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.A number, x, is truncated (not rounded) to 1 decimal place and the result is 6.2. Write down the error interval for x.
- 2.Simplify (y³)⁴, giving your answer as a single power of y.
- 3.A length, L cm, has the error interval 24.5 ≤ L < 25.5. Write down the degree of accuracy to which the length was measured.
- 4.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.
- 5.3/5 of a 5/6 litre bottle of juice is poured out. Work out the exact volume poured out, in litres.
- 6.Work out (5 + 2) × 3²
- 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 90 as a product of its prime factors.
- 9.A company's turnover is 5.6 × 10⁷ pounds, spread evenly across 3.5 × 10² shops. Work out the average turnover per shop, in standard form.
- 10.Simplify x⁷ × x⁴, giving your answer as a single power of x.
- 11.Work out an estimate for 8,900 ÷ 29, by rounding each number to 1 significant figure.
- 12.Rationalise the denominator of 10/(4 − √6), giving your answer in its simplest form.
- 13.Work out 35% of 180.
- 14.By considering fourth powers, work out which two consecutive integers ⁴√200 lies between.
- 15.At a bus station one bus leaves every 6 minutes and another leaves every 9 minutes. The two buses leave together at the start of the day. Work out how long it is until they next leave together.
Answer key
- (b) 6.2 ≤ x < 6.3 — Truncating simply cuts off the digits after the required decimal place instead of rounding them, so every value from 6.2 up to (but not reaching) 6.3 truncates to 6.2. This gives the error interval 6.2 ≤ x < 6.3, with no allowance made on the lower side because truncation never rounds a smaller value up into this interval. Using 6.15 ≤ x < 6.25 applies the rounding rule of going half a unit either side, which does not apply to truncation. Writing 6.1 < x ≤ 6.2 puts the interval below 6.2 instead of above it. Writing 6.2 ≤ x ≤ 6.3 wrongly includes 6.3, which truncates down to itself, not to 6.2.
- (a) y¹² — Method: when a power is raised to another power, multiply the two indices. Working: (y³)⁴ means y³ × y³ × y³ × y³, which is four lots of three y's multiplied together, so the index is 3 × 4 = 12 and (y³)⁴ = y¹². y⁷ comes from adding the indices, 3 + 4 = 7, which is the rule for multiplying two separate powers, not for raising a power to a power. y⁸¹ comes from working out 3⁴ = 81 and using that as the index, raising the inner index to the outer power instead of multiplying the two indices. 12y comes from multiplying the indices to make 12 but then treating y as a coefficient instead of a power. Answer: y¹².
- (a) to the nearest centimetre — Method: the error interval of a rounded measurement runs from half a unit below the stated value to half a unit above it, so the width of the interval is one whole unit of the accuracy used. Working: the interval runs from 24.5 to 25.5, a width of 25.5 − 24.5 = 1, so the unit of accuracy is 1 cm; the stated value is the midpoint, 25 cm, and 25 correct to the nearest centimetre is exactly what gives 24.5 ≤ L < 25.5. Answer: to the nearest centimetre. To the nearest 0.5 cm comes from reading the half-unit, 0.5, as the accuracy itself instead of doubling it back to the full unit. To 1 decimal place comes from seeing the bounds written with one decimal place and taking that as the accuracy, but the bounds of a value given to 1 decimal place would be only 0.05 either side. To the nearest 10 cm comes from confusing the size of the value, about 25, with the unit it was rounded to; rounding to the nearest 10 cm would give an interval 5 cm either side of the stated value.
- (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.
- (a) 1/2 — To find a fraction of an amount, multiply the fractions together: 3/5 × 5/6 = 15/30, which simplifies to 1/2 litre. Adding the fractions instead of multiplying them, using a common denominator of 30, gives 18/30 + 25/30 = 43/30, a value greater than the whole bottle. Dividing by 5/6 instead of multiplying by it, using its reciprocal 6/5, gives 3/5 × 6/5 = 18/25. Multiplying 5/6 by itself instead of by 3/5 gives 25/36.
- (b) 63 — 5 + 2 = 7, then 3² = 9, then 7 × 9 = 63. Ignoring the brackets and applying BIDMAS as if the expression were unbracketed gives 3² = 9, then 2 × 9 = 18, then 5 + 18 = 23. Squaring the bracket instead of the 3 gives 7² = 49, then 49 × 3 = 147 — the power belongs to the 3 alone. Multiplying by 3 before squaring the whole product gives 7 × 3 = 21, then 21² = 441.
- (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) 2 × 3² × 5 — Method: divide repeatedly by the smallest prime number until only prime factors remain. Working: 90 ÷ 2 = 45, 45 ÷ 3 = 15, 15 ÷ 3 = 5, and 5 is prime, so 90 = 2 × 3 × 3 × 5, written as 2 × 3² × 5. 2 × 3 × 15 stops before the 15 is broken down into 3 × 5, so it is not fully factorised. 3 × 3 × 10 stops before the 10 is broken down into 2 × 5. 2 × 45 stops after only one division. Answer: 2 × 3² × 5.
- (d) 1.6 × 10⁵ — 5.6 ÷ 3.5 = 1.6, and 7 − 2 = 5, so the average turnover per shop is 1.6 × 10⁵ pounds. Multiplying the exponents instead of subtracting them gives 7 × 2 = 14, so 1.6 × 10¹⁴. Adding the exponents instead of subtracting them gives 7 + 2 = 9, so 1.6 × 10⁹. Subtracting the coefficients instead of dividing them gives 5.6 − 3.5 = 2.1, so 2.1 × 10⁵.
- (a) x¹¹ — Method: when multiplying powers of the same base, add the indices. Working: 7 + 4 = 11, so x⁷ × x⁴ = x¹¹. x²⁸ comes from multiplying the indices, 7 × 4 = 28, instead of adding them. x³ comes from working out 7 − 4 = 3, which is the rule for dividing powers, not multiplying them. 11x comes from adding the indices to make 11 but then treating x as a coefficient instead of a power. Answer: x¹¹.
- (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) 4 + √6 — Multiply top and bottom by the conjugate, 4 + √6. The denominator becomes (4 − √6)(4 + √6) = 4² − (√6)² = 16 − 6 = 10. The numerator becomes 10 × (4 + √6) = 40 + 10√6. So the fraction is (40 + 10√6)/10 = 4 + √6, since both terms in the numerator divide by 10. Distributing the conjugate to only the whole-number term of the numerator, and forgetting the surd term entirely, leaves just 4. Rationalising by multiplying the numerator by the conjugate but leaving the ORIGINAL denominator's sign unchanged instead of squaring it lands on 4 − √6, with the surd's sign never actually flipping to positive. Dividing only the whole-number part of the numerator by 10 and forgetting to divide the surd term too leaves 4 + 10√6.
- (b) 63 — 10% of 180 = 18, so 5% = 9. 35% = (3 × 18) + 9 = 54 + 9 = 63. A candidate who uses 25% instead of 35% gets 45. A candidate who doubles 35% to get 70% by mistake gets 126. A candidate who subtracts 35 from 180 instead of finding a percentage gets 145.
- (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.
- (a) 18 minutes — Method: the buses leave together again after a number of minutes that is a multiple of both intervals, and the first such time is the lowest common multiple. Working: the multiples of 6 are 6, 12, 18, 24 … and the multiples of 9 are 9, 18, 27 … The first value in both lists is 18, which is 6 × 3 and 9 × 2. Answer: 18 minutes. The distractors: 54 minutes comes from multiplying 6 by 9, which does give a common multiple but not the lowest one; 3 minutes is the highest common factor of 6 and 9 rather than their lowest common multiple; 15 minutes comes from adding the two intervals together.
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