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.
Number worksheet — GCSE Higher
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- 1.The mass of a radioactive sample, in grams, n years after it was first weighed is modelled by M = 200 × (1/2)ⁿ. Work out the mass the model gives after 3 years.
- 2.Simplify 5⁻² × 5⁴, giving your answer as a single power of 5.
- 3.A photograph uses 4 × 10⁶ bytes of storage. A memory card holds 3.2 × 10¹⁰ bytes. Work out how many of these photographs the card can hold. Give your answer in standard form.
- 4.A number is multiplied by 4, then 8 is added, giving the result 40. Work out the number.
- 5.Write these numbers in order, starting with the largest: −0.6, 0.45, −0.15, 0.5, −0.09
- 6.Given that 5³ = 125 and 6³ = 216, use a midpoint test to estimate ∛130 to 1 decimal place.
- 7.A van has a mass of 2,000 kg, correct to 1 significant figure. Using m for the mass of the van in kilograms, write down the error interval for m.
- 8.Write 12 as a product of its prime factors.
- 9.Robert is converting 0.999... (with the 9s recurring forever) into a fraction. He lets x = 0.999... . Multiplying by 10 gives 10x = 9.999... . Subtracting x from 10x gives 9x = 9, so x = 1. Which statement correctly explains this result?
- 10.Five different books are placed in a row on a shelf. Work out how many different orders the five books can be placed in.
- 11.Write 0.325 as a fraction in its simplest form.
- 12.A café buys 18 boxes of teabags at £3.45 each, and sells all the teabags for £108 in total. Work out the café's profit.
- 13.A set of kitchen scales displays the mass of a bag of sugar as 0.63 recurring kilograms, meaning 0.636363... kg with the block '63' repeating forever. Convert this mass to a fraction of a kilogram, then work out the mass in grams, giving your answer to the nearest gram.
- 14.A square petri dish has an area of 121 mm². A scientist wants to know its side length. Work out the side length of the dish.
- 15.The highest common factor of two numbers is 4 and their lowest common multiple is 60. One of the numbers is 20. Work out the other number.
Answer key
- (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.
- (b) 5² — Method: multiplying two powers of the same base adds their indices, and a negative index is added as a negative number. Working: −2 + 4 = 2, so 5⁻² × 5⁴ = 5². Answer: 5². The distractors: 5⁶ comes from adding the sizes of the indices, 2 + 4, and ignoring the minus sign; 5⁻⁸ comes from multiplying the indices, −2 × 4, instead of adding them; 5⁻⁶ comes from subtracting the indices, −2 − 4, as though the powers were being divided.
- (a) 8 × 10³ — Method: divide the capacity of the card by the size of one photograph, dividing the coefficients and subtracting the indices, then bring the coefficient back into the range 1 to 10. Working: 3.2 ÷ 4 = 0.8 and 10 − 6 = 4, which gives 0.8 × 10⁴; a coefficient of 0.8 is smaller than 1, so the decimal point moves one place to the right and the index falls by 1. Answer: 8 × 10³. The distractors: 8 × 10⁴ comes from correcting 0.8 to 8 without reducing the index, which makes the answer ten times too large; 1.28 × 10¹⁷ comes from multiplying the two numbers instead of dividing them, since 3.2 × 4 = 12.8 and 10 + 6 = 16; 8 × 10¹⁵ comes from dividing the coefficients but adding the indices instead of subtracting them.
- (a) 8 — To undo 'multiply by 4, then add 8', reverse the operations in reverse order: subtract 8 first, then divide by 4. 40 − 8 = 32, and 32 ÷ 4 = 8, so the number is 8. A candidate who added 8 again instead of subtracting worked out 40 + 8 = 48, then 48 ÷ 4 = 12. A candidate who divided before subtracting, doing the inverse operations in the wrong order, worked out 40 ÷ 4 = 10, then 10 − 8 = 2. A candidate who multiplied instead of dividing at the last step worked out (40 − 8) × 4 = 32 × 4 = 128.
- (c) 0.5, 0.45, −0.09, −0.15, −0.6 — Method: compare the decimals by their position on a number line, remembering that with negative decimals the one closer to zero is larger. Working: 0.5 and 0.45 are positive, so they come first, with 0.5 the larger of the two. Among the negatives, −0.09 is closest to zero, then −0.15, then −0.6 is furthest from zero and so the smallest. Answer: 0.5, 0.45, −0.09, −0.15, −0.6. 0.5, 0.45, −0.15, −0.09, −0.6 swaps −0.09 and −0.15, treating the negative decimal with more digits after the point as closer to zero. −0.6, −0.15, −0.09, 0.45, 0.5 lists the numbers from smallest to largest instead of largest to smallest. 0.5, 0.45, −0.6, −0.15, −0.09 orders the negative decimals by the size of the digit (0.6 > 0.15 > 0.09) as if they were positive, instead of recognising that a bigger negative decimal is further from zero and so smaller.
- (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.
- (a) 1,500 ≤ m < 2,500 — Method: a four-digit figure written to 1 significant figure has been rounded to the nearest 1,000, so the mass lies within half of 1,000, that is 500, of the figure given. Working: 2,000 − 500 = 1,500 and 2,000 + 500 = 2,500. The lower limit is included, because 1,500 kg rounds up to 2,000 kg to 1 significant figure, while 2,500 kg rounds up to 3,000 kg, so the upper limit is not. Answer: 1,500 ≤ m < 2,500. The distractors: 1,950 ≤ m < 2,050 comes from rounding to the nearest 100 instead of to 1 significant figure; 1,000 ≤ m < 3,000 goes a whole 1,000 either side instead of half of it; 1,500 < m ≤ 2,500 has the two limits the wrong way round.
- (a) 2² × 3 — Method: divide repeatedly by the smallest prime that goes in, until 1 is reached, then write the primes used as a product with indices. Working: 12 ÷ 2 = 6, 6 ÷ 2 = 3 and 3 ÷ 3 = 1, so the primes used are 2, 2 and 3, which is written as 2² × 3. Answer: 2² × 3. The distractors: 2 × 6 comes from stopping at the first factor pair without splitting the 6, which is not prime; 2 × 3 comes from listing each prime once and losing the repeat, and it multiplies to 6 rather than 12; 2 × 3² puts the index on the wrong prime and multiplies to 18.
- (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) 120 — Method: fill the positions on the shelf one at a time; each book placed leaves one fewer book available for the next position, and the product rule multiplies the choices. Working: there are 5 books for the first position, 4 for the second, 3 for the third, 2 for the fourth and 1 for the last, so the number of orders is 5 × 4 × 3 × 2 × 1 = 120. Answer: 120. The distractors: 25 comes from multiplying the 5 books by the 5 positions rather than multiplying the shrinking number of choices at each position; 60 comes from halving the correct product, as though each order had been counted twice in the way that pairs are; 720 comes from carrying the product one factor too far and working out 6 × 5 × 4 × 3 × 2 × 1, as though there were six books.
- (a) 13/40 — Method: write the decimal over 1000 using its three decimal places, then simplify. Working: 0.325 = 325/1000 = 13/40 (dividing both numerator and denominator by 25). Answer: 13/40. 13/4 comes from writing the decimal over 100 instead of 1000, as if there were only two decimal places. 8/25 comes from rounding 0.325 down to 0.32 before converting. 3/8 comes from recalling the learned conversion 3/8 = 0.375 and matching it to 0.325 because both are three-place decimals beginning with 3, instead of converting the decimal given.
- (b) £45.90 — The cost is 18 × £3.45 = £62.10. Profit = £108 − £62.10 = £45.90. A candidate who does not borrow in the tenths column, doing 1 − 0 = 1 instead of borrowing to make 10 − 1 = 9 and so leaving the units as 8 − 2 = 6, gets £46.10. A candidate who adds the cost to the selling price instead of subtracting gets £108 + £62.10 = £170.10. A candidate who gives the cost instead of the profit gets £62.10.
- (d) 636 g — Let x = 0.636363... . Since two digits repeat, multiply by 100: 100x = 63.636363... . Subtracting removes the recurring part exactly: 100x − x = 63.636363... − 0.636363... = 63, so 99x = 63, giving x = 63/99 = 7/11 kg. Converting to grams: 7/11 × 1000 = 7000/11 = 636.3636... g, which rounds to 636 g. Treating the decimal as if it terminated, writing 0.63 as 63/100 kg, gives 630 g when multiplied by 1000 — this drops the recurring part entirely. Subtracting 10x instead of x, using 100x − 10x = 90x = 63, is the wrong power of ten for a two-digit block, giving x = 63/90 = 7/10 kg, which is 700 g. A numerator slip in the subtraction, 63 − 1 = 62 instead of 63, gives x = 62/99 kg, which is 62000/99 = 626.26... g, rounding to 626 g.
- (d) 11 mm — Method: for a square, the side length is the square root of the area. Working: 11 × 11 = 121, so the side length is 11 mm. 60.5 mm comes from working out 121 ÷ 2 = 60.5, halving the area instead of finding its square root. 242 mm comes from working out 121 × 2 = 242, doubling the area instead of finding its square root. 22 mm comes from working out 11 × 2 = 22, doubling the correct side length. Answer: 11 mm.
- (a) 12 — Method: for any two numbers, their highest common factor multiplied by their lowest common multiple equals the product of the two numbers. This holds because the HCF collects every prime factor the two numbers share, and the LCM collects every prime factor that appears in either number, so between them they use each prime factor of the two numbers exactly once — the same primes as the product. Working: 4 × 60 = 240, and 240 ÷ 20 = 12. 15 comes from working out 60 ÷ 4 = 15, dividing the wrong pair of numbers. 16 comes from working out 20 − 4 = 16, subtracting the highest common factor instead of using the product rule. 240 is 4 × 60, the product of the highest common factor and the lowest common multiple, left un-divided by 20. Answer: 12.
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