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.A necklace is made from red beads and gold beads in the ratio 7 : 3. What fraction of the beads are gold?
- 2.Last year a company made a profit of £5,200,000. Write this amount in standard form.
- 3.Maya buys 4 plants at £3.20 each and a bag of compost for £6.75. She pays with a £20 note. Work out her change.
- 4.Which statement about the number 91 is correct?
- 5.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?
- 6.A courier's van has a weight limit of 850 kg for its parcels. The driver's display shows the total mass of the parcels loaded as 850 kg, correct to the nearest 5 kg. Decide whether the parcels are definitely within the weight limit.
- 7.Work out (4 × 10³) × (3 × 10¹). Give your answer in standard form.
- 8.Work out 3.7 × 24.
- 9.Two lighthouses flash at the start of the same minute. The first lighthouse flashes every 8 minutes and the second flashes every 12 minutes. Work out how many minutes it will be until they next flash together.
- 10.Which of these statements about −0.7 and −0.25 is true?
- 11.A van has a mass of 2,000 kg, correct to 1 significant figure. Using m for the mass of the van in kilograms, write down the error interval for m.
- 12.Two allotments are combined. Allotment A grows vegetables and flowers in the ratio 3 : 2, and has 20 plots in total. Allotment B grows vegetables and flowers in the ratio 1 : 4, and has 20 plots in total. Work out the fraction of all the plots, in both allotments combined, that grow vegetables.
- 13.During one night on a mountain, four hikers each recorded the temperature they felt: Ben −6 °C, Priya −2 °C, Sam −9 °C, Alex 1 °C. Work out the difference between the coldest and the warmest of these four temperatures.
- 14.By listing systematically, work out how many two-digit multiples of 5 can be made using the digits 0, 3 and 5, if each digit can be used at most once and the number cannot start with 0.
- 15.In a class, 1/3 of the pupils are girls. There are 12 girls in the class. Work out how many pupils are in the class.
Answer key
- (d) 3/10 — Total parts = 7 + 3 = 10. Gold beads are 3 of those 10 parts, so the fraction is 3/10. 7/10 comes from finding the fraction of red beads instead of gold beads. 7/3 comes from writing the ratio itself as a fraction, without adding the parts to find the total. 3/7 comes from comparing gold beads with red beads instead of with the total number of beads.
- (b) 5.2 × 10⁶ — Method: write the digits as a coefficient that is at least 1 and less than 10, then count the places the decimal point moves to reach that position. Working: the digits give a coefficient of 5.2, and the decimal point travels from the end of 5,200,000 until it sits between the 5 and the 2, a move of 6 places. Answer: 5.2 × 10⁶. The distractors: 52 × 10⁵ is the same amount but not in standard form, because 52 is not less than 10; 5.2 × 10⁵ comes from counting the five zeros in 5,200,000 rather than the six places the decimal point moves; 5.2 × 10⁻⁶ comes from making the index negative because the decimal point was carried to the left.
- (c) £0.45 — Method: find the total cost, then subtract from £20. Working: 4 × £3.20 = £12.80. £12.80 + £6.75 = £19.55. Change = £20.00 − £19.55 = £0.45. Answer: £0.45. (£7.20 comes from forgetting to include the compost and subtracting only the plants' cost from £20. £1.45 comes from dropping the carry when adding the pence: 80p + 75p = £1.55, but only the 55p is written down, giving £18.55 instead of £19.55. £10.05 comes from buying only one plant instead of four, using £3.20 + £6.75.)
- (a) 91 is not prime, because 91 = 7 × 13. — Check 91 for prime factors up to its square root, which is just under 10: 91 ÷ 7 = 13, and both 7 and 13 are prime, so 91 = 7 × 13 and 91 is not a prime number. Checking only 2, 3 and 5 misses that 7 also needs to be tried — 91 is odd, its digits do not sum to a multiple of 3 (9 + 1 = 10), and it does not end in 0 or 5, so those three checks alone wrongly suggest it is prime. Assuming any odd number ending in 1 must be prime ignores that 91 = 7 × 13 is a counterexample. Misapplying the digit-sum test for 3 by miscounting 9 + 1 as a multiple of 3 wrongly concludes 91 is divisible by 3, when the correct digit sum, 10, is not a multiple of 3. So 91 is not prime, because 91 = 7 × 13.
- (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.)
- (b) No, the true mass could be as high as 852.5 kg — Method: find the upper bound of the true mass and compare it with the weight limit. Working: the display is correct to the nearest 5 kg, so half of 5 kg is 2.5 kg, and the true mass, m kg, satisfies 847.5 ≤ m < 852.5. Part of that interval lies above 850 kg, so the parcels are not definitely within the limit. Answer: the true mass could be as high as 852.5 kg, which is above the limit. ("Yes, the display reads 850 kg, which is not above the limit" compares the limit with the displayed value instead of with the largest value the true mass could take. "Yes, the true mass is at least 847.5 kg and at most 850 kg" uses the correct half unit below but caps the interval at the limit instead of at 852.5 kg. "No, 850 kg on the display rounds up to 855 kg" wrongly treats the displayed value as if it rounds again.)
- (d) 1.2 × 10⁵ — 4 × 3 = 12, and 3 + 1 = 4, giving 12 × 10⁴ — but 12 is not between 1 and 10, so this must be rewritten as 1.2 × 10⁵. Stopping at 12 × 10⁴ without rewriting it leaves the coefficient out of range. Rewriting 12 as 1.2 but leaving the exponent at 4 instead of increasing it to 5 gives 1.2 × 10⁴, which is ten times too small. Adding the coefficients instead of multiplying them gives 4 + 3 = 7, so 7 × 10⁴.
- (b) 88.8 — Multiply as whole numbers first, ignoring the decimal point: 37 × 24. Split it as 37 × 20 = 740 and 37 × 4 = 148, so 37 × 24 = 740 + 148 = 888. 3.7 has 1 decimal place and 24 has none, so the answer needs 1 decimal place: 88.8. Counting the 2 digits in "3.7" as though that were the number of decimal places gives 8.88 instead of 1 decimal place. Leaving the decimal point out altogether gives 888. Misreading 37 × 4 as 138 rather than 148 gives a running total of 878, placed with 1 decimal place as 87.8. So 3.7 × 24 = 88.8.
- (b) 24 — List multiples of 8 and of 12: multiples of 8 are 8, 16, 24, 32; multiples of 12 are 12, 24, 36. The lowest number in both lists is 24, so the lighthouses next flash together after 24 minutes. Multiplying the two numbers together, 8 × 12, gives 96, which double-counts the common factor of 4 shared by 8 and 12. Working out the highest common factor instead of the lowest common multiple gives 4, far too soon a time for both lighthouses to line up again. Adding the two numbers, 8 + 12, gives 20, which is not even a multiple of either 8 or 12. So the lighthouses next flash together after 24 minutes.
- (b) −0.7 < −0.25 — Method: compare the two negative decimals by their distance from zero on a number line. Working: −0.7 is 0.7 away from zero and −0.25 is 0.25 away from zero, so −0.7 is further from zero in the negative direction, making it the smaller number. Answer: −0.7 < −0.25 is true. "−0.7 > −0.25" comes from comparing 0.7 and 0.25 as if both numbers were positive, ignoring the negative signs. "−0.7 = −0.25" comes from assuming the two numbers are equal because they are both negative decimals. "−0.7 ≥ −0.25" combines the false statement "−0.7 > −0.25" with the false statement "−0.7 = −0.25".
- (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.
- (a) 2/5 — Method: find the number of vegetable plots in each allotment, add them, then divide by the total number of plots in both allotments. Working: Allotment A: 3/5 × 20 = 12 vegetable plots. Allotment B: 1/5 × 20 = 4 vegetable plots. Total vegetable plots = 12 + 4 = 16. Total plots = 20 + 20 = 40. Fraction = 16/40 = 2/5. Answer: 2/5. 3/10 comes from using only Allotment A's 12 vegetable plots over the combined total of 40 plots, forgetting to add Allotment B's vegetable plots. 1/5 comes from using only Allotment B's ratio (1:4) as the fraction of vegetables, ignoring Allotment A altogether. 3/5 comes from working out the fraction of the combined plots that grow flowers instead of vegetables.
- (a) 10 °C — Method: work out the coldest and warmest of the four temperatures, then subtract to find the difference. Working: the coldest temperature is Sam's, −9 °C, and the warmest is Alex's, 1 °C. The difference is 1 − (−9) = 1 + 9 = 10. Answer: 10 °C. 8 °C comes from working out 1 − 9 = −8 and reporting 8, dropping the negative sign on −9 instead of turning the subtraction into an addition. 3 °C comes from comparing the wrong pair, Sam's −9 °C and Ben's −6 °C, instead of the coldest and the warmest: −6 − (−9) = 3. 7 °C comes from comparing Ben's −6 °C with Alex's 1 °C, mistakenly treating Ben's reading as the coldest instead of Sam's.
- (b) 3 — Method: list all valid two-digit numbers that can be made without starting with 0, then keep only the ones that are multiples of 5. Working: the two-digit numbers possible are 30, 35, 50 and 53. A number is a multiple of 5 only if it ends in 0 or 5: 30 ends in 0, 35 ends in 5, 50 ends in 0, but 53 ends in 3. So there are 3 multiples of 5. Answer: 3. 4 comes from including 53 as a multiple of 5 without checking that its last digit is not 0 or 5. 2 comes from leaving out 50, wrongly assuming 0 cannot be used as the second digit either. 6 comes from listing every two-digit arrangement of the three digits, including ones that start with 0, without applying either restriction.
- (d) 36 — Method: the fraction is acting as an operator on the whole class, so one third of the class equals 12; the operation has to be reversed, and the inverse of dividing by 3 is multiplying by 3. Working: 1/3 × (number of pupils) = 12, so the number of pupils = 12 × 3 = 36. Answer: 36 pupils. The distractors: 4 comes from applying the operator instead of reversing it, working out 12 ÷ 3 = 4; 18 comes from reading the 12 girls as two thirds of the class, giving 12 ÷ 2 × 3 = 18; 24 comes from working out the number of boys, the other two thirds, as 2 × 12 = 24 and giving that instead of the size of the class.
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