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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Answer key: Number worksheet — GCSE Higher
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- (d) x⁴ — Method: dividing two powers of the same letter subtracts the index of the divisor from the index of the term being divided. Working: six factors of x on the top and two on the bottom cancel in pairs, leaving 6 − 2 = 4 factors of x. Answer: x⁴. The distractors: x³ comes from dividing the indices, 6 ÷ 2, instead of subtracting them; x⁸ comes from adding the indices, 6 + 2, as though the powers were being multiplied; x¹² comes from multiplying the indices, 6 × 2, as though a power were being raised to a power.
- (b) 120 — Method: fill the positions on the shelf one at a time; each book placed leaves one fewer book available for the next position, and the product rule multiplies the choices. Working: there are 5 books for the first position, 4 for the second, 3 for the third, 2 for the fourth and 1 for the last, so the number of orders is 5 × 4 × 3 × 2 × 1 = 120. Answer: 120. The distractors: 25 comes from multiplying the 5 books by the 5 positions rather than multiplying the shrinking number of choices at each position; 60 comes from halving the correct product, as though each order had been counted twice in the way that pairs are; 720 comes from carrying the product one factor too far and working out 6 × 5 × 4 × 3 × 2 × 1, as though there were six books.
- (a) 7 — Dividing by 0.1 is the same as multiplying by 10, so 0.7 ÷ 0.1 = 7. Meera's answer of 0.07 comes from dividing 0.7 by 10 instead of by 0.1, the wrong way round. A candidate who confuses 0.1 with 0.01 multiplies by 100 instead of 10 and gets 70. A candidate who thinks dividing by a number less than 1 does not change the value gets 0.7.
- (b) 3/50 — Method: write the decimal over the power of ten that matches the number of digits after the point, counting every digit including a zero, then divide the numerator and the denominator by their highest common factor. Working: 0.06 has two digits after the point, so it is 6 hundredths and can be written as 6/100; the highest common factor of 6 and 100 is 2, and 6 ÷ 2 = 3 with 100 ÷ 2 = 50. Answer: 3/50. The distractors: 3/5 comes from ignoring the zero straight after the point and converting 0.6 instead, giving 6/10, which cancels to 3/5; 3/500 comes from counting three decimal places instead of two and writing 6/1000, which cancels to 3/500; 1/6 comes from putting 1 over the digits after the point, as though 0.06 meant one sixth.
- (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.
- (c) 360 — There are 5 choices for the letter. The first digit can be any of the 9 digits from 1 to 9, giving 9 choices, and the second digit must differ from the first, leaving 8 choices. By the product rule, the number of PINs is 5 × 9 × 8 = 360. Allowing the second digit to repeat the first, ignoring the 'no digit twice' rule, gives 5 × 9 × 9 = 405. Adding the numbers of choices instead of multiplying them, 5 + 9 + 8, gives 22. Treating the pair of digits as an unordered choice, rather than as a first digit followed by a second digit in a fixed order, halves the digit count: 5 × (9 × 8 ÷ 2) = 180.
- (d) 6 + 2√3 — Multiply √3 by each term in the bracket separately. First term: √3 × 2 = 2√3. Second term: √3 × √12 = √(3 × 12) = √36 = 6. Adding the two results in the order they were found, and writing the whole-number term first, gives 6 + 2√3. Adding the numbers under the root for the second term instead of multiplying them (3 + 12 = 15) gives √15 in place of 6, leading to √15 + 2√3. Multiplying √3 by the 2 but never distributing to the √12 term at all leaves just 2√3. Treating √3 × 2 as if the 3 were multiplied by the 2 inside the root, √3 × 2 → √6, while still getting the second term correct, gives 6 + √6.
- (a) 10 — √49 = 7 and ∛27 = 3, so √49 + ∛27 = 7 + 3 = 10. Treating the cube root as dividing by 3 instead of finding the cube root gives 27 ÷ 3 = 9, then 7 + 9 = 16. Multiplying the two roots instead of adding them gives 7 × 3 = 21. Ignoring the cube root symbol and using 27 as it stands gives 7 + 27 = 34.
- (b) 12 — Method: round the number of pupils to the nearest 10, divide by the number of passengers each minibus can carry, then round up because a part-full minibus still needs a whole vehicle. Working: 179 rounds to 180 (nearest 10); 180 ÷ 16 = 11.25; 11 minibuses only carry 176 passengers, so a 12th minibus is needed for the rest. Answer: 12. 11 comes from rounding 11.25 to the nearest whole number in the usual way, without checking that the leftover pupils still need transporting. 10 comes from rounding 179 down to 170 instead of to the nearest 10, 180. 180 comes from stopping after rounding the number of pupils, without dividing by the number of passengers each minibus carries at all.
- (c) 23.375 — The error intervals are 7.5 ≤ base < 8.5 and 4.5 ≤ height < 5.5. The upper bound of the area uses the upper bound of both the base and the height, then halves the product: 8.5 × 5.5 ÷ 2 = 23.375 cm². Using the lower bound of both dimensions instead, 7.5 × 4.5 ÷ 2 = 16.875, gives the lower bound of the area rather than the upper one. Multiplying the two upper bounds together but forgetting to halve for the triangle formula, 8.5 × 5.5 = 46.750, treats the triangle as if it were a rectangle. Using the given values directly without applying any bound at all, 8 × 5 ÷ 2 = 20.000, ignores that each rounded measurement has its own range of possible values.
- (d) 3/5 — The box is 8 equal shares, of which 5 are milk, so the dark chocolates take 8 − 5 = 3 shares and dark : milk = 3 : 5. The comparison asked for is dark with milk, so the milk share count is the denominator and the fraction is 3/5. 5/3 compares milk with dark, 3/8 compares the dark chocolates with the whole box rather than with the milk ones, and 8/5 comes from reading 5/8 as the ratio milk : dark.
- (d) 6 — Method: the square root of a product can be found either by multiplying first and then rooting, or by rooting each factor and multiplying the two roots together. Working: 4 × 9 = 36, and 6 × 6 = 36, so the root is 6; the same value comes from √4 × √9 = 2 × 3. Answer: 6. The distractors: 36 comes from multiplying inside the root and then leaving the root untaken; 5 comes from rooting each factor and adding the results, 2 + 3, instead of multiplying them; 18 comes from rooting the 4 only and leaving the 9 untouched, giving 2 × 9.
- (b) 1/9 — Method: let x stand for the recurring decimal, multiply by the power of ten that moves exactly one repeating block past the decimal point, subtract the original equation so that the recurring tail cancels, and solve the equation that is left. Working: let x = 0.111...; the repeating block is one digit long, so multiply by 10 to give 10x = 1.111...; subtracting the first equation from the second gives 10x − x = 1.111... − 0.111..., that is 9x = 1; dividing both sides by 9 gives x = 1/9. Answer: 1/9. The distractors: 1/10 comes from dividing by the multiplier 10 at the last step instead of by the 9 that is left in front of x; 1/11 comes from recalling the elevenths family instead of the ninths, although 1/11 = 0.0909... has a two-digit repeating block rather than a one-digit one; 11/100 comes from stopping the decimal after two digits and converting 0.11 into hundredths.
- (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) 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.
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