Printable · GCSE Higher · ages 14-16
Powers and roots worksheet — GCSE Higher
Fifteen questions on "powers and roots" — DfE statement N6. Print it, or print three versions so neighbours cannot copy by letter; the key gives the letter for each version.
part Higher
Answer key: Powers and roots worksheet — GCSE Higher
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- (c) 31 — Method: work out each power separately, remembering that any non-zero base raised to the power 0 is 1 and a base raised to the power 1 is itself, then add the three values. Working: 5⁰ = 1, 5¹ = 5 and 5² = 25, so the total is 1 + 5 + 25 = 31. Answer: 31. The distractors: 30 comes from taking 5⁰ as 0 instead of 1; 35 comes from taking 5⁰ as 5, treating a zero index as leaving the base unchanged; 125 comes from adding the indices first, as though the three terms were being multiplied, and working out 5³.
- (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) −28 — 6² = 36 and 4³ = 64. Work out 36 − 64 = −28. A candidate who subtracts in the wrong order gets 64 − 36 = 28. A candidate who adds instead of subtracting gets 36 + 64 = 100. A candidate who multiplies the base by the exponent instead of raising the power (6 × 2 − 4 × 3 = 12 − 12) gets 0.
- (c) 43 — Method: work out each power separately before adding. Working: 3³ = 27 and 2⁴ = 16, so 3³ + 2⁴ = 27 + 16 = 43. Answer: 43. (25 comes from using 3² instead of 3³, giving 9 + 16. 35 comes from working out 2⁴ as 2 × 4 = 8 instead of 2 × 2 × 2 × 2, giving 27 + 8. 432 comes from multiplying the two powers together instead of adding them.)
- (c) 5 — Method: work out the volume of one box, divide the total volume by it, then round down since a partial box cannot fit. Working: volume of one box = 7³ = 343 cm³. 2000 ÷ 343 = 5.83 (2 d.p.). Since only whole boxes fit, the greatest number is 5. Answer: 5. (6 comes from rounding 5.83 up to the nearest whole number instead of rounding down to the number of boxes that actually fit. 343 comes from giving the volume of one box instead of the number of boxes. 5.8 comes from leaving the division as a decimal instead of rounding down to a whole number of boxes.)
- (c) −1 — Method: find each cube root separately, keeping its sign, and then add the two results. Working: (−3) × (−3) × (−3) = −27, so ∛(−27) = −3, and 2 × 2 × 2 = 8, so ∛8 = 2. Adding gives −3 + 2 = −1. Answer: −1. The distractors: 5 comes from taking the cube root of a negative number as positive, giving 3 + 2; −5 comes from reading the minus sign as applying to the whole sum and working out −(3 + 2); −6 comes from multiplying the two roots, −3 × 2, instead of adding them.
- (b) 25 g — Method: substitute the number of years into the model, raise the fraction to that power first, then multiply by the starting mass. Working: with n = 3 the model gives M = 200 × (1/2)³. Since (1/2)³ = 1/8, the mass is 200 ÷ 8 = 25. Answer: 25 g. The distractors: 12.5 g comes from halving four times instead of three, counting the first weighing as a year; 300 g comes from multiplying by 1/2 × 3 = 1.5 instead of raising 1/2 to the power 3; 0.125 g comes from working out (1/2)³ = 0.125 and stopping there, without multiplying by the starting mass.
- (a) 3 — Method: the fourth root of a number is the positive value that gives that number when it is multiplied by itself four times. Working: 2 × 2 × 2 × 2 = 16, which is too small, and 3 × 3 × 3 × 3 = 9 × 9 = 81. Answer: 3. The distractors: 9 comes from taking the square root of 81 instead of its fourth root; 4.5 comes from taking the square root and then halving it, as though a fourth root were half a square root; 20.25 comes from dividing 81 by 4, treating the root's index as a divisor.
- (a) 2a²b³ — Method: a fourth root applies to every factor inside it, and taking the fourth root of a power divides that power's index by 4. Working: 2 × 2 × 2 × 2 = 16, so the fourth root of 16 is 2; 8 ÷ 4 = 2 gives a², and 12 ÷ 4 = 3 gives b³. Answer: 2a²b³. The distractors: 2a⁴b⁶ comes from halving both indices, treating every root sign as a square root; 4a²b³ comes from taking the square root of 16 while dividing the letters' indices by 4; 2a²b⁴ comes from dividing b's index by 3 instead of by 4, as though b sat under a cube root.
- (d) √13 — Method: everything under a root sign is worked out first, because a square root cannot be taken term by term across an addition. Working: 2² = 4 and 3² = 9, so the expression under the root is 4 + 9 = 13. As 13 is not a square number, the exact value is left in root form as √13. Answer: √13. The distractors: 5 comes from rooting each square separately and adding, 2 + 3, which treats the root of a sum of squares as the sum of the numbers; 13 comes from working out the sum under the root correctly and then forgetting to take the root; √5 comes from subtracting the two squares, 9 − 4, instead of adding them.
- (b) 5.1 — ∛130 lies between 5 and 6, since 125 < 130 < 216, and closer to 5 because 130 is much nearer 125 than 216. To pin down the first decimal place, test the midpoint of the tenth, 5.05: 5.05³ = 5.05 × 5.05 × 5.05 ≈ 128.79. Since 130 is greater than 128.79, ∛130 lies above 5.05, so it rounds to 5.1 rather than 5.0. Rounding down to 5.0, on the assumption that a value close to the lower bound 125 must round down, ignores that 5.05³ is already less than 130. Estimating 5.2 overshoots the true root: 5.2³ = 140.608, which is well above 130, so ∛130 cannot round to 5.2. Taking 6.0, the upper of the two whole numbers the root lies between, ignores that 130 is far nearer to 5³ = 125 than to 6³ = 216, so the root sits just above 5, not just below 6.
- (d) 15 — Use √a × √b = √(ab): √20 × √12 = √(20 × 12) = √240. Since 15² = 225 and 16² = 256, and 240 is a little closer to 225 than to 256, √240 is a little under 15.5 — in fact √240 ≈ 15.49, which rounds to 15. Adding the two roots instead of multiplying them, √20 + √12 ≈ 4.47 + 3.46 ≈ 7.94, rounds to 8, but the question asks for the product, not the sum. Multiplying 20 by 12 and stopping there, without ever taking a square root, leaves 240, which is the number under the root, not its value. Rounding each root to the nearest whole number BEFORE multiplying — √20 ≈ 4 and √12 ≈ 3 — gives 4 × 3 = 12, a cruder estimate that loses accuracy by rounding twice instead of once.
- (d) 64 — Method: each time the power increases by 1, the value doubles, since one more 2 is multiplied in. Working: 2⁶ = 2⁵ × 2 = 32 × 2 = 64. Answer: 64. (37 comes from adding the power to the value, 32 + 5, instead of multiplying by the base. 192 comes from multiplying 32 by the new power, 6, instead of by the base, 2. 34 comes from adding the base to the value, 32 + 2, instead of multiplying.)
- (a) 71 — 2.9⁴ = (2.9²)². First, 2.9² = 8.41. Then square that: 8.41² = 70.7281, since 841² = 707281 and the decimal point moves four places. To 2 significant figures this rounds to 71, because the figure after the first two significant figures (7 and 0) is a 7, which rounds the 0 up to 1. Rounding the working down instead of up — taking 70.7281 to 70 — ignores that the next figure is 5 or more. Rounding 2.9 up to 3 before doing any working at all, then computing 3⁴ = 81, uses a much cruder approximation and overshoots the true value. Rounding 8.41 all the way down to 8 before squaring, 8² = 64, rounds far too aggressively and loses the accuracy needed for 2 significant figures.
- (b) Chloe's estimate — √70 = 8.3666... to 4 decimal places. Comparing each estimate against this: Ben's 8.3 is 0.0666 away; Chloe's 8.4 is only 0.0334 away; Dan's 8.9 is 0.5334 away; Ella's 8.5 is 0.1334 away. Chloe's estimate is closest, because her method tests an actual calculation, 8.5² = 72.25, sees that it overshoots 70, and corrects slightly downward from it, rather than only comparing which perfect square is nearer. Ben's sharing-out is not a bad idea in itself — 70 sits 6 of the way along the 17 from 64 to 81, so 'about a third of the way' from 8 to 9 points at roughly 8.35 — but he then rounds that position down to 8.3, and it is that rounding, not the sharing-out, that leaves him twice as far from √70 as Chloe. Dan's claim that 81 is closer to 70 than 64 is is backwards: 70 − 64 = 6, while 81 − 70 = 11, so 64 is in fact the nearer square, which makes his estimate of 8.9 the furthest from the truth of all four. Ella's plain midpoint of 8 and 9 tests nothing at all: √70 does not sit halfway between 8 and 9, and her 8.5 lands further from √70 than Chloe's checked estimate does.
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