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.
Non-calculator
Answer key: Number worksheet — GCSE Higher
MathsUKwww.geekhero.co.uk
- (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.
- (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) 2/5 — The ratio 2:3 has 2 + 3 = 5 parts in total, and the larger plot is 3 of those parts. Since the larger plot is 18 m², each part is 18 ÷ 3 = 6 m², so the total area is 5 × 6 = 30 m² and the smaller plot is 2 × 6 = 12 m². The fraction of the total area taken up by the smaller plot is 12/30, which simplifies to 2/5. Giving the fraction for the larger plot instead of the smaller one gives 3/5. Comparing the smaller plot to the larger plot instead of to the total area gives 2/3. Assuming the two plots split the area evenly, ignoring the given ratio altogether, gives 1/2.
- (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.
- (b) 5.2 × 10⁶ — Method: write the digits as a coefficient that is at least 1 and less than 10, then count the places the decimal point moves to reach that position. Working: the digits give a coefficient of 5.2, and the decimal point travels from the end of 5,200,000 until it sits between the 5 and the 2, a move of 6 places. Answer: 5.2 × 10⁶. The distractors: 52 × 10⁵ is the same amount but not in standard form, because 52 is not less than 10; 5.2 × 10⁵ comes from counting the five zeros in 5,200,000 rather than the six places the decimal point moves; 5.2 × 10⁻⁶ comes from making the index negative because the decimal point was carried to the left.
- (b) 3/50 — Method: write the decimal over the power of ten that matches the number of digits after the point, counting every digit including a zero, then divide the numerator and the denominator by their highest common factor. Working: 0.06 has two digits after the point, so it is 6 hundredths and can be written as 6/100; the highest common factor of 6 and 100 is 2, and 6 ÷ 2 = 3 with 100 ÷ 2 = 50. Answer: 3/50. The distractors: 3/5 comes from ignoring the zero straight after the point and converting 0.6 instead, giving 6/10, which cancels to 3/5; 3/500 comes from counting three decimal places instead of two and writing 6/1000, which cancels to 3/500; 1/6 comes from putting 1 over the digits after the point, as though 0.06 meant one sixth.
- (d) 1 7/12 — Method: convert both mixed numbers to improper fractions with a common denominator, then subtract. Working: 3 1/4 = 39/12 and 1 2/3 = 20/12, so 39/12 − 20/12 = 19/12 = 1 7/12. Answer: 1 7/12. Chloe's method, subtracting whole numbers (3−1=2) and fraction parts (2/3−1/4=5/12) separately without exchanging, gives 2 5/12. 1 3/4 comes from converting 2/3 to twelfths incorrectly as 6/12 instead of 8/12, then subtracting. 8 comes from converting both mixed numbers to improper fractions correctly (13/4 and 5/3) but then subtracting numerators and denominators separately: (13−5)/(4−3) = 8/1.
- (b) £1,000, so £900 is not enough — Method: round each number to 1 significant figure, multiply to estimate the total cost, then compare the estimate with the money available. Working: 187 rounds to 200 and £4.85 rounds to £5, so the estimate is 200 × 5 = 1,000, and £1,000 is more than the £900 the school has. Answer: £1,000, so £900 is not enough. The distractors: £800 comes from cutting £4.85 down to £4 instead of rounding it up to £5, giving 200 × 4 = 800, and that estimate wrongly suggests the money stretches; £935 comes from rounding the price only and keeping 187 lunches, giving 187 × 5 = 935; £950 comes from rounding 187 to the nearest 10 rather than to 1 significant figure, giving 190 × 5 = 950.
- (c) 3/11 — Method: let a letter stand for the recurring decimal, multiply by the power of ten that shifts exactly one repeating block past the point, subtract the original equation so that the recurring tail cancels, then solve and cancel. Working: let x = 0.272727...; the repeating block is two digits long, so multiply by 100 to give 100x = 27.272727...; subtracting gives 99x = 27, so x = 27/99; the highest common factor of 27 and 99 is 9, and 27 ÷ 9 = 3 with 99 ÷ 9 = 11. Answer: 3/11. The distractors: 27/100 comes from writing the repeating block over 100 instead of over 99, forgetting that subtracting x leaves 99x rather than 100x; 3/10 comes from rounding the decimal to one place and converting 0.3; 2/9 comes from treating only the 2 as recurring and converting 0.222... instead.
- (a) £600 — Method: round the number of cakes and the price of each cake to 1 significant figure, then multiply the rounded values. Working: 187 rounds to 200, and £2.95 rounds to £3 (the digit after the first, 9, rounds the 2 up to 3), so the estimate is 200 × £3 = £600. £400 comes from rounding £2.95 down to £2 instead of up to £3, giving 200 × £2 = £400. £561 comes from rounding only the price and using the exact number of cakes, 187 × £3 = £561. £570 comes from rounding 187 to the nearest 10 as 190 instead of to 1 significant figure as 200, giving 190 × £3 = £570. Answer: £600.
- (b) £72 — One fifth of £60 = £12. New price = £60 + £12 = £72. A candidate who gives the increase instead of the new price gets £12. A candidate who subtracts the increase instead of adding it gets £60 − £12 = £48. A candidate who uses 1/4 instead of 1/5 gets £60 + £15 = £75.
- (c) 2/5 — To find a fraction of a fraction, multiply them together: 3/5 × 2/3 = 6/15, which simplifies to 2/5. Multiplying only the numerators, 3 × 2 = 6, but adding the denominators, 5 + 3 = 8, instead of multiplying them gives 6/8, which simplifies to 3/4. Using only the fraction who study French, 3/5, and ignoring that a further fraction of them also study Spanish gives 3/5. Dividing by 2/3 instead of multiplying by it, using its reciprocal 3/2, gives 3/5 × 3/2 = 9/10.
- (c) 0.0479 — Method: round each option to 2 significant figures and check which one gives 0.048. Working: for 0.0479, the first two significant figures are 4 and 7; the next digit is 9, so 7 rounds up to 8, giving 0.048. For 0.0485, the first two significant figures are 4 and 8; the next digit is 5, so 8 rounds up to 9, giving 0.049, not 0.048. 0.052 already has exactly 2 significant figures, 5 and 2, so it stays as 0.052 and does not round to 0.048 at all. 0.04 has only 1 significant figure, so it is already less precise than the 2 significant figures asked for. Answer: 0.0479.
- (a) the nearest whole number — The error intervals are 23.5 ≤ p < 24.5 and 5.5 ≤ q < 6.5. The minimum of p ÷ q is 23.5 ÷ 6.5 ≈ 3.615, and the maximum is 24.5 ÷ 5.5 ≈ 4.455. Both of these round to 4 at the nearest whole number, so the answer is guaranteed correct to the nearest whole number — but not to the nearest 0.1, since 3.615 rounds to 3.6 while 4.455 rounds to 4.5, which do not agree. Claiming the nearest 0.1 assumes every figure a calculator shows is trustworthy, without checking whether the bounds actually agree that far. Claiming only the nearest 10 badly understates how much can be guaranteed here, since both bounds already round to 4, not merely to 0. Saying no degree of accuracy can be guaranteed gives up before checking whether the bounds agree at any level at all.
- (a) −22 — Method: both multiplications are carried out before the addition, and a positive multiplied by a negative is negative. Working: (−3) × 4 = −12 and 2 × (−5) = −10, so the calculation becomes −12 + (−10) = −22. Answer: −22. The distractors: 22 comes from ignoring the minus signs and working out 3 × 4 + 2 × 5 = 22; 50 comes from working from left to right with no priority at all, giving −12 + 2 = −10 and then −10 × (−5) = 50; −2 comes from taking 2 × (−5) as +10, so that −12 + 10 = −2.
Build your own mix at the worksheet builder.