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) 4.1 × 10³, 3.2 × 10⁴, 2.9 × 10⁵ — The exponent decides the size first: 10³ is smaller than 10⁴, which is smaller than 10⁵, so the order is 4.1 × 10³, then 3.2 × 10⁴, then 2.9 × 10⁵. Reversing the whole list gives largest to smallest instead of smallest to largest. Comparing 3.2 × 10⁴ and 4.1 × 10³ by their coefficients alone, 3.2 against 4.1, and swapping them ignores that 10⁴ is bigger than 10³ regardless of the coefficient. Comparing 2.9 × 10⁵ and 3.2 × 10⁴ by their coefficients alone and swapping them makes the same mistake at the top of the list.
- (c) Exactly 1: the subtraction has no rounding at any step. — 10x − x removes the recurring part completely, because the digits after the decimal point in 10x and in x are identical from the tenths place onward, so they cancel exactly: 9.999... − 0.999... = 9.000... = 9. Nothing was rounded to reach 9x = 9, so x = 1 is an exact equality, not an approximation, and the statement that the value is exactly 1, with no rounding at any step, is the correct one. Calling it only approximately 1, on the ground that a recurring decimal can never reach a whole number, misunderstands what the subtraction has just shown: the recurring tail cancels completely, leaving no gap to approximate away. Claiming the method only works because the recurring digit is 9 is also wrong — the same subtraction cancels the recurring part for any repeating digit, not just 9; it is the choice of multiplier (10, matching the one-digit repeat) that makes the cancellation exact, not the digit itself. Saying 10x minus x gives 8.999... rather than 9 misreads the subtraction: 9.999... − 0.999... has no digit to borrow from, since every decimal digit in the two numbers matches, so the result is exactly 9, not 8.999... .
- (b) 23 kg — Method: the rounded value sits exactly in the middle of the error interval. Working: the interval 22.5 ≤ m < 23.5 stretches 0.5 either side of the rounded value, so the rounded value is 23. Answer: 23 kg. (22 kg comes from rounding the lower bound down instead of finding the middle of the interval. 22.5 kg comes from giving the lower bound itself rather than the rounded value. 23.5 kg comes from giving the upper bound itself rather than the rounded value.)
- (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.
- (c) 20 — Method: work out each power separately before subtracting. Working: 6² = 36 and 4² = 16, so 6² − 4² = 36 − 16 = 20. Answer: 20. (4 comes from subtracting first, 6 − 4 = 2, and then squaring that result, instead of squaring each number first. 52 comes from adding the two squares, 36 + 16, instead of subtracting them. 2 comes from subtracting the two numbers, 6 − 4, and forgetting to square at all.)
- (d) 2 — Method: a fourth root undoes raising to the power 4, so look for the number that gives 16 when it is multiplied by itself four times. Working: 2 × 2 = 4, 4 × 2 = 8 and 8 × 2 = 16, which uses four factors of 2. Answer: 2. The distractors: 4 comes from taking the square root of 16 instead of its fourth root; 8 comes from halving 16, treating any root as a halving; 64 comes from multiplying 16 by 4 instead of taking a fourth root.
- (b) 2/5 — The statement compares girls with boys, so girls : boys = 2 : 3. The club is therefore 2 + 3 = 5 equal shares and the girls take 2 of them, so 2/5 of the members are girls. 2/3 repeats the part-to-part fraction the question gives, 3/5 is the fraction that are boys, and 1/3 comes from treating 2/3 as a fraction of the whole club and taking what is left.
- (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.
- (b) −5 — Using the order of operations, work out the multiplication first: 4 × (−2) = −8. Then 3 + (−8) = −5. A candidate who adds before multiplying gets (3 + 4) × (−2) = −14. A candidate who drops the negative sign on the multiplication gets 3 + 4 × 2 = 11. A candidate who works out the multiplication correctly but gives that as the final answer, forgetting to combine it with the 3, gets −8.
- (c) 7 + 4√3 — Expand the brackets fully: (2 + √3)² = 2² + 2 × 2 × √3 + (√3)² = 4 + 4√3 + 3. Adding the two whole-number terms, 4 + 3 = 7, gives 7 + 4√3. Using (a + b)² = a² + b² and skipping the middle cross term entirely gives just 4 + 3 = 7, with no surd term at all. Treating (√3)² as if it stayed √3 rather than becoming 3, then merging it with the existing surd term, gives 4 + 5√3. Squaring only the surd term correctly but carrying the whole-number term as 2 instead of squaring it to 4 gives 2 + 3 + 4√3 = 5 + 4√3.
- (b) 93.9975 — Each measurement was rounded to 1 decimal place, so the error is half of 0.1: length is 12.35 ≤ L < 12.45, and width is 7.45 ≤ W < 7.55. The upper bound for the area comes from multiplying the upper bounds of both dimensions: 12.45 × 7.55 = 93.9975 m². Using the lower bound of both dimensions instead, 12.35 × 7.45 = 92.0075 m², gives the lower bound of the area rather than the upper one. Multiplying the two given rounded values directly, 12.4 × 7.5 = 93, forgets that a rounded measurement is not exact and needs its own error interval. Bounding only the length and leaving the width at its given value, 12.45 × 7.5 = 93.375, misses that the width also has an upper bound of its own.
- (c) Sam is correct — The leading zeros in 0.070268 are not significant, so the first three significant figures are 7, 0 and 2. The next digit along is 6, and since 6 is 5 or more, the third significant figure rounds up from 2 to 3, giving 0.0703. This means Sam's answer is correct. Writing 0.070 keeps only 2 significant figures, one short of what was asked. Writing 0.0702 ignores the digit 6 that follows and leaves the third figure unrounded. Writing 0.0704 rounds the third figure up twice, as if a later digit had also pushed it up.
- (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.
- (a) to the nearest centimetre — Method: the error interval of a rounded measurement runs from half a unit below the stated value to half a unit above it, so the width of the interval is one whole unit of the accuracy used. Working: the interval runs from 24.5 to 25.5, a width of 25.5 − 24.5 = 1, so the unit of accuracy is 1 cm; the stated value is the midpoint, 25 cm, and 25 correct to the nearest centimetre is exactly what gives 24.5 ≤ L < 25.5. Answer: to the nearest centimetre. To the nearest 0.5 cm comes from reading the half-unit, 0.5, as the accuracy itself instead of doubling it back to the full unit. To 1 decimal place comes from seeing the bounds written with one decimal place and taking that as the accuracy, but the bounds of a value given to 1 decimal place would be only 0.05 either side. To the nearest 10 cm comes from confusing the size of the value, about 25, with the unit it was rounded to; rounding to the nearest 10 cm would give an interval 5 cm either side of the stated value.
- (d) £50.00 — The sale price is 85% of the original, so the original price = £42.50 ÷ 0.85 = £50.00. 15% of £42.50 is £6.375. A candidate who finds 15% of £42.50 and subtracts it from the sale price gets £42.50 − £6.375 = £36.125, which is £36.13 to the nearest penny. A candidate who adds 15% of £42.50 instead of reversing the decrease gets £42.50 + £6.375 = £48.875, which is £48.88 to the nearest penny. A candidate who divides by 0.15 instead of 0.85 gets £283.33.
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