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.
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Answer key: Number worksheet — GCSE Higher
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- (d) 1/4 — Method: deal with the fractional index first, then the negative sign. Working: $8^{2/3} = (\sqrt[3]{8})^2 = 2^2 = 4$. A negative index means take the reciprocal of that result, so $8^{-2/3} = \frac{1}{8^{2/3}} = \frac{1}{4}$. Answer: 1/4. A candidate who evaluates $8^{2/3}$ correctly but forgets the negative sign entirely gets 4 — they have dropped the instruction to take a reciprocal. A candidate who takes the reciprocal step but applies it as a sign change to the finished number instead of inverting it gets −4. A candidate who multiplies 8 by −2/3, treating the index as an ordinary factor rather than a power, gets −16/3.
- (a) 10 — √49 = 7 and ∛27 = 3, so √49 + ∛27 = 7 + 3 = 10. Treating the cube root as dividing by 3 instead of finding the cube root gives 27 ÷ 3 = 9, then 7 + 9 = 16. Multiplying the two roots instead of adding them gives 7 × 3 = 21. Ignoring the cube root symbol and using 27 as it stands gives 7 + 27 = 34.
- (c) 590 — To round to the nearest 10, decide which multiple of 10 the number is nearer to. 592.5 lies between 590 and 600. It is 592.5 − 590 = 2.5 above 590, but 600 − 592.5 = 7.5 below 600, so it is much nearer to 590. Equivalently, the units digit is 2, and 2 is less than 5, so round down: 592.5 rounds to 590. A candidate who wrote 600 rounded up because of the 5 in the tenths place, but that digit decides rounding to the nearest whole number, not to the nearest 10 — the units digit is what matters here. A candidate who wrote 595 rounded to the nearest 5 instead of the nearest 10. A candidate who wrote 500 cut the number down to its hundreds digit instead of rounding to the nearest 10.
- (c) 150 — Since 90 students represent 3 of the 5 equal parts, one part is 90 ÷ 3 = 30, and the whole year group is five parts: 30 × 5 = 150. Applying the fraction forwards to 90 instead of reversing it, 90 × 3/5 = 54, treats the given number as the whole rather than as three fifths of it. Finding one part correctly as 30 but forgetting to scale up to the whole year group leaves 30 as the final answer. Treating 90 as the whole year group and adding on 2/5 of 90 for the students who do not walk, 90 + (90 × 2/5) = 126, applies the missing fraction to the wrong base amount.
- (c) 150 m — Method: a fraction acts as an operator, so finding 3/7 of a length means dividing by the denominator and multiplying by the numerator. Working: 350 ÷ 7 = 50, so one seventh of the route is 50 m, and three sevenths is 50 × 3 = 150 m. Answer: 150 m. The distractors: 50 m comes from finding one seventh and stopping there instead of multiplying by 3; 1050 m comes from multiplying by the numerator without dividing by the denominator, giving 350 × 3 = 1050; 200 m comes from working out the stretch of the route the footpath does not run alongside, which is 4/7 of 350 m, instead of the stretch it does.
- (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) 18 — The error intervals are 35 ≤ distance < 45 and 2.5 ≤ time < 3.5. Average speed is distance ÷ time, and to make a quotient as large as possible you divide the largest possible numerator by the SMALLEST possible denominator: 45 ÷ 2.5 = 18 km/h. Using the upper bound of time as well as the upper bound of distance, 45 ÷ 3.5, gives roughly 12.9 km/h — dividing by a bigger number produces a smaller result, so this actually finds a value smaller than the true maximum. Using the lower bound of distance with the lower bound of time, 35 ÷ 2.5 = 14, mixes up which bound belongs to a maximum calculation. Using the lower bound of distance with the upper bound of time, 35 ÷ 3.5 = 10, is in fact the correct method for the MINIMUM speed, not the maximum.
- (c) 1/3 — Method: find non-fiction's fraction of the whole, then multiply by the fraction of non-fiction that is history. Working: total parts = 4 + 5 = 9, so non-fiction is 5/9 of all books. History books are 3/5 of the non-fiction books: 3/5 × 5/9 = 15/45 = 1/3. Answer: 1/3. 3/5 comes from giving the fraction of non-fiction books that are history, without relating it to all the books at the fair. 5/9 comes from stopping after finding the fraction of all books that are non-fiction, without finding the history books within that. 4/15 comes from multiplying 3/5 by the fraction that is fiction (4/9) instead of the fraction that is non-fiction (5/9).
- (c) 6 — 2 × 3 = 6, then 36 ÷ 6 = 6. Ignoring the brackets and working left to right gives 36 ÷ 2 = 18, then 18 × 3 = 54. Multiplying by the bracket instead of dividing by it gives 2 × 3 = 6, then 36 × 6 = 216. Dividing by only the 2 inside the bracket, and ignoring the × 3, gives 36 ÷ 2 = 18.
- (a) 5√2 — Simplify each surd first: √8 = √4 × √2 = 2√2, and √18 = √9 × √2 = 3√2. Both terms are now multiples of √2, so they are like terms: 2√2 + 3√2 = 5√2. Adding the numbers under the two roots first, 8 + 18 = 26, and writing √26 treats unlike surds as if they combine under one root — they only combine once they share the same radicand, which is not how addition of surds works. Writing 9√2 for √18 instead of 3√2 (forgetting to root the 9) and then adding gives 2√2 + 9√2 = 11√2. Writing 4√2 for √8 instead of 2√2 (forgetting to root the 4) and adding gives 4√2 + 3√2 = 7√2.
- (c) 2 — First write 1 1/2 as an improper fraction, 3/2. To divide by 3/4, multiply by its reciprocal, 4/3: 3/2 × 4/3 = 12/6 = 2. Dropping the whole number and dividing only the fractional part, 1/2 ÷ 3/4 = 1/2 × 4/3, gives 2/3. Multiplying by 3/4 directly instead of using its reciprocal, 3/2 × 3/4, gives 9/8. Using the reciprocal of the first fraction instead of the second, 2/3 × 3/4, gives 1/2.
- (c) 9 — Method: factors come in pairs that multiply to give the number, so work through the pairs in order; a factor paired with itself is counted only once. Working: the pairs are 1 × 100, 2 × 50, 4 × 25, 5 × 20 and 10 × 10. The first four pairs give eight different factors, and the last pair adds only one more, so the factors are 1, 2, 4, 5, 10, 20, 25, 50 and 100. Answer: 9. The distractors: 10 comes from counting the pair 10 × 10 as two separate factors; 8 comes from leaving 1 out of the list, on the view that 1 is not a proper factor; 4 comes from writing 100 = 2² × 5² and multiplying the two indices together instead of adding 1 to each index first.
- (a) 2, 3, 4, 5 — Method: work out which whole numbers satisfy both parts of the inequality. Working: n ≥ 2 means n can be 2 or more; n < 6 means n must be less than 6, so 6 itself is not included. The whole numbers that fit both conditions are 2, 3, 4 and 5. Answer: 2, 3, 4, 5. 2, 3, 4, 5, 6 treats < 6 as ≤ 6 and wrongly includes 6. 3, 4, 5 treats ≥ 2 as > 2 and wrongly leaves out 2. 1, 2, 3, 4, 5 wrongly includes 1, which does not satisfy n ≥ 2.
- (b) 1 — Method: BIDMAS deals with the index before the subtraction, and a negative number multiplied by itself gives a positive result. Working: (−2)² = (−2) × (−2) = 4, so the calculation becomes 4 − 3 = 1. Answer: 1. The distractors: −7 comes from squaring only the 2 and leaving the minus sign outside the index, giving −(2²) − 3 = −4 − 3 = −7; −1 comes from subtracting the square from 3 instead of 3 from the square, giving 3 − 4 = −1; 25 comes from carrying out the subtraction before the index, giving (−2 − 3)² = (−5)² = 25.
- (a) 4 + √6 — Multiply top and bottom by the conjugate, 4 + √6. The denominator becomes (4 − √6)(4 + √6) = 4² − (√6)² = 16 − 6 = 10. The numerator becomes 10 × (4 + √6) = 40 + 10√6. So the fraction is (40 + 10√6)/10 = 4 + √6, since both terms in the numerator divide by 10. Distributing the conjugate to only the whole-number term of the numerator, and forgetting the surd term entirely, leaves just 4. Rationalising by multiplying the numerator by the conjugate but leaving the ORIGINAL denominator's sign unchanged instead of squaring it lands on 4 − √6, with the surd's sign never actually flipping to positive. Dividing only the whole-number part of the numerator by 10 and forgetting to divide the surd term too leaves 4 + 10√6.
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