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) 2⁷ — When multiplying powers of the same base, the indices add: 3 + 4 = 7, so 2³ × 2⁴ = 2⁷. Multiplying the indices instead of adding them gives 3 × 4 = 12, so 2¹². Subtracting the indices instead of adding them gives 4 − 3 = 1, so 2¹. Multiplying the bases together as well as adding the indices gives 2 × 2 = 4, so 4⁷.
- (c) 3 — Method: the index counts how many times the coefficient has been multiplied by 10, which is the number of places the decimal point moves from the end of the number to just after the first significant digit. Working: 2,000 = 2 × 1,000, and 1,000 = 10 × 10 × 10, which is three tens. Answer: 3. The distractors: 4 comes from counting the four digits of 2,000 rather than the three places the decimal point moves; 2 comes from copying the coefficient 2 into the index; −3 comes from making the index negative, which would describe a number smaller than 1 rather than two thousand.
- (c) £196.35 — Method: convert both prices to pounds, multiply each by its quantity, then add the two totals. Working: 340 rolls at £0.24 each = £81.60; 85 cakes at £1.35 each = £114.75; £81.60 + £114.75 = £196.35. Answer: £196.35. £81.60 comes from working out the cost of the rolls only and forgetting to add the cost of the cakes. £114.75 comes from working out the cost of the cakes only and forgetting to add the cost of the rolls. £122.91 comes from converting 24p to £0.024 instead of £0.24, a place value error of a factor of 10 in the price of the rolls, before adding the correctly worked out cost of the cakes.
- (d) 10 — Method: picking 3 flowers from 5 leaves 2 flowers behind, so counting the different pairs that could be left out counts the bunches, and those pairs can be listed systematically. Working: number the flowers 1 to 5; the first flower can be left out alongside any of the 4 flowers after it, the second alongside any of the 3 after it, the third alongside any of the 2 after it and the fourth alongside the last one, so the number of pairs left out is 4 + 3 + 2 + 1 = 10. Answer: 10. The distractors: 60 comes from working out 5 × 4 × 3 and treating the three picks as an ordered selection when the order does not matter; 30 comes from dividing that product by 2 instead of by the 6 orders in which three chosen flowers could have been picked; 15 comes from multiplying the 5 flowers by the 3 flowers picked instead of counting the selections.
- (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.
- (b) Chloe's estimate — √70 = 8.3666... to 4 decimal places. Comparing each estimate against this: Ben's 8.3 is 0.0666 away; Chloe's 8.4 is only 0.0334 away; Dan's 8.9 is 0.5334 away; Ella's 8.5 is 0.1334 away. Chloe's estimate is closest, because her method tests an actual calculation, 8.5² = 72.25, sees that it overshoots 70, and corrects slightly downward from it, rather than only comparing which perfect square is nearer. Ben's sharing-out is not a bad idea in itself — 70 sits 6 of the way along the 17 from 64 to 81, so 'about a third of the way' from 8 to 9 points at roughly 8.35 — but he then rounds that position down to 8.3, and it is that rounding, not the sharing-out, that leaves him twice as far from √70 as Chloe. Dan's claim that 81 is closer to 70 than 64 is is backwards: 70 − 64 = 6, while 81 − 70 = 11, so 64 is in fact the nearer square, which makes his estimate of 8.9 the furthest from the truth of all four. Ella's plain midpoint of 8 and 9 tests nothing at all: √70 does not sit halfway between 8 and 9, and her 8.5 lands further from √70 than Chloe's checked estimate does.
- (a) 1/16 — Method: terms can only be subtracted once they share a denominator, so write every term over the largest denominator, 16, and then subtract the numerators in order from left to right. Working: 1 = 16/16, 1/2 = 8/16, 1/4 = 4/16 and 1/8 = 2/16, so the numerators give 16 − 8 − 4 − 2 − 1 = 1, over a denominator of 16. Answer: 1/16. The distractors: 1/8 comes from stopping one term early, after 16 − 8 − 4 − 2 = 2; 3/16 comes from a sign slip on the last term, adding it instead of subtracting it, which gives 2 + 1 = 3; 15/16 comes from working from the right-hand end as though the last four terms were bracketed together, so that only a single sixteenth is taken away from 1.
- (a) 7 — Method: BIDMAS deals with the index first, then the multiplication, then the subtraction. Working: (−2)² = (−2) × (−2) = 4, then 3 × 4 = 12, and finally 12 − 5 = 7. Answer: 7. The distractors: −17 comes from squaring only the 2 and keeping the minus sign, giving 3 × (−4) = −12 and then −12 − 5 = −17; 31 comes from multiplying before applying the index, giving (3 × (−2))² = (−6)² = 36 and then 36 − 5 = 31; −3 comes from carrying out the subtraction before the multiplication, giving 3 × (4 − 5) = 3 × (−1) = −3.
- (c) 45 — Method: count the ordered pairings with the product rule and then correct for the fact that a game between two players is the same game whichever player it is counted from. Working: each of the 10 players meets 9 opponents, so 10 × 9 = 90 pairings are counted; every game has been counted twice, once from each player's side, so the number of games is 90 ÷ 2 = 45. Answer: 45. The distractors: 90 comes from stopping at 10 × 9 and never halving, so that each game is counted once for each of its two players; 55 comes from adding 10 + 9 + 8 + ... + 1 instead of 9 + 8 + ... + 1, which counts one extra round of games; 20 comes from multiplying the 10 players by the 2 players in each game rather than pairing the players with one another.
- (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.
- (a) 900 kg/m³ — Both parts of the compound unit change. There are 100 cm in a metre, so 1 m³ = 100³ = 1 000 000 cm³, and there are 1000 g in a kilogram. So 1 g/cm³ = 1 000 000 ÷ 1000 = 1000 kg/m³, and the plastic is 0.9 × 1000 = 900 kg/m³. 900 000 kg/m³ converts the volume but leaves the mass in grams, 0.0009 kg/m³ converts the mass but leaves the volume in cm³, and 0.09 kg/m³ uses the 100 cm in a metre without cubing it.
- (a) 2/11 — Let x = 0.18 recurring, so x = 0.181818... . Since two digits repeat, multiply by 100: 100x = 18.181818... . Subtracting the original x removes the recurring part, because the digits line up exactly: 100x − x = 18.181818... − 0.181818... = 18, so 99x = 18, giving x = 18/99 = 2/11. Treating the decimal as if it terminated at two places gives 18/100 = 9/50, which is only 0.18 and drops the repeating part entirely. Subtracting 10x instead of x — using 100x − 10x = 90x = 18 — is the wrong power of ten for a two-digit repeating block, and gives x = 18/90 = 1/5. Making an arithmetic slip in the numerator, 18 − 1 = 17 instead of 18, gives 17/99.
- (d) 1.2 × 10⁵ — 4 × 3 = 12, and 3 + 1 = 4, giving 12 × 10⁴ — but 12 is not between 1 and 10, so this must be rewritten as 1.2 × 10⁵. Stopping at 12 × 10⁴ without rewriting it leaves the coefficient out of range. Rewriting 12 as 1.2 but leaving the exponent at 4 instead of increasing it to 5 gives 1.2 × 10⁴, which is ten times too small. Adding the coefficients instead of multiplying them gives 4 + 3 = 7, so 7 × 10⁴.
- (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.
- (b) 0.3, 32%, 7/20 — Method: convert every number to a decimal so they can be compared on the same scale. Working: 7/20 = 0.35, 0.3 stays as 0.3, and 32% = 0.32. Comparing 0.3, 0.32 and 0.35 in size gives the order 0.3, then 0.32, then 0.35. Answer: 0.3, 32%, 7/20. 7/20, 32%, 0.3 lists the numbers from largest to smallest instead of smallest to largest. 0.3, 7/20, 32% swaps 32% and 7/20, treating the fraction 7/20 as smaller even though 7/20 = 0.35 is bigger than 32% = 0.32. 32%, 0.3, 7/20 comes from moving the digits one place too far when converting the percentage, giving 0.032 instead of 0.32, which makes 32% look far smaller than it really is.
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