Printable · GCSE Foundation · ages 14-16
Number worksheet — GCSE Foundation
Fifteen questions across the number statements at Foundation 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 Foundation
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- (c) 2000 mm — Put both lengths into the same unit first. There are 1000 mm in a metre, so the ribbon is 2.4 × 1000 = 2400 mm, and there are 10 mm in a centimetre, so the piece cut off is 40 × 10 = 400 mm. The length left is 2400 − 400 = 2000 mm. 2360 mm subtracts 40 mm instead of 400 mm, 200 mm works in centimetres and then labels the result as millimetres, and 2800 mm adds the piece that was cut off instead of subtracting it.
- (a) 11 — Method: list every possible total from the smallest to the largest, and count how many different values there are. Working: the smallest total is 1+1=2 and the largest is 6+6=12, and every whole number total from 2 to 12 is possible: 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 — that is 11 different totals. Answer: 11. 36 comes from counting the number of possible dice outcomes (6×6) instead of the number of different totals. 10 comes from listing the totals but missing one from the ends of the list, for example starting at 3 instead of 2. 6 comes from counting only the number of different scores on one die, not the totals of both dice together.
- (a) 1,500 ≤ m < 2,500 — Method: a four-digit figure written to 1 significant figure has been rounded to the nearest 1,000, so the mass lies within half of 1,000, that is 500, of the figure given. Working: 2,000 − 500 = 1,500 and 2,000 + 500 = 2,500. The lower limit is included, because 1,500 kg rounds up to 2,000 kg to 1 significant figure, while 2,500 kg rounds up to 3,000 kg, so the upper limit is not. Answer: 1,500 ≤ m < 2,500. The distractors: 1,950 ≤ m < 2,050 comes from rounding to the nearest 100 instead of to 1 significant figure; 1,000 ≤ m < 3,000 goes a whole 1,000 either side instead of half of it; 1,500 < m ≤ 2,500 has the two limits the wrong way round.
- (b) 5/9 — Add the parts of the ratio to find how many equal shares make up the whole: 4 + 5 = 9 shares. The blue counters take 5 of those 9 shares, so 5/9 of the counters are blue. 4/9 is the fraction that are red, 4/5 is the ratio copied straight down as a fraction, and 5/4 compares the blue counters with the red counters instead of with the whole box.
- (c) (12 + 3π) cm — The perimeter of a quarter-circle is made up of two straight radii plus a quarter of the circumference. The two radii give 2 × 6 = 12 cm, and a quarter of the circumference is (1/4) × 2 × π × 6 = 3π cm, so the total perimeter is (12 + 3π) cm. Giving only the curved part, 3π cm, forgets the two straight edges entirely. Using the full circumference, 2 × π × 6 = 12π, instead of a quarter of it gives (12 + 12π) cm. Including only one radius instead of two gives (6 + 3π) cm.
- (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.
- (d) 17 — Without the restriction there would be 5 × 4 = 20 combinations. The dragon piece can only be paired with the gold token, so of the 4 tokens, 3 are not allowed with the dragon piece, giving 20 − 3 = 17 valid combinations. 20 comes from ignoring the restriction completely. 19 comes from subtracting only 1 of the 3 invalid dragon combinations instead of all 3, 20 − 1 = 19. 16 comes from multiplying only the 4 non-dragon pieces by the 4 tokens, 4 × 4 = 16, and forgetting to add back the one valid combination of the dragon piece with the gold token.
- (b) 1/2 — To multiply fractions, multiply the numerators together and multiply the denominators together: 2 × 3 = 6 and 3 × 4 = 12, giving 6/12, which simplifies to 1/2. Adding the fractions instead of multiplying them, using a common denominator of 12, gives 8/12 + 9/12 = 17/12. Dividing by 3/4 instead of multiplying by it, so multiplying by its reciprocal 4/3, gives 2/3 × 4/3 = 8/9. Multiplying only the numerators, 2 × 3 = 6, and keeping the first denominator, 3, unchanged gives 6/3 = 2.
- (b) 11 — Method: roots and the multiplication are worked out before the addition and subtraction, and what is left is then worked through from left to right. Working: √144 = 12, √25 = 5 and 2 × 3 = 6, so the calculation becomes 12 − 6 + 5, which gives 6 + 5 = 11. Answer: 11. The distractors: 1 comes from carrying out the addition before the subtraction, giving 12 − (6 + 5) = 12 − 11 = 1; 35 comes from working from left to right with no priority, giving 12 − 2 = 10, then 10 × 3 = 30 and 30 + 5 = 35; 7 comes from combining the two roots as √(144 + 25) = √169 = 13 and then subtracting the product, giving 13 − 6 = 7.
- (a) 3200 — Method: round each number to 1 significant figure, then multiply the rounded numbers. Working: 37 rounds to 40 (the digit after the first, 7, rounds the 3 up to 4), and 84 rounds to 80, so the estimate is 40 × 80 = 3200. 2400 comes from rounding 37 down to 30, keeping the first digit as it is instead of letting the 7 round it up, giving 30 × 80 = 2400. 3108 comes from multiplying the exact numbers, 37 × 84, without rounding either of them first. 120 comes from adding the rounded numbers, 40 + 80 = 120, instead of multiplying them. Answer: 3200.
- (c) 3 : 5 — The club is 8 equal shares and the girls take 3 of them, so the boys take 8 − 3 = 5 shares. A ratio compares one part with the other part, so girls : boys = 3 : 5. Writing 3 : 8 compares the girls with the whole club rather than with the boys, 5 : 3 puts the two parts in the wrong order, and 8 : 3 compares the whole club with the girls.
- (a) 6 — Method: list the factors of each number and compare them. Working: the factors of 18 are 1, 2, 3, 6, 9, 18; the factors of 24 are 1, 2, 3, 4, 6, 8, 12, 24. The only option that appears in both lists is 6. 8 is a factor of 24 but not of 18. 9 is a factor of 18 but not of 24. 12 is a factor of 24 but not of 18. Answer: 6.
- (b) 9/4 — Method: write the whole part as a fraction with the same denominator, then add the fraction part to it. Working: there are 4 quarters in 1 whole, so 2 wholes are 2 × 4 = 8 quarters; adding the 1 quarter that is already there gives 8 + 1 = 9 quarters over a denominator of 4. Answer: 9/4. The distractors: 3/4 comes from adding the whole number to the numerator, as 2 + 1, instead of multiplying it by the denominator first; 7/4 comes from multiplying correctly but then subtracting the numerator, as 2 × 4 − 1; 5/4 comes from multiplying the numerator by the denominator instead of the whole number, as 1 × 4 + 1.
- (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) 3 — Method: solving an index equation like this means finding how many factors of the base multiply together to give the number on the right. Working: 4¹ = 4, 4² = 16 and 4³ = 64, so three factors of 4 are needed. Answer: 3. The distractors: 4 comes from listing 4, 16 and 64 and counting the base itself as a step, which gives one more than the index; 6 comes from solving the equation with 2 as the base instead of 4, since 2⁶ = 64; 16 comes from dividing 64 by 4, treating the index as an instruction to divide.
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