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.
Answer key: Number worksheet — GCSE Higher
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- (d) 2/25 — Method: write the decimal over 100 using its two decimal places, then simplify. Working: 0.08 = 8/100 = 2/25 (dividing both numerator and denominator by 4). Answer: 2/25. The student's fraction, 8/10, comes from ignoring the zero in the tenths column and reading 0.08 as though it were 0.8; it simplifies to 4/5. 1/125 comes from writing the decimal over 1000 instead of 100, as if there were three decimal places. 25/2 comes from flipping the correct fraction upside down.
- (c) Sam is correct — The leading zeros in 0.070268 are not significant, so the first three significant figures are 7, 0 and 2. The next digit along is 6, and since 6 is 5 or more, the third significant figure rounds up from 2 to 3, giving 0.0703. This means Sam's answer is correct. Writing 0.070 keeps only 2 significant figures, one short of what was asked. Writing 0.0702 ignores the digit 6 that follows and leaves the third figure unrounded. Writing 0.0704 rounds the third figure up twice, as if a later digit had also pushed it up.
- (c) 7/30 — Let x = 0.2333... . Because only the 3 recurs, use two multiples of x that line up the recurring part exactly: 10x = 2.333... and 100x = 23.333... . Subtracting removes the recurring tail completely: 100x − 10x = 23.333... − 2.333... = 21, so 90x = 21, giving x = 21/90 = 7/30. Treating the decimal as if it terminated after two places, writing 0.23 as 23/100, ignores that the 3 carries on forever. Misreading which digits recur — treating 0.2333... as if the block '23' repeated, giving 0.232323... — leads to x = 23/99, which is a different, larger recurring decimal from the one given. A numerator slip in the subtraction, computing 22 instead of 21, gives x = 22/90 = 11/45.
- (d) 360 km — Method: first find the kilometres per litre by dividing distance by fuel used, then multiply this rate by the new tank size. Working: 180 ÷ 6 = 30 km per litre; 30 × 12 = 360 km. Answer: 360 km. 30 km comes from finding the correct fuel consumption but stopping there, without scaling it up to the full tank. 2160 km comes from multiplying the original distance (180) by the tank size (12) directly, skipping the unit rate. 90 km comes from pairing the numbers the wrong way round: dividing the distance by the new tank size, 180 ÷ 12 = 15, and then multiplying by the original 6 litres, 15 × 6 = 90.
- (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) £45.90 — The cost is 18 × £3.45 = £62.10. Profit = £108 − £62.10 = £45.90. A candidate who does not borrow in the tenths column, doing 1 − 0 = 1 instead of borrowing to make 10 − 1 = 9 and so leaving the units as 8 − 2 = 6, gets £46.10. A candidate who adds the cost to the selling price instead of subtracting gets £108 + £62.10 = £170.10. A candidate who gives the cost instead of the profit gets £62.10.
- (d) 6,000,000 — Method: round each number to 1 significant figure, then subtract. Working: 8,340,000 rounds to 8,000,000 (1 s.f.); 1,950,000 rounds to 2,000,000 (1 s.f.); 8,000,000 − 2,000,000 = 6,000,000. Answer: 6,000,000. 6,390,000 is the exact difference, found without rounding the numbers first. 8,000,000 comes from rounding the population correctly but forgetting to subtract the capital's population at all. 6,300,000 comes from rounding 8,340,000 to the nearest hundred thousand, 8,300,000, instead of to 1 significant figure, then subtracting the correctly rounded 2,000,000.
- (d) 1/16 — Method: a negative index means the reciprocal of the power, so 2⁻⁴ is 1 divided by 2⁴. Working: 2⁴ = 2 × 2 × 2 × 2 = 16, so the value is 1/16. Answer: 1/16. The distractors: −16 comes from reading the negative index as a minus sign on the result; 16 comes from ignoring the minus sign and working out 2⁴; 1/8 comes from multiplying the base by the index, 2 × 4, and writing 1 over that product.
- (a) 7/8 — Method: write the decimal over the power of ten that matches the number of digits after the point, then divide the numerator and the denominator by their highest common factor. Working: 0.875 has three digits after the point, so it is 875 thousandths and can be written as 875/1000; the highest common factor of 875 and 1000 is 125, and 875 ÷ 125 = 7 with 1000 ÷ 125 = 8. Answer: 7/8. The distractors: 8/7 comes from cancelling correctly but writing the two parts the wrong way round; 9/10 comes from rounding 0.875 to one decimal place as 0.9 before converting; 7/80 comes from counting four decimal places instead of three and using a denominator of 10000, giving 875/10000.
- (b) 30 — Method: 12% of an amount is 12/100 of it, so find 1% by dividing by 100 and then multiply by 12. Working: 1% of 250 is 250 ÷ 100 = 2.5, and 12% is 2.5 × 12 = 30. Answer: 30 seats. The distractors: 3 comes from writing 12% as 0.012 instead of 0.12, giving 0.012 × 250 = 3; 25 comes from finding 10% of the seats and stopping there; 24 comes from counting 12 seats for each whole hundred, 12 + 12 = 24, and ignoring the remaining 50 seats.
- (b) 1.45 ≤ m < 1.55 — The flour's mass is labelled 1.5 kg, correct to the nearest 0.1 kg, so half of 0.1 kg is added to and subtracted from 1.5 kg to find the interval: 1.5 − 0.05 = 1.45 and 1.5 + 0.05 = 1.55, giving 1.45 ≤ m < 1.55. '1.4 ≤ m < 1.6' comes from taking the whole 0.1 kg as the margin either side, instead of half of it. '1.45 < m ≤ 1.55' comes from writing the inequality signs the wrong way round — the lower bound should be included and the upper bound excluded, not the other way round. '1.45 ≤ m ≤ 1.55' comes from including the upper bound, when the convention is that the upper bound is never actually reached.
- (a) the nearest whole number — The error intervals are 23.5 ≤ p < 24.5 and 5.5 ≤ q < 6.5. The minimum of p ÷ q is 23.5 ÷ 6.5 ≈ 3.615, and the maximum is 24.5 ÷ 5.5 ≈ 4.455. Both of these round to 4 at the nearest whole number, so the answer is guaranteed correct to the nearest whole number — but not to the nearest 0.1, since 3.615 rounds to 3.6 while 4.455 rounds to 4.5, which do not agree. Claiming the nearest 0.1 assumes every figure a calculator shows is trustworthy, without checking whether the bounds actually agree that far. Claiming only the nearest 10 badly understates how much can be guaranteed here, since both bounds already round to 4, not merely to 0. Saying no degree of accuracy can be guaranteed gives up before checking whether the bounds agree at any level at all.
- (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.
- (b) 18.5 ≤ T < 18.7 — Method: the error interval reaches half the rounding unit either side of the recorded value. Working: half of 0.2 is 0.1, so the interval runs from 18.6 − 0.1 to 18.6 + 0.1. Answer: 18.5 ≤ T < 18.7. (18.4 ≤ T < 18.8 comes from using the full rounding unit, 0.2, either side instead of half of it. 18.5 ≤ T ≤ 18.7 comes from including the upper bound with ≤ instead of excluding it with <. 18.6 ≤ T < 18.8 comes from treating the recorded value as the start of the interval and adding the whole rounding unit, 0.2, above it.)
- (a) 8 × 10² — Method: to multiply numbers written in standard form, multiply the coefficients and add the indices. Working: 4 × 2 = 8 for the coefficients, and −3 + 5 = 2 for the indices; 8 already lies between 1 and 10, so no adjustment is needed. Answer: 8 × 10². The distractors: 6 × 10² comes from adding the coefficients, 4 + 2, instead of multiplying them; 8 × 10⁸ comes from ignoring the minus sign and adding 3 + 5; 8 × 10⁻¹⁵ comes from multiplying the indices, −3 × 5, instead of adding them.
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