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.Work out the value of .
- 2.A padlock code is formed from 3 different digits chosen from 1, 2, 3, 4, 5 and 6 (no digit may be used twice in the same code). Work out how many different codes can be made.
- 3.A three-digit code is made using the digits 1, 2 and 3, and each digit may be used more than once. Work out how many different three-digit codes can be made.
- 4.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.
- 5.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.
- 6.Find the missing number: ▢ ÷ 15 = 24
- 7.A bag holds 5 different sweets. Noah takes 2 of the sweets out of the bag together, so the order in which he takes them does not matter. Work out how many different pairs of sweets he could take.
- 8.During one night on a mountain, four hikers each recorded the temperature they felt: Ben −6 °C, Priya −2 °C, Sam −9 °C, Alex 1 °C. Work out the difference between the coldest and the warmest of these four temperatures.
- 9.Tickets for a fairground ride are sold in packs of 6. Tokens for the dodgems are sold in packs of 10. Yusuf wants to buy the smallest number of packs of each so that he ends up with the same number of ride tickets as dodgem tokens. Work out how many ride tickets that is.
- 10.A lorry travels 105 miles in 1 hour 45 minutes. Work out its average speed in miles per hour.
- 11.Write 0.005 in standard form.
- 12.Four friends share a restaurant bill of £53.90 equally. A calculator gives each share as 13.475. Work out how much each friend should pay.
- 13.A circle has a circumference of 8π cm. Work out the exact area of the circle in terms of π.
- 14.Write 0.325 as a fraction in its simplest form.
- 15.Write ∛(x²) as a single power of x.
Answer key
- (d) 8 — Method: write $16^{3/4}$ as $(\sqrt[4]{16})^3$ — the denominator of the index gives the root, the numerator gives the power. Working: $\sqrt[4]{16} = 2$, so $16^{3/4} = 2^3 = 8$. Answer: 8. A candidate who multiplies 16 by 3/4 is treating the index as an ordinary factor and gets 12 — a fractional index is not a multiplier. A candidate who takes the square root instead of the fourth root and then cubes it works out $(\sqrt{16})^3 = 4^3$ and gets 64; the denominator 4 names a fourth root, not a square root. A candidate who takes the fourth root of 16 correctly but stops there, without cubing it, gets 2.
- (a) 120 — There are 6 choices for the first digit. The second digit must be different from the first, leaving 5 choices, and the third digit must differ from both of the first two, leaving 4 choices. By the product rule, the number of codes is 6 × 5 × 4 = 120. Allowing every digit to repeat, ignoring the 'no digit twice' rule entirely, gives 6 × 6 × 6 = 216. Adding the number of choices at each position instead of multiplying them, 6 + 5 + 4, gives 15. Treating the three chosen digits as one unordered set, rather than as digits in a fixed order on the padlock, divides by the 3! = 6 ways of arranging them: 120 ÷ 6 = 20.
- (c) 27 — Each of the 3 digits can be chosen independently for each of the 3 positions, so multiply: 3 × 3 × 3 = 27. 9 comes from multiplying only two of the three positions, 3 × 3, and forgetting the third. 6 comes from working out 3 × 2 × 1 = 6, which counts codes where digits do not repeat, but the question allows repeated digits. 3 comes from considering only one digit position.
- (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.
- (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.
- (a) 360 — The inverse of ÷ 15 is × 15, so the missing number is 24 × 15 = 360. Subtracting instead of multiplying gives 24 − 15 = 9. Dividing by 15 again instead of multiplying gives 24 ÷ 15 = 1.6. Adding instead of multiplying gives 24 + 15 = 39.
- (d) 10 — Method: list the pairs systematically, taking each sweet in turn and pairing it only with the sweets that come after it, so that no pair is written down twice. Working: numbering the sweets 1 to 5, the first sweet pairs with 4 others, the second pairs with 3 sweets that come after it, the third with 2 and the fourth with 1, so the total is 4 + 3 + 2 + 1 = 10. Answer: 10. The distractors: 20 comes from working out 5 × 4 and never halving, which counts each pair twice, once in each order; 25 comes from working out 5 × 5, which allows the same sweet to be chosen twice; 9 comes from adding the 5 choices and the 4 remaining choices instead of combining them as a selection of two.
- (a) 10 °C — Method: work out the coldest and warmest of the four temperatures, then subtract to find the difference. Working: the coldest temperature is Sam's, −9 °C, and the warmest is Alex's, 1 °C. The difference is 1 − (−9) = 1 + 9 = 10. Answer: 10 °C. 8 °C comes from working out 1 − 9 = −8 and reporting 8, dropping the negative sign on −9 instead of turning the subtraction into an addition. 3 °C comes from comparing the wrong pair, Sam's −9 °C and Ben's −6 °C, instead of the coldest and the warmest: −6 − (−9) = 3. 7 °C comes from comparing Ben's −6 °C with Alex's 1 °C, mistakenly treating Ben's reading as the coldest instead of Sam's.
- (b) 30 — Method: the smallest matching total is the lowest common multiple of the two pack sizes. Working: multiples of 6 are 6, 12, 18, 24, 30 …; multiples of 10 are 10, 20, 30 …. The lowest common multiple is 30. 60 comes from working out 6 × 10 = 60, the product of the pack sizes rather than their lowest common multiple. 16 comes from working out 6 + 10 = 16, which is not a common multiple at all. 2 is the highest common factor of 6 and 10, not a total of tickets. Answer: 30.
- (d) 60 mph — Average speed = distance ÷ time, with the time measured in hours. 45 minutes is 45/60 of an hour, which is 0.75 of an hour, so the journey takes 1.75 hours. Speed = 105 ÷ 1.75 = 60 mph. 52.5 mph rounds the time up to 2 hours, 72.4 mph writes 1 hour 45 minutes as 1.45 hours, and 183.75 mph multiplies the distance by the time instead of dividing.
- (c) 5 × 10⁻³ — Method: in standard form the coefficient must be at least 1 and less than 10, and for a number smaller than 1 the index is negative and counts the places the decimal point moves to the right. Working: the only significant digit is 5, so the coefficient is 5; moving the decimal point in 0.005 three places to the right gives 5, so the index is −3. Answer: 5 × 10⁻³. The distractors: 0.5 × 10⁻² is the same value written the wrong way, because 0.5 is smaller than 1 and so is not an allowed coefficient; 5 × 10⁻² comes from counting the two zeros after the decimal point instead of the three places the point moves; 5 × 10³ comes from taking the index as positive, which describes a number in the thousands rather than one smaller than 1.
- (b) £13.48 — Method: an amount of money is written to the nearest penny, which is 2 decimal places, so the calculator display has to be rounded to 2 decimal places. Working: 53.90 ÷ 4 = 13.475, and the digit in the third decimal place is 5, so the penny digit goes up from 7 to 8. Answer: £13.48. The distractors: £13.47 comes from chopping the third decimal place off instead of rounding with it; £13.50 comes from rounding to the nearest 10p rather than to the nearest penny; £13.40 comes from cutting the display short at 1 decimal place, which is both the wrong degree of accuracy and a truncation rather than a rounding.
- (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) 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.
- (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.
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