 0.1 kg. The true mass of the flour is m kg. Which inequality gives all the possible values of m?
Answer key
- (a) 1/2 — The ratio 1:2:3 has 1 + 2 + 3 = 6 parts in total. Amir and Bo together receive 1 + 2 = 3 of those parts, so together they receive 3/6 of the £60, which simplifies to 1/2. Using only Amir's single part, 1/6, ignores Bo's share entirely. Adding Bo's and Chen's parts instead of Amir's and Bo's, 2 + 3 = 5, gives 5/6. Comparing Amir and Bo's 3 parts to Chen's 3 parts, rather than to the total of 6 parts, gives 3/3 = 1.
- (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.
- (b) 0.83333... — Divide 5 by 6 using long division. 5.000... ÷ 6: 50 ÷ 6 = 8 remainder 2, giving the first decimal digit 8. Bring down a 0 to make 20, and 20 ÷ 6 = 3 remainder 2 — the remainder 2 has reappeared, so from here the digit 3 repeats forever. This gives 5/6 = 0.83333... . Stopping after two decimal places and writing 0.83 treats the division as if it terminated, when the remainder never reaches zero. Shifting the decimal point one place too far to the left gives 0.083333..., the same digits divided by an extra power of ten. A slip in the long division itself, misreading a remainder, can produce the wrong repeating digit, 0.85555... .
- (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) 0.024 cm — Method: significant figures are counted from the first non-zero digit; the zeros in front of it only fix the place value and are not significant. Working: in 0.02384 the first significant figure is 2 and the second is 3, so the rounding is decided by the next digit, 8. As 8 is 5 or more, the second significant figure goes up from 3 to 4, in the same place value. Answer: 0.024 cm. The distractors: 0.023 cm comes from chopping the 84 off instead of rounding it; 0.02 cm comes from counting the leading zeros as significant figures, so the 2 is taken as the second figure and the rounding stops there; 0.0238 cm is 0.02384 correct to 3 significant figures, one figure too many.
- (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.
- (c) 3:4 — The jacket costs 3/7 of £84, which is 3 × (84 ÷ 7) = 3 × 12 = £36. The bag then costs the rest of the money, £84 − £36 = £48. The ratio of the jacket to the bag is 36:48, which simplifies to 3:4. Writing the fraction spent on the jacket, 3/7, directly as the ratio, without working out that the bag's share is the remaining 4/7, gives 3:7. Giving the ratio the wrong way round, bag to jacket instead of jacket to bag, gives 4:3. Assuming the jacket and bag cost the same, ignoring the fraction given, gives 1:1.
- (a) The tape can only give the length to the nearest centimetre — Method: a measurement should never be written to a finer degree of accuracy than the instrument used can read. Working: the tape is marked in centimetres, so the smallest division Leah can read is 1 cm, which is 0.01 m and two decimal places in metres; writing 7.3157 m claims the length to the nearest tenth of a millimetre, four decimal places, which the markings cannot support. A record of 7.32 m, to the nearest centimetre, is what this tape justifies. Answer: The tape can only give the length to the nearest centimetre. The distractors: the nearest millimetre contradicts the markings described in the question, which are centimetres, and would still claim more accuracy than the tape offers; the rule that a length in metres must be written to 2 decimal places borrows the habit of writing money to the penny, when the accuracy of a length depends on the instrument; rounding to the nearest metre would throw away accuracy the tape genuinely provides.
- (d) 1/4 — Method: deal with the fractional index first, then the negative sign. Working: $8^{2/3} = (\sqrt[3]{8})^2 = 2^2 = 4$. A negative index means take the reciprocal of that result, so $8^{-2/3} = \frac{1}{8^{2/3}} = \frac{1}{4}$. Answer: 1/4. A candidate who evaluates $8^{2/3}$ correctly but forgets the negative sign entirely gets 4 — they have dropped the instruction to take a reciprocal. A candidate who takes the reciprocal step but applies it as a sign change to the finished number instead of inverting it gets −4. A candidate who multiplies 8 by −2/3, treating the index as an ordinary factor rather than a power, gets −16/3.
- (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) the nearest whole number — The error intervals are 23.5 ≤ p < 24.5 and 5.5 ≤ q < 6.5. The minimum of p ÷ q is 23.5 ÷ 6.5 ≈ 3.615, and the maximum is 24.5 ÷ 5.5 ≈ 4.455. Both of these round to 4 at the nearest whole number, so the answer is guaranteed correct to the nearest whole number — but not to the nearest 0.1, since 3.615 rounds to 3.6 while 4.455 rounds to 4.5, which do not agree. Claiming the nearest 0.1 assumes every figure a calculator shows is trustworthy, without checking whether the bounds actually agree that far. Claiming only the nearest 10 badly understates how much can be guaranteed here, since both bounds already round to 4, not merely to 0. Saying no degree of accuracy can be guaranteed gives up before checking whether the bounds agree at any level at all.
- (c) Sam is correct — The leading zeros in 0.070268 are not significant, so the first three significant figures are 7, 0 and 2. The next digit along is 6, and since 6 is 5 or more, the third significant figure rounds up from 2 to 3, giving 0.0703. This means Sam's answer is correct. Writing 0.070 keeps only 2 significant figures, one short of what was asked. Writing 0.0702 ignores the digit 6 that follows and leaves the third figure unrounded. Writing 0.0704 rounds the third figure up twice, as if a later digit had also pushed it up.
- (d) £70 — Method: the discount is a percentage of the original price only, so work it out, subtract it, and add the fixed delivery charge afterwards. Working: 25% of £80 is £80 ÷ 4 = £20, so the discounted price is £80 − £20 = £60, and the total is £60 + £10 = £70. Answer: £70. The distractors: £60 comes from working out the discounted price and stopping there, leaving the delivery charge out of the total; £65 comes from taking £25 off the price instead of 25% of it, giving £80 − £25 = £55 and then £55 + £10 = £65; £67.50 comes from adding the delivery charge before the discount and reducing the whole amount, giving 75% of £90 = £67.50.
- (d) 2⁷ — When multiplying powers of the same base, the indices add: 3 + 4 = 7, so 2³ × 2⁴ = 2⁷. Multiplying the indices instead of adding them gives 3 × 4 = 12, so 2¹². Subtracting the indices instead of adding them gives 4 − 3 = 1, so 2¹. Multiplying the bases together as well as adding the indices gives 2 × 2 = 4, so 4⁷.
- (b) 1.45 ≤ m < 1.55 — The flour's mass is labelled 1.5 kg, correct to the nearest 0.1 kg, so half of 0.1 kg is added to and subtracted from 1.5 kg to find the interval: 1.5 − 0.05 = 1.45 and 1.5 + 0.05 = 1.55, giving 1.45 ≤ m < 1.55. '1.4 ≤ m < 1.6' comes from taking the whole 0.1 kg as the margin either side, instead of half of it. '1.45 < m ≤ 1.55' comes from writing the inequality signs the wrong way round — the lower bound should be included and the upper bound excluded, not the other way round. '1.45 ≤ m ≤ 1.55' comes from including the upper bound, when the convention is that the upper bound is never actually reached.
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