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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Number worksheet — GCSE Higher
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- 1.Priya adds 3.6 and 0.45 on paper and writes down 3.65 as her answer. Work out the correct value of 3.6 + 0.45.
- 2.Work out an estimate for 8,900 ÷ 29, by rounding each number to 1 significant figure.
- 3.Write 0.325 as a fraction in its simplest form.
- 4.A florist has 60 red roses and 84 white roses. She wants to make identical bunches using all the flowers, with the greatest possible number of bunches. Work out how many red roses will be in each bunch.
- 5.Work out an estimate for 588 ÷ 31, by rounding each number to 1 significant figure.
- 6.Work out 2 × 3 × 5 + 1 and decide whether the result is a prime number.
- 7.Simplify (3a²)² × (2a)³
- 8.A rectangular patio measures 90 cm by 120 cm. Ben wants to cover it exactly with identical square tiles, as large as possible, with no tiles cut. Work out the side length of the largest square tile he can use.
- 9.Simplify x⁽³⁄⁴⁾ ÷ x⁽¹⁄⁴⁾
- 10.A car manufacturer offers a car in 6 colours and 4 trim levels. Two colour-and-trim combinations are not available: red with sport trim, and white with sport trim. Work out how many different colour-and-trim combinations are available.
- 11.The population of a planet is 8 × 10⁹. Write this population as an ordinary number.
- 12.Simplify (5² × 2³)³ ÷ (5³ × 2²)²
- 13.n is a whole number and 2 ≤ n < 6. Write down all the possible values of n.
- 14.Oliver drives 95 km at an average speed of 50 km/h. Work out an estimate for the time the journey takes, by rounding the distance to the nearest 100 km.
- 15.Priya has 24 red beads and 36 blue beads. She makes identical bracelets, using every bead and with none left over. Work out the greatest number of bracelets she can make.
Answer key
- (b) 4.05 — Method: line up the decimal points (or place value columns) before adding. Working: 3.60 + 0.45 = 4.05. Answer: 4.05. 3.65 is Priya's answer, from adding the digits without lining up the place value columns, which effectively treats 0.45 as 0.05. 4.5 comes from rounding both numbers up first, 3.6 to 4 and 0.45 to 0.5, and adding those instead of adding the exact values. 0.81 comes from adding the digits 36 and 45 together to get 81, then placing the decimal point in the wrong position.
- (a) 300 — Method: round each number to 1 significant figure, then divide the rounded values. Working: 8,900 rounds to 9,000 and 29 rounds to 30; cancelling a zero from each gives 900 ÷ 3. Answer: 300. The distractors: 450 comes from rounding 29 down to 20 instead of to the nearest ten, giving 9,000 ÷ 20; 3,000 comes from rounding 29 to 3 rather than to 30, a place-value slip that divides by a number ten times too small; 307 is the exact quotient rounded to the nearest whole number, worked out in full when the question asks for an estimate.
- (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.
- (d) 5 — Method: the greatest number of identical bunches is the highest common factor of the two flower totals; then divide the red roses by that number of bunches. Working: 60 = 2² × 3 × 5 and 84 = 2² × 3 × 7, so their highest common factor is 2² × 3 = 12. That means 12 bunches, and 60 ÷ 12 = 5 red roses in each. 7 is the number of white roses in each bunch, since 84 ÷ 12 = 7, not red roses. 12 is the number of bunches itself, not the number of red roses in one bunch. 20 comes from working out 60 ÷ 3 = 20, dividing by only part of the highest common factor. Answer: 5.
- (a) 20 — Method: round each number to 1 significant figure, then divide the rounded values. Working: 588 rounds to 600 (1 s.f.) and 31 rounds to 30 (1 s.f.). 600 ÷ 30 = 20. Answer: 20. 17 comes from cutting 588 down to 500, keeping the leading digit as it stands instead of rounding it up to 1 significant figure, 600, then dividing by the correctly rounded 30. 200 comes from misreading the rounded divisor 30 as 3, giving 600 ÷ 3 instead of 600 ÷ 30. 19 is the exact value of 588 ÷ 31 rounded to the nearest whole number, found without rounding the numbers first.
- (d) 31, which is prime — Method: work out the value, remembering that multiplication comes before addition, then test it for primality by dividing by each prime up to its square root. Working: 2 × 3 × 5 = 30, so the value is 30 + 1 = 31. Since 6² = 36 is larger than 31, only 2, 3 and 5 need testing: 31 is odd, 31 ÷ 3 leaves a remainder of 1, and 31 does not end in 0 or 5. It therefore has exactly two factors, 1 and itself. Answer: 31, which is prime. The distractors: 30, which is not prime comes from working out 2 × 3 × 5 and forgetting to add the 1; the claim that 31 = 1 × 31 makes it non-prime comes from treating any factor pair as proof, forgetting that a prime is allowed the pair 1 and itself; the claim that 31 is a multiple of 3 comes from assuming that a number containing the digit 3 divides by 3, when in fact 31 ÷ 3 leaves a remainder.
- (d) 72a⁷ — Method: a power outside brackets applies to every factor inside them, and multiplying two powers of the same letter adds their indices. Working: (3a²)² = 3² × a⁴ = 9a⁴, and (2a)³ = 2³ × a³ = 8a³. Multiplying the two results gives 9 × 8 = 72 for the number and 4 + 3 = 7 for the index of a. Answer: 72a⁷. The distractors: 36a⁷ comes from squaring the 2 in (2a)³ instead of cubing it, giving 4a³ and then 9 × 4; 72a¹² comes from multiplying the indices 4 and 3 when the two terms are multiplied, instead of adding them; 17a⁷ comes from adding the coefficients 9 and 8 rather than multiplying them.
- (a) 30 cm — The tile's side length must be a common factor of 90 and 120. The factors of 90 include 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90; the factors of 120 include 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120. The highest number common to both lists is 30, so the largest square tile has a side length of 30 cm. Picking 15 cm, a common factor but not the largest, gives tiles that are smaller than necessary. Picking 10 cm, also a common factor but smaller still, wastes even more of the possible tile size. Working out the lowest common multiple instead of the highest common factor gives 360 cm, a length far bigger than either side of the patio. So the largest square tile Ben can use has a side length of 30 cm.
- (d) x⁽¹⁄²⁾ — Method: dividing two powers of the same letter subtracts the index of the divisor from the index of the term being divided, and fractional indices are subtracted like any other fractions. Working: 3/4 − 1/4 = 2/4, which simplifies to 1/2, so the result is x⁽¹⁄²⁾. Answer: x⁽¹⁄²⁾. The distractors: x comes from adding the indices, 3/4 + 1/4 = 1, as though the powers were being multiplied; x³ comes from dividing the indices, so that 3/4 divided by 1/4 gives 3; x⁽³⁄¹⁶⁾ comes from multiplying the indices, 3/4 × 1/4.
- (c) 22 — Without restriction there are 6 × 4 = 24 combinations. Two specific combinations are not available, so subtract 2: 24 − 2 = 22. 24 comes from ignoring the restriction completely. 23 comes from subtracting only 1 of the 2 excluded combinations. 18 comes from removing the whole sport trim level, 6 × 3 = 18, instead of removing just the two excluded combinations.
- (c) 8,000,000,000 — Method: a power of ten is 1 followed by that many zeros, so 8 × 10⁹ is 8 multiplied by 1 followed by nine zeros. Working: 10⁹ = 1,000,000,000, and 8 × 1,000,000,000 puts an 8 in front of those nine zeros. Answer: 8,000,000,000. The distractors: 800,000,000 comes from writing nine digits altogether instead of nine zeros after the 8; 720 comes from reading 10⁹ as 10 × 9 = 90 and then working out 8 × 90; 0.000000008 comes from treating the index as negative and carrying the decimal point nine places to the left.
- (b) 2⁵ — Method: a power outside brackets multiplies the index of every factor inside them, and dividing powers of the same base subtracts their indices. Working: (5² × 2³)³ = 5⁶ × 2⁹ and (5³ × 2²)² = 5⁶ × 2⁴. Dividing gives 5⁶⁻⁶ × 2⁹⁻⁴, and since 5⁰ = 1 the whole expression reduces to 2⁵. Answer: 2⁵. The distractors: 2² comes from adding the outside index to each inside index instead of multiplying, which gives 5⁵ × 2⁶ over 5⁵ × 2⁴; 2¹³ comes from adding the indices of 2 when dividing, 9 + 4, instead of subtracting them; 2 comes from applying the outside power to the first factor inside each bracket only, leaving 5⁶ × 2³ over 5⁶ × 2².
- (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.
- (d) 2 hours — Method: the time for a journey is the distance divided by the speed, so round the distance first and then divide by the speed. Working: 95 km rounds to 100 km, and 100 ÷ 50 = 2; the speed is in kilometres per hour, so the answer is a number of hours. Answer: 2 hours. The distractors: 1 hour comes from rounding the distance down to 50 km to match the speed, so that the journey looks like a single hour of driving; 30 minutes comes from dividing the speed by the distance, 50 ÷ 100, instead of the distance by the speed; 1 hour 54 minutes is the exact time, 95 ÷ 50 = 1.9 hours, worked out in full when the question asks for an estimate.
- (d) 12 — Method: if the bracelets are identical and no beads are left over, the number of bracelets must divide exactly into both totals, so it is the highest common factor of 24 and 36. Working: 24 = 2³ × 3 and 36 = 2² × 3²; taking the lower index of each shared prime gives 2² × 3 = 4 × 3 = 12. Each bracelet then has 2 red beads and 3 blue beads. Answer: 12. The distractors: 6 comes from taking each shared prime once rather than at its lower index, giving 2 × 3, which is a common factor but not the highest; 72 is the lowest common multiple of 24 and 36, from taking the higher index of each prime instead of the lower; 60 comes from adding the two bead totals instead of looking for a common factor.
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