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.
Answer key: Number worksheet — GCSE Higher
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- (d) 7/20 — Method: write the decimal over the power of ten that matches the number of digits after the point, then divide the numerator and the denominator by their highest common factor. Working: 0.35 has two digits after the point, so it is 35 hundredths and can be written as 35/100; the highest common factor of 35 and 100 is 5, and 35 ÷ 5 = 7 with 100 ÷ 5 = 20. Answer: 7/20. The distractors: 3/10 comes from reading only the first digit after the point and converting 0.3; 7/25 comes from dividing the numerator by 5 but the denominator by 4, using a different factor on the top and on the bottom; 35/10 comes from counting one decimal place instead of two and writing the digits over 10.
- (b) 9 g/cm³ — Method: density = mass ÷ volume. Working: 540 ÷ 60 = 9. Answer: 9 g/cm³. (0.11 g/cm³ comes from dividing the volume by the mass instead of the mass by the volume. 480 g/cm³ comes from subtracting the volume from the mass instead of dividing. 32400 g/cm³ comes from multiplying the mass by the volume instead of dividing.)
- (d) −1.7 — Since the numbers have different signs, find the difference between their sizes: 4.5 − 2.8 = 1.7, then keep the sign of the number further from zero. So −4.5 + 2.8 = −1.7. A candidate who drops the negative sign gets 1.7. A candidate who adds the magnitudes instead of finding the difference gets −(4.5 + 2.8) = −7.3. A candidate who takes the smaller digit from the larger in the tenths column, doing 8 − 5 = 3 instead of borrowing to make 15 − 8 = 7, gets 2.3 and so −2.3.
- (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.
- (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.
- (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.
- (b) 400 — Method: round each number to the nearest 100, then subtract the rounded values. Working: 812 rounds to 800 (nearest 100) and 397 rounds to 400 (nearest 100). 800 − 400 = 400. Answer: 400. 500 comes from rounding 397 down to 300 instead of up to the nearest 100, 400. 300 comes from rounding 812 down to 700 instead of up to the nearest 100, 800. 415 is the exact value of 812 − 397, found without rounding first, so it is not an estimate — the spreadsheet's answer of 315 is too far from the estimate of 400 to be correct.
- (d) 30 kg — Round 0.485 kg to 1 significant figure: 0.5 kg. Multiply by the 60 cakes: 0.5 × 60 = 30 kg. A candidate who rounded to 2 significant figures instead of 1 used 0.49 kg, giving 0.49 × 60 = 29.4 kg. A candidate who used the unrounded amount instead of the estimate worked out 0.485 × 60 = 29.1 kg. A candidate who rounded 0.485 down to 0.4 kg instead of up to 0.5 kg worked out 0.4 × 60 = 24 kg.
- (d) £120 — Total parts = 5 + 3 = 8, so one part is worth 320 ÷ 8 = 40 pounds. The cake sale is 3 parts, so it raised 3 × 40 = 120 pounds. £200 comes from working out the raffle's share, 5 × 40, instead of the cake sale's share. £40 comes from finding the value of one part but forgetting to multiply by 3. £192 comes from dividing the total by 5 instead of 8 to find the value of one part, 320 ÷ 5 = 64, then multiplying by 3, 3 × 64 = 192.
- (b) 2² × 3 × 5 — Repeatedly divide 60 by prime numbers: 60 ÷ 2 = 30, 30 ÷ 2 = 15, 15 ÷ 3 = 5, and 5 is itself prime. So 60 is 2 × 2 × 3 × 5, which in index notation is 2² × 3 × 5. Stopping the factor tree after only three divisions and writing 2 × 3 × 5 misses that the 2 divides in twice, and gives only 30, not 60. Squaring the 3 as well as the 2 gives 2² × 3² × 5, which comes to 180, far too big. Squaring the 5 instead of the 2 gives 2 × 3 × 5², which comes to 150, also too big. So 60 = 2² × 3 × 5.
- (c) 36 — Method: set the outcomes out in a grid with one die along the top and the other down the side, so that every cell of the grid is one outcome, and count the cells. Working: the red die can land in 6 ways, so the grid has 6 columns, and the blue die can also land in 6 ways, so the grid has 6 rows; the number of cells is 6 × 6 = 36. Answer: 36. The distractors: 12 comes from adding 6 and 6 instead of multiplying them; 6 comes from counting the outcomes of a single die and forgetting that the second die also has to land; 21 comes from treating the two dice as indistinguishable, so that a red 2 with a blue 3 and a red 3 with a blue 2 are counted as one outcome.
- (c) 7 × 10⁻⁸ — '100 times smaller' means dividing by 100 = 10². Dividing 7 × 10⁻⁶ by 10² means subtracting 2 from the exponent: −6 − 2 = −8, giving 7 × 10⁻⁸. A candidate who multiplied by 100 instead of dividing added 2 to the exponent, getting 7 × 10⁻⁴. A candidate who divided by 10 instead of 100 subtracted only 1 from the exponent, getting 7 × 10⁻⁵. A candidate who did not apply the scale factor at all left the diameter as 7 × 10⁻⁶, the same as the red blood cell.
- (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.
- (c) 3 — Method: the index counts how many times the coefficient has been multiplied by 10, which is the number of places the decimal point moves from the end of the number to just after the first significant digit. Working: 2,000 = 2 × 1,000, and 1,000 = 10 × 10 × 10, which is three tens. Answer: 3. The distractors: 4 comes from counting the four digits of 2,000 rather than the three places the decimal point moves; 2 comes from copying the coefficient 2 into the index; −3 comes from making the index negative, which would describe a number smaller than 1 rather than two thousand.
- (b) 6.2 ≤ x < 6.3 — Truncating simply cuts off the digits after the required decimal place instead of rounding them, so every value from 6.2 up to (but not reaching) 6.3 truncates to 6.2. This gives the error interval 6.2 ≤ x < 6.3, with no allowance made on the lower side because truncation never rounds a smaller value up into this interval. Using 6.15 ≤ x < 6.25 applies the rounding rule of going half a unit either side, which does not apply to truncation. Writing 6.1 < x ≤ 6.2 puts the interval below 6.2 instead of above it. Writing 6.2 ≤ x ≤ 6.3 wrongly includes 6.3, which truncates down to itself, not to 6.2.
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