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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- (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.
- (a) 1,850 — Method: find the error interval, then check which value falls outside it. Working: half of 100 is 50, so the actual number of visitors, v, satisfies 1,750 ≤ v < 1,850. 1,850 sits exactly on the excluded upper boundary, since a value of 1,850 would round to 1,900, not 1,800. Answer: 1,850. (1,750 is a genuine possible value — it sits at the included lower boundary. 1,799 is a genuine possible value, just below the upper boundary. 1,760 is a genuine possible value, well inside the interval.)
- (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.
- (c) 3 and 4 — Find the two consecutive fourth powers either side of 200: 3⁴ = 81 and 4⁴ = 256. Since 81 < 200 < 256, ⁴√200 lies between 3 and 4. Answering 14 and 15 comes from taking a square root instead of a fourth root: √200 ≈ 14.14, which does lie between 14 and 15, but that is not the root the question asks for. Answering 5 and 6 comes from taking a cube root instead of a fourth root: ∛200 ≈ 5.85, which lies between 5 and 6. Answering 4 and 5 comes from working 4⁴ as though it were 4 × 4 × 4 = 64 and stopping a factor short, then concluding that 200 is already past 4⁴ and the root must be above 4.
- (b) £13.48 — Method: an amount of money is written to the nearest penny, which is 2 decimal places, so the calculator display has to be rounded to 2 decimal places. Working: 53.90 ÷ 4 = 13.475, and the digit in the third decimal place is 5, so the penny digit goes up from 7 to 8. Answer: £13.48. The distractors: £13.47 comes from chopping the third decimal place off instead of rounding with it; £13.50 comes from rounding to the nearest 10p rather than to the nearest penny; £13.40 comes from cutting the display short at 1 decimal place, which is both the wrong degree of accuracy and a truncation rather than a rounding.
- (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.
- (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) 72a⁷ — Method: a power outside brackets applies to every factor inside them, and multiplying two powers of the same letter adds their indices. Working: (3a²)² = 3² × a⁴ = 9a⁴, and (2a)³ = 2³ × a³ = 8a³. Multiplying the two results gives 9 × 8 = 72 for the number and 4 + 3 = 7 for the index of a. Answer: 72a⁷. The distractors: 36a⁷ comes from squaring the 2 in (2a)³ instead of cubing it, giving 4a³ and then 9 × 4; 72a¹² comes from multiplying the indices 4 and 3 when the two terms are multiplied, instead of adding them; 17a⁷ comes from adding the coefficients 9 and 8 rather than multiplying them.
- (b) 14.5 ≤ l < 15.5 — A measurement given to the nearest metre could have been rounded from anywhere up to half a metre below or above it: 15 − 0.5 = 14.5 and 15 + 0.5 = 15.5. Every value from 14.5 up to (but not reaching) 15.5 rounds to 15, so the error interval is 14.5 ≤ l < 15.5, with the lower bound included and the upper bound excluded. Making both ends strict, 14.5 < l < 15.5, wrongly excludes 14.5 itself, even though 14.5 does round to 15. Making both ends inclusive, 14.5 ≤ l ≤ 15.5, wrongly includes 15.5, which actually rounds up to 16, not 15. Using a whole metre either side instead of half a metre, giving 14 ≤ l < 16, comes from forgetting that the error is only half the rounding unit.
- (b) 63 — 5 + 2 = 7, then 3² = 9, then 7 × 9 = 63. Ignoring the brackets and applying BIDMAS as if the expression were unbracketed gives 3² = 9, then 2 × 9 = 18, then 5 + 18 = 23. Squaring the bracket instead of the 3 gives 7² = 49, then 49 × 3 = 147 — the power belongs to the 3 alone. Multiplying by 3 before squaring the whole product gives 7 × 3 = 21, then 21² = 441.
- (b) £45 — Since 2/3 of the amount is £30, one third is £30 ÷ 2 = £15, and the whole amount is three thirds: £15 × 3 = £45. Applying the fraction forwards to £30 instead of reversing it, £30 × 2/3 = £20, treats the given amount as the whole rather than as two thirds of it. Finding one third correctly as £15 but forgetting to multiply by 3 to get the whole amount leaves £15 as the final answer. Reading £30 as one third of the amount rather than as two thirds, and so multiplying straight by 3, gives £30 × 3 = £90.
- (a) 0.625 — Method: convert the fraction to a decimal so it can be compared properly with 0.6. Working: 5/8 = 0.625, and since 0.625 > 0.6, the larger value is 0.625. Answer: 0.625. 0.6 repeats Sam's incorrect claim, made by comparing single digits rather than full place value. 0.58 comes from converting 5/8 incorrectly, treating it as if it read 5 tenths and 8 hundredths. 0.85 comes from turning the fraction upside down and writing its digits straight after the decimal point, 8 then 5, instead of dividing.
- (a) 37.5 — The error intervals are 45 ≤ c < 55 and 17.5 ≤ d < 18.5. The maximum possible value of a difference comes from the largest possible value being reduced by the smallest amount: use the upper bound of c together with the LOWER bound of d, since subtracting less gives a bigger result: 55 − 17.5 = 37.5. Using the upper bound for both quantities, 55 − 18.5 = 36.5, forgets that subtracting a bigger number gives a smaller answer, not a bigger one. Using the lower bounds for both, 45 − 17.5 = 27.5, gives the lower bound of the difference instead of the upper one. Using the lower bound of c with the upper bound of d, 45 − 18.5 = 26.5, combines the two bounds the wrong way round entirely.
- (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) 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.
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