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
MathsUKwww.geekhero.co.uk
- 1.Amelia estimates 48 × 21 by working out 50 × 20 = 1,000. Work out whether her estimate is an under-estimate or an over-estimate, and by how much.
- 2.For any two whole numbers, the product of the numbers is equal to the product of their highest common factor and their lowest common multiple. The highest common factor of 6 and 8 is 2, and 6 × 8 = 48. Work out the lowest common multiple of 6 and 8.
- 3.Postage on a parcel is calculated as £4.60, correct to the nearest 20p. Which of these could not be the actual cost of the postage?
- 4.Grace works out 7 × 99 by writing 99 as 100 − 1. Use her method to work out 7 × 99.
- 5.Work out the value of .
- 6.Simplify x⁷ × x⁴, giving your answer as a single power of x.
- 7.A semicircle has a diameter of 8 cm. Work out the exact area of the semicircle, in terms of π.
- 8.A number is multiplied by 4, then 8 is added, giving the result 40. Work out the number.
- 9.A box stands on a table. The box exerts a force of 240 N on the table, and the base of the box has an area of 0.8 m². Work out the pressure the box exerts on the table, in N/m².
- 10.A netball team has 8 players. Two of them are chosen to be captains, and the two captains have equal standing. Work out how many different pairs of captains could be chosen.
- 11.A choir has sopranos, altos and tenors in the ratio 6 : 4 : 5. What fraction of the choir is not tenors?
- 12.Work out (−3) × 4 + 2 × (−5)
- 13.A machine fills bags of sugar and shows the mass of each bag to the nearest 10 g. A checker rejects any bag whose actual mass is less than 996 g. One bag shows a mass of 1,000 g on the machine. Decide whether this bag could be rejected, and give a reason for your answer.
- 14.The recurring decimal 0.181818... can be written as 0.18 recurring, where both digits repeat forever. Let x = 0.18 recurring. Work out x as a fraction in its simplest form.
- 15.A recipe for one cake needs 2/3 of a cup of sugar. Priya has 3 1/2 cups of sugar. Work out how many complete cakes she can make.
Answer key
- (c) An under-estimate, by 8 — Method: work out the exact product, then compare it with the estimate; an estimate that is smaller than the exact value is an under-estimate, and the difference between them is the size of the error. Working: 48 × 21 = 48 × 20 + 48 = 960 + 48 = 1,008, and 1,008 − 1,000 = 8, so the estimate falls short. Answer: an under-estimate, by 8. The distractors: an over-estimate by 8 has the size of the error right but the direction wrong, and comes from assuming that rounding 48 up to 50 must push the estimate above the exact value, without allowing for 21 being rounded down; an over-estimate by 19 comes from working out 48 × 21 as 48 × 20 + 21 = 981, adding a 21 where another 48 belongs; the claim that the estimate is exactly right comes from arguing that one number was rounded up and the other down, so the two changes must cancel.
- (a) 24 — Method: rearrange the relationship so that the lowest common multiple stands alone; it is the product of the two numbers divided by their highest common factor. Working: 48 = 2 × the lowest common multiple, so the lowest common multiple is 48 ÷ 2 = 24. Checking, 24 is in the 6 times table and in the 8 times table. Answer: 24. The distractors: 48 comes from giving the product of the two numbers and never dividing by the highest common factor; 96 comes from multiplying by the highest common factor instead of dividing by it; 12 comes from dividing by the highest common factor twice, once for each of the two numbers.
- (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.)
- (c) 693 — Method: multiplying a bracket by a number multiplies every term inside it, so 7 × (100 − 1) = 7 × 100 − 7 × 1. Working: 7 × 100 = 700 and 7 × 1 = 7, so the calculation becomes 700 − 7 = 693. Answer: 693. The distractors: 699 comes from subtracting the 1 itself rather than 7 lots of it, giving 700 − 1 = 699; 707 comes from adding the second product instead of subtracting it, giving 700 + 7 = 707; 700 comes from rounding 99 up to 100 and then offering the estimate 7 × 100 as an exact value.
- (c) 0.001 — Method: a negative index means 'one over' the positive power, so $10^{-3}$ means one over $10^{3}$. Working: ten cubed is 1000, and one over 1000 is 0.001. 1000 comes from ignoring the negative sign and working out ten cubed instead of its reciprocal. −1000 comes from ignoring what the negative index does to the size, while still writing a negative sign on the large value. −0.001 comes from correctly finding the size, 0.001, but wrongly keeping a negative sign because the index was negative. Answer: 0.001.
- (a) x¹¹ — Method: when multiplying powers of the same base, add the indices. Working: 7 + 4 = 11, so x⁷ × x⁴ = x¹¹. x²⁸ comes from multiplying the indices, 7 × 4 = 28, instead of adding them. x³ comes from working out 7 − 4 = 3, which is the rule for dividing powers, not multiplying them. 11x comes from adding the indices to make 11 but then treating x as a coefficient instead of a power. Answer: x¹¹.
- (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) 8 — To undo 'multiply by 4, then add 8', reverse the operations in reverse order: subtract 8 first, then divide by 4. 40 − 8 = 32, and 32 ÷ 4 = 8, so the number is 8. A candidate who added 8 again instead of subtracting worked out 40 + 8 = 48, then 48 ÷ 4 = 12. A candidate who divided before subtracting, doing the inverse operations in the wrong order, worked out 40 ÷ 4 = 10, then 10 − 8 = 2. A candidate who multiplied instead of dividing at the last step worked out (40 − 8) × 4 = 32 × 4 = 128.
- (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.
- (b) 28 — Method: count the ordered choices with the product rule and then correct for the double counting, because the two captains have equal standing and so a pair is the same pair whichever captain is named first. Working: there are 8 players who could be named first and 7 who could be named second, giving 8 × 7 = 56 ordered choices; each pair has been counted twice, once in each order, so the number of pairs is 56 ÷ 2 = 28. Answer: 28. The distractors: 56 comes from stopping at 8 × 7 and never halving, which counts each pair of captains twice; 64 comes from working out 8 × 8, which allows the same player to be chosen as both captains; 16 comes from multiplying the 8 players by the 2 captaincies instead of pairing the players with one another.
- (a) 2/3 — Total parts = 6 + 4 + 5 = 15. Sopranos and altos together are not tenors: 6 + 4 = 10 parts, so the fraction is 10/15, which simplifies to 2/3. 1/3 comes from finding the fraction of tenors instead of the fraction that is not tenors. 2/5 comes from counting only the sopranos as not tenors and leaving the altos out. 4/9 comes from leaving sopranos out of the total, 4 + 5 = 9, and then using only the altos as the fraction that is not tenors.
- (a) −22 — Method: both multiplications are carried out before the addition, and a positive multiplied by a negative is negative. Working: (−3) × 4 = −12 and 2 × (−5) = −10, so the calculation becomes −12 + (−10) = −22. Answer: −22. The distractors: 22 comes from ignoring the minus signs and working out 3 × 4 + 2 × 5 = 22; 50 comes from working from left to right with no priority at all, giving −12 + 2 = −10 and then −10 × (−5) = 50; −2 comes from taking 2 × (−5) as +10, so that −12 + 10 = −2.
- (b) Yes — the actual mass could be as low as 995 g — Method: a mass shown to the nearest 10 g lies within half of 10 g, that is 5 g, of the figure on the display, so compare the smallest mass the bag can have with the checker's limit of 996 g. Working: 1,000 − 5 = 995, so the actual mass of the bag can be as low as 995 g, and 995 g is below the 996 g limit, so a bag showing 1,000 g on the machine can still be rejected. Answer: Yes — the actual mass could be as low as 995 g. The distractors: 990 g comes from going a whole 10 g below the display instead of half of it; 999.5 g comes from treating the display as being to the nearest gram, when it is to the nearest 10 g; the claim that the mass is exactly 1,000 g treats a rounded display as an exact measurement.
- (a) 2/11 — Let x = 0.18 recurring, so x = 0.181818... . Since two digits repeat, multiply by 100: 100x = 18.181818... . Subtracting the original x removes the recurring part, because the digits line up exactly: 100x − x = 18.181818... − 0.181818... = 18, so 99x = 18, giving x = 18/99 = 2/11. Treating the decimal as if it terminated at two places gives 18/100 = 9/50, which is only 0.18 and drops the repeating part entirely. Subtracting 10x instead of x — using 100x − 10x = 90x = 18 — is the wrong power of ten for a two-digit repeating block, and gives x = 18/90 = 1/5. Making an arithmetic slip in the numerator, 18 − 1 = 17 instead of 18, gives 17/99.
- (b) 5 — Method: divide the total amount of sugar by the amount needed for one cake, then round down because a part-used amount of sugar cannot make an extra whole cake. Working: 3 1/2 ÷ 2/3 = 7/2 × 3/2 = 21/4 = 5.25; only 5 complete cakes can be made, since the leftover 0.25 of a portion is not enough for a 6th cake. Answer: 5. 5.25 gives the exact result of the division without rounding down to a whole number of cakes. 7 comes from multiplying 3.5 by 2 and ignoring the need to also divide by 3 as part of dividing by the fraction 2/3. 6 comes from rounding 5.25 up to the nearest whole number instead of down, wrongly assuming a 6th cake could be made from the leftover sugar.
Build your own mix at the worksheet builder.