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.A grain of fine sand has a mass of 0.001 grams. Write this mass in standard form.
- 2.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.
- 3.A cycle route is 350 m long. A footpath runs alongside it for 3/7 of that length. Work out the length of the footpath.
- 4.A shop buys boxes of pens for £4 each. It sells each box on to a school for £3 more than it paid, and charges an extra £2 delivery fee for the whole order. A school orders 5 boxes. Work out the total cost of the order.
- 5.Write ∛(x²) as a single power of x.
- 6.Ava measures a metal rod and records its length as 15 cm, correct to the nearest centimetre. Ben measures the same rod and records its length as 14.8 cm, correct to the nearest 0.1 cm. Decide whether both records can be correct, and give a reason for your answer.
- 7.Rice is sold in a 400 g bag for £1.12 and in a 1.5 kg bag for £3.90. Work out how much less the rice in the larger bag costs per kilogram.
- 8.Work out the reciprocal of (2 + 3)
- 9.Without using a calculator, estimate √20 × √12, giving your answer to the nearest whole number.
- 10.Write down the symbol that makes this statement true: −2/3 ___ −5/6
- 11.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.
- 12.Work out 1 − 1/2 − 1/4 − 1/8 − 1/16. Give your answer as a fraction.
- 13.Work out an estimate for 2.9² + 3.1², by rounding each number to the nearest whole number.
- 14.Light travels at 2.998 × 10⁸ metres per second. A distant object in space is 3.1 × 10¹⁵ metres from Earth. Work out an estimate for the number of seconds light takes to travel from the object to Earth, by rounding each number to 1 significant figure.
- 15.Work out (−36) ÷ (−6) × 2
Answer key
- (d) 1 × 10⁻³ — Method: for a number smaller than 1 the index is negative, and it counts the places the decimal point moves to the right to leave a coefficient between 1 and 10. Working: the only significant digit is 1, so the coefficient is 1; the decimal point in 0.001 moves three places to the right to reach that 1, so the index is −3. Answer: 1 × 10⁻³. The distractors: 1 × 10³ comes from taking the index as positive, which describes one thousand grams rather than one thousandth of a gram; 0.1 × 10⁻² is the same mass written with a coefficient of 0.1, which is smaller than 1 and so is not standard form; 1 × 10⁻⁴ comes from counting the zero in front of the decimal point as well as the places after it.
- (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.
- (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.
- (b) £37 — One box costs £4 + £3 = £7. Five boxes cost 5 × £7 = £35. Adding the single £2 delivery fee gives £35 + £2 = £37. A candidate who added the £2 delivery fee to each box instead of once for the whole order worked out 5 × (£7 + £2) = 5 × £9 = £45. A candidate who forgot the £3 markup and used the shop's buying price worked out 5 × £4 + £2 = £22. A candidate who added the £3 markup only once, after multiplying the buying price by 5, worked out 5 × £4 + £3 + £2 = £25.
- (c) x⁽²⁄³⁾ — Method: a root can be written as a fractional index, with the root's index as the denominator and the power inside the root as the numerator. Working: the cube root gives a denominator of 3 and the square inside gives a numerator of 2, so ∛(x²) = x⁽²⁄³⁾. Answer: x⁽²⁄³⁾. The distractors: x⁽³⁄²⁾ comes from writing the fraction upside down, with the root's index on top; x⁽¹⁄⁶⁾ comes from treating the square as a second root and multiplying 1/3 by 1/2; x⁶ comes from multiplying the root's index by the power, 3 × 2, and keeping the result as a whole-number index.
- (b) Yes, because 14.8 cm rounds to 15 cm to the nearest cm — Method: a recorded measurement is not an exact length; it stands for every length that rounds to it, so the two records agree if one rod can produce both. Working: Ben's record of 14.8 cm to the nearest 0.1 cm means the rod is between 14.75 cm and 14.85 cm, and 14.8 is nearer to 15 than to 14, so a rod of that length is recorded as 15 cm to the nearest centimetre. Both records can therefore come from the same rod. Answer: Yes, because 14.8 cm rounds to 15 cm to the nearest cm. The distractors: the claim that 14.8 cm rounds to 15.0 cm to 1 decimal place is false, since 14.8 cm is already written to 1 decimal place and stays 14.8 cm; the claim that it rounds to 14 cm is false, because 14.8 is 0.2 away from 15 and 0.8 away from 14; the claim that the two lengths are not the same treats each record as an exact length, when each is only a rounded record of one rod.
- (c) 20p — Turn each price into the same rate before comparing. The small bag is 400 g = 0.4 kg, so it costs £1.12 ÷ 0.4 = £2.80 per kg. The large bag costs £3.90 ÷ 1.5 = £2.60 per kg. The saving is £2.80 − £2.60 = £0.20, which is 20p per kg. 2p compares the prices per 100 g rather than per kilogram, £2.78 subtracts one bag price from the other without turning either into a rate, and £2.60 is the large bag's price per kilogram rather than the saving.
- (a) 1/5 — Work out the bracket first: 2 + 3 = 5. The reciprocal of 5 is 1/5. A candidate who forgot to take the reciprocal and just gave the value of the bracket wrote 5. A candidate who took the reciprocal but made a sign error wrote −1/5. A candidate who found the reciprocal of each number separately and added them, treating reciprocal as if it distributes over addition, worked out 1/2 + 1/3 = 5/6.
- (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.
- (c) > — Method: write both fractions with a common denominator so they can be compared directly. Working: −2/3 = −4/6. Comparing −4/6 and −5/6, −4 is greater than −5, so −4/6 is greater than −5/6. Answer: −2/3 > −5/6, so the symbol is >. < comes from treating the fraction with the larger numerator and denominator, 5/6, as the greater number, ignoring that both fractions are negative. = comes from assuming the fractions must be equal because they look similar in size. ≤ combines < (false) with = (false).
- (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) 1/16 — Method: terms can only be subtracted once they share a denominator, so write every term over the largest denominator, 16, and then subtract the numerators in order from left to right. Working: 1 = 16/16, 1/2 = 8/16, 1/4 = 4/16 and 1/8 = 2/16, so the numerators give 16 − 8 − 4 − 2 − 1 = 1, over a denominator of 16. Answer: 1/16. The distractors: 1/8 comes from stopping one term early, after 16 − 8 − 4 − 2 = 2; 3/16 comes from a sign slip on the last term, adding it instead of subtracting it, which gives 2 + 1 = 3; 15/16 comes from working from the right-hand end as though the last four terms were bracketed together, so that only a single sixteenth is taken away from 1.
- (a) 18 — Method: round each number to the nearest whole number, then square each rounded number and add the results. Working: 2.9 rounds to 3 and 3.1 rounds to 3, so the estimate is 3² + 3² = 9 + 9. Answer: 18. The distractors: 36 comes from adding before squaring, working out (3 + 3)² instead of 3² + 3²; 12 comes from doubling each rounded number instead of squaring it, adding 6 and 6; 6 comes from adding the two rounded numbers and forgetting to square them at all.
- (d) 1 × 10⁷ seconds — Method: the time for a journey is the distance divided by the speed, so round each number to 1 significant figure and then divide; dividing numbers in standard form means dividing the coefficients and subtracting the indices. Working: 3.1 × 10¹⁵ rounds to 3 × 10¹⁵ and 2.998 × 10⁸ rounds to 3 × 10⁸; 3 ÷ 3 = 1 for the coefficients, and 15 − 8 = 7 for the indices. Answer: about 1 × 10⁷ seconds. The distractors: 3 × 10⁷ seconds comes from subtracting the indices correctly but leaving the coefficient as 3 instead of dividing 3 by 3; 1 × 10⁻⁷ seconds comes from dividing the speed by the distance instead of the distance by the speed; 9 × 10²³ seconds comes from multiplying the two quantities instead of dividing them, since 3 × 3 = 9 and 15 + 8 = 23.
- (c) 12 — Method: division and multiplication have equal priority, so they are carried out in the order they are written, from left to right; a negative divided by a negative is positive. Working: (−36) ÷ (−6) = 6, and then 6 × 2 = 12. Answer: 12. The distractors: 3 comes from carrying out the multiplication first, (−6) × 2 = −12 followed by (−36) ÷ (−12) = 3; −12 comes from treating a negative divided by a negative as negative, giving −6 and then −6 × 2 = −12; 6 comes from stopping at the division and never carrying out the multiplication by 2.
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