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.Three-digit numbers are made using the digits 2, 3, 4, 6, 8 and 9. No digit may be used twice in the same number. Work out how many of these three-digit numbers are odd.
- 2.A number, x, is truncated (not rounded) to 1 decimal place and the result is 6.2. Write down the error interval for x.
- 3.A cheetah runs at a steady speed of 25 metres per second. Work out this speed in kilometres per hour.
- 4.Given that 2⁵ = 32, work out 2⁶.
- 5.Light travels at 3 × 10⁸ metres per second. Work out how far light travels in 2 × 10⁻⁶ seconds. Give your answer in standard form.
- 6.Work out 100 − 4 × 5²
- 7.A spreadsheet shows that 812 − 397 = 315. Work out an estimate for 812 − 397, by rounding each number to the nearest 100, to check whether the spreadsheet's answer is reasonable.
- 8.A recipe uses 0.625 kg of flour. Write this mass as a fraction of a kilogram, in its simplest form.
- 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.A paint mix is made from white paint and blue paint in the ratio 5 : 3. Priya uses 1.2 litres of blue paint and the matching amount of white paint. Work out the total volume of the mix.
- 11.A circle has a circumference of 8π cm. Work out the exact area of the circle in terms of π.
- 12.A crowd of 8,400 people is recorded correct to the nearest 100. Work out the smallest possible number of people in the crowd.
- 13.A train journey takes 45 minutes, correct to the nearest 5 minutes. Using t for the actual time of the journey in minutes, write down the error interval for t.
- 14.Simplify (y³)⁴, giving your answer as a single power of y.
- 15.A cyclist travels 40 km, correct to the nearest 10 km, in a time of 3 hours, correct to the nearest hour. Work out the maximum possible average speed, in km/h.
Answer key
- (a) 40 — Method: a number is odd exactly when its units digit is odd, so the restricted position is filled first and the two free positions are then filled from the digits that are left, multiplying the number of choices at each stage. Working: of the six digits only 3 and 9 are odd, so there are 2 choices for the units digit; once that digit has been used, 5 digits remain for the hundreds position and then 4 remain for the tens position, so the count is 2 × 5 × 4 = 40. Answer: 40. The distractors: 120 comes from ignoring the word odd altogether and counting every three-digit number that can be made from the six digits, 6 × 5 × 4; 60 comes from filling the hundreds and tens positions first, 6 then 5, and only then allowing 2 odd digits for the units position, which overcounts because one of 3 and 9 may already have been used, giving 6 × 5 × 2; 72 comes from restricting the units digit to 3 or 9 correctly but overlooking the condition that no digit may be used twice, so all six digits are still counted as available for each of the other two positions, giving 2 × 6 × 6.
- (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.
- (b) 90 km/h — To convert metres per second to kilometres per hour, multiply by 3.6 (there are 3600 seconds in an hour and 1000 metres in a kilometre, and 3600 ÷ 1000 = 3.6): 25 × 3.6 = 90 km/h. Dividing by 3.6 instead of multiplying gives 25 ÷ 3.6 = 6.9 km/h (to 1 d.p.). Multiplying by 60 instead of 3.6, confusing the conversion from seconds to minutes with the conversion to hours, gives 25 × 60 = 1500 km/h. Multiplying by 3600 to convert seconds to hours but forgetting to convert metres to kilometres gives 25 × 3600 = 90000, which is a speed in metres per hour, not kilometres per hour.
- (d) 64 — Method: each time the power increases by 1, the value doubles, since one more 2 is multiplied in. Working: 2⁶ = 2⁵ × 2 = 32 × 2 = 64. Answer: 64. (37 comes from adding the power to the value, 32 + 5, instead of multiplying by the base. 192 comes from multiplying 32 by the new power, 6, instead of by the base, 2. 34 comes from adding the base to the value, 32 + 2, instead of multiplying.)
- (a) 6 × 10² metres — Method: distance = speed × time, so multiply the coefficients and add the indices. Working: 3 × 2 = 6 for the coefficients, and 8 + (−6) = 2 for the indices; 6 lies between 1 and 10, so the coefficient needs no adjustment. Answer: 6 × 10² metres, which is 600 metres. The distractors: 5 × 10² metres comes from adding the coefficients, 3 + 2, instead of multiplying them; 6 × 10¹⁴ metres comes from subtracting the indices, 8 − (−6), which is the rule for dividing rather than for multiplying; 6 × 10⁻⁴⁸ metres comes from multiplying the indices, 8 × (−6), instead of adding them.
- (a) 0 — Method: BIDMAS works through the index first, then the multiplication, then the subtraction. Working: 5² = 25, then 4 × 25 = 100, and finally 100 − 100 = 0. Answer: 0. The distractors: 2400 comes from working from left to right and subtracting first, giving (100 − 4) × 25 = 96 × 25 = 2400; −300 comes from multiplying before applying the index, giving (4 × 5)² = 20² = 400 and then 100 − 400 = −300; 60 comes from reading 5² as 5 × 2 = 10, so that 4 × 10 = 40 and 100 − 40 = 60.
- (b) 400 — Method: round each number to the nearest 100, then subtract the rounded values. Working: 812 rounds to 800 (nearest 100) and 397 rounds to 400 (nearest 100). 800 − 400 = 400. Answer: 400. 500 comes from rounding 397 down to 300 instead of up to the nearest 100, 400. 300 comes from rounding 812 down to 700 instead of up to the nearest 100, 800. 415 is the exact value of 812 − 397, found without rounding first, so it is not an estimate — the spreadsheet's answer of 315 is too far from the estimate of 400 to be correct.
- (c) 5/8 — Method: write the decimal over 1000 using its three decimal places, then simplify. Working: 0.625 = 625/1000 = 5/8 (dividing both numerator and denominator by 125). Answer: 5/8. 25/4 comes from writing the decimal over 100 instead of 1000, as if there were only two decimal places. 31/50 comes from rounding 0.625 to 0.62 before converting. 8/5 comes from simplifying correctly to 5/8 and then writing the fraction upside down.
- (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.
- (c) 3.2 litres — The blue paint is 3 of the 8 equal shares in the mix, that is 3/8 of the total. Three shares are 1.2 litres, so one share is 1.2 ÷ 3 = 0.4 litres. The whole mix is 8 shares: 8 × 0.4 = 3.2 litres. 2 litres is the volume of white paint, 1.92 litres divides the blue paint by the white paint's 5 shares instead of its own 3, and 9.6 litres treats the 1.2 litres as a single share.
- (b) 16π — Use the circumference formula C = 2πr to find the radius: 8π = 2πr, so r = 8π ÷ 2π = 4 cm. Then use the area formula A = πr²: A = π × 4² = 16π cm². Using C = πr instead of C = 2πr gives r = 8, and squaring that gives 64π. Finding r = 4 correctly but then substituting it back into the circumference formula instead of the area formula gives 2π × 4 = 8π. Finding r = 4 correctly but forgetting to square it in the area formula, using A = πr instead of A = πr², gives 4π.
- (a) 8,350 — Rounding to the nearest 100 means the true number can be up to half of 100, which is 50, below the recorded figure before it would round down to a lower hundred. The smallest possible number is therefore 8,400 − 50 = 8,350. Adding 50 instead of subtracting it gives 8,450, which is the upper end of the interval rather than the smallest value, and 8,450 is not itself possible because it would round up to 8,500. Subtracting a whole 100 instead of half of it gives 8,300, going too far below the recorded value. Subtracting 10 instead of half of the rounding unit gives 8,390, treating the rounding unit as 100 but the tolerance as only 10.
- (a) 42.5 ≤ t < 47.5 — Rounding to the nearest 5 minutes means the actual time can be up to half of 5 minutes, 2.5 minutes, below or above 45 before it would round to a different multiple of 5. The lower bound is 45 − 2.5 = 42.5 and the upper bound is 45 + 2.5 = 47.5. A time of exactly 47.5 minutes would round up to 50, not 45, so 47.5 is excluded while 42.5 does still round to 45. Writing 42.5 ≤ t ≤ 47.5 wrongly includes 47.5. Writing 40 ≤ t < 50 uses a whole rounding unit, 5, either side instead of half of it. Writing 44.5 ≤ t < 45.5 treats the rounding unit as 1 minute instead of 5 minutes.
- (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.
- (c) 18 — The error intervals are 35 ≤ distance < 45 and 2.5 ≤ time < 3.5. Average speed is distance ÷ time, and to make a quotient as large as possible you divide the largest possible numerator by the SMALLEST possible denominator: 45 ÷ 2.5 = 18 km/h. Using the upper bound of time as well as the upper bound of distance, 45 ÷ 3.5, gives roughly 12.9 km/h — dividing by a bigger number produces a smaller result, so this actually finds a value smaller than the true maximum. Using the lower bound of distance with the lower bound of time, 35 ÷ 2.5 = 14, mixes up which bound belongs to a maximum calculation. Using the lower bound of distance with the upper bound of time, 35 ÷ 3.5 = 10, is in fact the correct method for the MINIMUM speed, not the maximum.
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