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
- (d) £4.80 — Method: find the error interval, then check which value falls outside it. Working: half of 20p is 10p, so the actual cost, c, satisfies £4.50 ≤ c < £4.70. £4.80 is above £4.70, so it could not be the actual cost. Answer: £4.80. (£4.50 is a genuine possible cost — it sits at the included lower boundary. £4.65 is a genuine possible cost, below the £4.70 upper boundary. £4.55 is a genuine possible cost, well inside the interval.)
- (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.
- (c) −1 — Method: find each cube root separately, keeping its sign, and then add the two results. Working: (−3) × (−3) × (−3) = −27, so ∛(−27) = −3, and 2 × 2 × 2 = 8, so ∛8 = 2. Adding gives −3 + 2 = −1. Answer: −1. The distractors: 5 comes from taking the cube root of a negative number as positive, giving 3 + 2; −5 comes from reading the minus sign as applying to the whole sum and working out −(3 + 2); −6 comes from multiplying the two roots, −3 × 2, instead of adding them.
- (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.
- (a) 36 — Method: find the lowest common multiple of 6 and 9, then move up the list of common multiples until one is greater than 20. Working: the common multiples of 6 and 9 are 18, 36, 54 …. 18 is not greater than 20, so the next one, 36, is the smallest value of n that is greater than 20. 18 is the lowest common multiple itself, but it fails the 'greater than 20' condition. 54 is the common multiple after 36, one step too far. 27 is a multiple of 9 but not of 6, since 27 ÷ 6 is not a whole number. Answer: 36.
- (a) 8 × 10³ — Method: divide the capacity of the card by the size of one photograph, dividing the coefficients and subtracting the indices, then bring the coefficient back into the range 1 to 10. Working: 3.2 ÷ 4 = 0.8 and 10 − 6 = 4, which gives 0.8 × 10⁴; a coefficient of 0.8 is smaller than 1, so the decimal point moves one place to the right and the index falls by 1. Answer: 8 × 10³. The distractors: 8 × 10⁴ comes from correcting 0.8 to 8 without reducing the index, which makes the answer ten times too large; 1.28 × 10¹⁷ comes from multiplying the two numbers instead of dividing them, since 3.2 × 4 = 12.8 and 10 + 6 = 16; 8 × 10¹⁵ comes from dividing the coefficients but adding the indices instead of subtracting them.
- (d) 4/3 — Method: the product of two negative numbers is positive, so work with 2/5 × 10/3 and then simplify. Multiply the numerators together and the denominators together. Working: 2 × 10 = 20 and 5 × 3 = 15, giving 20/15; both 20 and 15 divide by 5, so 20/15 = 4/3. Answer: 4/3. The distractors: −4/3 has the arithmetic right but keeps a minus sign, from treating negative × negative as negative; 3/25 comes from turning the second fraction upside down and multiplying, which divides instead of multiplying and gives 2/5 × 3/10 = 6/50; −56/15 comes from adding the two fractions instead of multiplying them, giving −6/15 − 50/15.
- (c) −28 — 6² = 36 and 4³ = 64. Work out 36 − 64 = −28. A candidate who subtracts in the wrong order gets 64 − 36 = 28. A candidate who adds instead of subtracting gets 36 + 64 = 100. A candidate who multiplies the base by the exponent instead of raising the power (6 × 2 − 4 × 3 = 12 − 12) gets 0.
- (c) 4.6 — Since 100 lies between 64 and 125, ∛100 lies between 4 and 5. Narrow it down: 4.6³ = 97.336, which is less than 100, so ∛100 is greater than 4.6. To decide how it rounds to 1 decimal place, test the midpoint: 4.65³ = 100.544, which is more than 100, so ∛100 is less than 4.65 and therefore rounds down to 4.6. Simply taking the midpoint of 4 and 5 without testing any cube gives 4.5. Going up to the next tenth because 4.6³ fell short of 100, without checking that 4.65³ already overshoots, gives 4.7. Comparing 100 with the two given cubes, 64 and 125, noticing that 100 is nearer to 125, and rounding straight to the nearest whole number gives 5.0 — but that comparison is between the cubes, not between the cube roots, and cubing stretches the gaps unevenly, so it says nothing about which value the cube root rounds to.
- (d) 60 mph — Average speed = distance ÷ time, with the time measured in hours. 45 minutes is 45/60 of an hour, which is 0.75 of an hour, so the journey takes 1.75 hours. Speed = 105 ÷ 1.75 = 60 mph. 52.5 mph rounds the time up to 2 hours, 72.4 mph writes 1 hour 45 minutes as 1.45 hours, and 183.75 mph multiplies the distance by the time instead of dividing.
- (a) They cannot both be describing the same path — Jon's measurement means the true length, l, satisfies 11.5 m ≤ l < 12.5 m. Mia's measurement means the true length satisfies 12.55 m ≤ l < 12.65 m. These two ranges do not overlap, so the two measurements cannot both be describing the same path. 'They must both be describing the same path' ignores that the two ranges do not overlap at all. 'Jon's measurement must be wrong' wrongly assumes Jon is the one at fault, when the mismatch does not show which measurement, if either, is wrong. 'Mia's measurement must be wrong' makes the same unjustified assumption in the other direction.
- (b) 4 — Method: the cube root of a number is the value that multiplies by itself three times to give that number. Working: 4 × 4 × 4 = 64, so ∛64 = 4. Answer: 4. (8 comes from finding the square root of 64 instead of the cube root. 192 comes from multiplying 64 by 3 instead of finding the number that cubes to 64. 21.3 comes from dividing 64 by 3 instead of finding its cube root.)
- (c) 5/11 — Let x = 0.45 recurring, so x = 0.454545... . Since two digits repeat, multiply by 100: 100x = 45.454545... . Subtracting the original x removes the recurring part exactly, because it lines up digit for digit: 100x − x = 45.454545... − 0.454545... = 45, so 99x = 45, giving x = 45/99 = 5/11. Treating the decimal as if it terminated at two places gives 45/100 = 9/20, which is only 0.45 and drops the repeating part entirely. Subtracting 10x instead of x — using 100x − 10x = 90x = 45 — is the wrong power of ten for a two-digit repeating block, and gives x = 45/90 = 1/2. Making an arithmetic slip in the numerator, 45 − 1 = 44 instead of 45, gives 44/99 = 4/9.
- (b) 30 — Method: 12% of an amount is 12/100 of it, so find 1% by dividing by 100 and then multiply by 12. Working: 1% of 250 is 250 ÷ 100 = 2.5, and 12% is 2.5 × 12 = 30. Answer: 30 seats. The distractors: 3 comes from writing 12% as 0.012 instead of 0.12, giving 0.012 × 250 = 3; 25 comes from finding 10% of the seats and stopping there; 24 comes from counting 12 seats for each whole hundred, 12 + 12 = 24, and ignoring the remaining 50 seats.
- (a) 1,500 ≤ m < 2,500 — Method: a four-digit figure written to 1 significant figure has been rounded to the nearest 1,000, so the mass lies within half of 1,000, that is 500, of the figure given. Working: 2,000 − 500 = 1,500 and 2,000 + 500 = 2,500. The lower limit is included, because 1,500 kg rounds up to 2,000 kg to 1 significant figure, while 2,500 kg rounds up to 3,000 kg, so the upper limit is not. Answer: 1,500 ≤ m < 2,500. The distractors: 1,950 ≤ m < 2,050 comes from rounding to the nearest 100 instead of to 1 significant figure; 1,000 ≤ m < 3,000 goes a whole 1,000 either side instead of half of it; 1,500 < m ≤ 2,500 has the two limits the wrong way round.
Build your own mix at the worksheet builder.