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.
Non-calculator
Answer key: Number worksheet — GCSE Higher
MathsUKwww.geekhero.co.uk
- (c) 1/x⁶ — Method: raising a power to another power multiplies the two indices, and a negative index means one over the matching positive power. Working: −2 × 3 = −6, so (x⁻²)³ = x⁻⁶, and x⁻⁶ written as a fraction is 1/x⁶. Answer: 1/x⁶. The distractors: x⁶ comes from multiplying the indices correctly but dropping the minus sign; 1/x⁵ comes from adding the sizes of the indices, 2 + 3, instead of multiplying them; −x⁶ comes from reading the negative index as a minus sign in front of the whole term.
- (a) 1/5 — Work out the bracket first: 2 + 3 = 5. The reciprocal of 5 is 1/5. A candidate who forgot to take the reciprocal and just gave the value of the bracket wrote 5. A candidate who took the reciprocal but made a sign error wrote −1/5. A candidate who found the reciprocal of each number separately and added them, treating reciprocal as if it distributes over addition, worked out 1/2 + 1/3 = 5/6.
- (c) 31 — Method: work out each power separately, remembering that any non-zero base raised to the power 0 is 1 and a base raised to the power 1 is itself, then add the three values. Working: 5⁰ = 1, 5¹ = 5 and 5² = 25, so the total is 1 + 5 + 25 = 31. Answer: 31. The distractors: 30 comes from taking 5⁰ as 0 instead of 1; 35 comes from taking 5⁰ as 5, treating a zero index as leaving the base unchanged; 125 comes from adding the indices first, as though the three terms were being multiplied, and working out 5³.
- (a) 0.0069 — Leading zeros are not significant, so the significant figures in 0.006852 start at 6: 6, 8, 5, 2. Rounding to 2 significant figures means keeping 6 and 8, and looking at the next digit, 5, to decide whether to round up. Since 5 rounds up, the second significant figure increases from 8 to 9: 0.006852 rounds to 0.0069. A candidate who rounded to 1 significant figure instead of 2 wrote 0.007. A candidate who rounded to 3 significant figures instead of 2 wrote 0.00685. A candidate who did not round up despite the next digit being 5 wrote 0.0068.
- (c) 24 — Method: round each number to the nearest whole number, then multiply the rounded values. Working: 6.4 rounds to 6 (nearest whole number) and 3.9 rounds to 4 (nearest whole number). 6 × 4 = 24. Answer: 24. 18 comes from rounding 3.9 down to 3 instead of up to the nearest whole number, 4, giving 6 × 3. 28 comes from rounding 6.4 up to 7 instead of down to the nearest whole number, 6, giving 7 × 4. 25 is the exact value of 6.4 × 3.9, which is 24.96, rounded to the nearest whole number after multiplying, rather than estimated by rounding first.
- (c) 45 — Method: count the ordered pairings with the product rule and then correct for the fact that a game between two players is the same game whichever player it is counted from. Working: each of the 10 players meets 9 opponents, so 10 × 9 = 90 pairings are counted; every game has been counted twice, once from each player's side, so the number of games is 90 ÷ 2 = 45. Answer: 45. The distractors: 90 comes from stopping at 10 × 9 and never halving, so that each game is counted once for each of its two players; 55 comes from adding 10 + 9 + 8 + ... + 1 instead of 9 + 8 + ... + 1, which counts one extra round of games; 20 comes from multiplying the 10 players by the 2 players in each game rather than pairing the players with one another.
- (a) 40 — Method: a number is odd exactly when its units digit is odd, so the restricted position is filled first and the two free positions are then filled from the digits that are left, multiplying the number of choices at each stage. Working: of the six digits only 3 and 9 are odd, so there are 2 choices for the units digit; once that digit has been used, 5 digits remain for the hundreds position and then 4 remain for the tens position, so the count is 2 × 5 × 4 = 40. Answer: 40. The distractors: 120 comes from ignoring the word odd altogether and counting every three-digit number that can be made from the six digits, 6 × 5 × 4; 60 comes from filling the hundreds and tens positions first, 6 then 5, and only then allowing 2 odd digits for the units position, which overcounts because one of 3 and 9 may already have been used, giving 6 × 5 × 2; 72 comes from restricting the units digit to 3 or 9 correctly but overlooking the condition that no digit may be used twice, so all six digits are still counted as available for each of the other two positions, giving 2 × 6 × 6.
- (d) 6 — By Pythagoras' theorem, the square of the hypotenuse equals the sum of the squares of the other two sides: (√12)² + (√24)² = 12 + 24 = 36. The hypotenuse is √36 = 6 cm. Adding the two side lengths directly instead of squaring them first, treating the theorem as if it were a straight sum of the sides, gives √12 + √24 = 2√3 + 2√6. Multiplying the two squared values, 12 × 24 = 288, instead of adding them, then taking the root, gives √288 = 12√2. Adding the squares correctly to get 36 but forgetting to take the square root at the end leaves 36 as the answer instead of the hypotenuse itself.
- (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.
- (c) £7.25 — Find the total cost of the books first: 3 × 4.25 = 12.75, so the books cost £12.75 in total. Subtract this from the £20 note: 20.00 − 12.75 = 7.25, so the change is £7.25. Stopping after finding the cost and not subtracting it from £20 gives £12.75, which is the amount spent, not the change. Borrowing correctly in the pence column but forgetting to reduce the pounds column by 1 gives £8.25 instead of £7.25. Multiplying 3 × 4.25 as 12.25 instead of 12.75, a multiplication slip, makes the change come out £0.50 too high, at £7.75. So Jack receives £7.25 change.
- (a) 36 — Method: for the highest common factor, take the LOWER power of each prime that appears in both numbers. Working: for 2, the lower power is 2² (from 108); for 3, the lower power is 3² (from 72), so the highest common factor is 2² × 3² = 4 × 9 = 36. 216 comes from taking the higher power of each prime instead, 2³ × 3³ = 8 × 27 = 216, which gives the lowest common multiple, not the highest common factor. 108 is simply one of the two numbers, not their highest common factor. 6 comes from multiplying the primes without any powers at all, 2 × 3 = 6. Answer: 36.
- (b) 5/27 — Method: multiply the numerators together and the denominators together, then simplify. Working: (5 × 2)/(6 × 9) = 10/54 = 5/27. Answer: 5/27. 7/15 comes from adding the fractions instead of multiplying: (5+2)/(6+9) = 7/15. 15/4 comes from flipping the second fraction, as if dividing: (5 × 9)/(6 × 2) = 45/12 = 15/4. 5/3 comes from cancelling the two denominators against each other, dividing both 6 and 9 by 3 to leave 5/2 × 2/3 = 10/6 = 5/3; cancelling is only valid between a numerator and a denominator, never between two denominators.
- (c) 20 — Method: work out each power separately before subtracting. Working: 6² = 36 and 4² = 16, so 6² − 4² = 36 − 16 = 20. Answer: 20. (4 comes from subtracting first, 6 − 4 = 2, and then squaring that result, instead of squaring each number first. 52 comes from adding the two squares, 36 + 16, instead of subtracting them. 2 comes from subtracting the two numbers, 6 − 4, and forgetting to square at all.)
- (c) 1/4 — A negative index means the reciprocal of the positive power, so 2⁻² = 1 ÷ 2² = 1/4. Treating the negative sign as making the answer negative instead gives −(2²) = −4. Ignoring the negative sign altogether gives just 2² = 4. Finding the reciprocal correctly but then also applying a negative sign gives −1/4.
- (b) 12 — Method: round the number of pupils to the nearest 10, divide by the number of passengers each minibus can carry, then round up because a part-full minibus still needs a whole vehicle. Working: 179 rounds to 180 (nearest 10); 180 ÷ 16 = 11.25; 11 minibuses only carry 176 passengers, so a 12th minibus is needed for the rest. Answer: 12. 11 comes from rounding 11.25 to the nearest whole number in the usual way, without checking that the leftover pupils still need transporting. 10 comes from rounding 179 down to 170 instead of to the nearest 10, 180. 180 comes from stopping after rounding the number of pupils, without dividing by the number of passengers each minibus carries at all.
Build your own mix at the worksheet builder.