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.
Answer key: Number worksheet — GCSE Foundation
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- (a) 6/11 — Method: add the parts of the ratio to find the total, then write the required part over the total. Working: 5 + 6 = 11 parts in total; lemons make up 6 of the 11 parts, so the fraction of lemons is 6/11, which is already in its simplest form. Answer: 6/11. 5/11 comes from finding the fraction of oranges instead of lemons. 5/6 comes from writing the ratio of oranges to lemons directly as a fraction instead of comparing lemons to the total. 6/5 comes from writing the ratio of lemons to oranges directly as a fraction instead of comparing lemons to the total.
- (c) 10 — Method: the multiplication is carried out before the subtraction. Working: 4 × 2 = 8, so the calculation becomes 18 − 8 = 10. Answer: 10. The distractors: 28 comes from working from left to right, giving (18 − 4) × 2 = 14 × 2 = 28; −10 comes from subtracting the 18 from the 8 rather than the 8 from the 18, giving 8 − 18 = −10; 12 comes from reading 4 × 2 as 4 + 2 = 6 and then working out 18 − 6 = 12.
- (a) 12 — Method: build the number one place at a time, listing systematically: fix the tens digit, then run through every units digit that is still available. Working: any of the 4 digits can go in the tens place, and once it has been used only 3 digits are left for the units place, so there are 4 × 3 = 12 numbers; listing the numbers that begin with 1 gives 12, 13 and 14, and each of the other three starting digits gives 3 numbers in the same way. Answer: 12. The distractors: 16 comes from working out 4 × 4, which allows a digit to be used twice; 8 comes from multiplying the 4 digits by the 2 places in the number instead of multiplying the choices available at each place; 6 comes from treating a number and its reverse as the same, counting only the unordered pairs of digits.
- (b) 23 kg — Method: the rounded value sits exactly in the middle of the error interval. Working: the interval 22.5 ≤ m < 23.5 stretches 0.5 either side of the rounded value, so the rounded value is 23. Answer: 23 kg. (22 kg comes from rounding the lower bound down instead of finding the middle of the interval. 22.5 kg comes from giving the lower bound itself rather than the rounded value. 23.5 kg comes from giving the upper bound itself rather than the rounded value.)
- (b) 2/5 — The statement compares girls with boys, so girls : boys = 2 : 3. The club is therefore 2 + 3 = 5 equal shares and the girls take 2 of them, so 2/5 of the members are girls. 2/3 repeats the part-to-part fraction the question gives, 3/5 is the fraction that are boys, and 1/3 comes from treating 2/3 as a fraction of the whole club and taking what is left.
- (c) 6/8 — Method: two fractions can only be compared directly when they share a denominator, so rewrite 3/4 in eighths and then compare the numerators. Working: 5/8 is already in eighths, and 3/4 = (3 × 2)/(4 × 2) = 6/8. Comparing the numerators, 6 > 5, so Oliver eats the larger share. Answer: 6/8. The distractors: 5/8 comes from skipping the conversion altogether and assuming that a bar cut into eighths must give the bigger share because it has more pieces; once both shares are written over the same denominator, 5 eighths is one eighth less than 6 eighths. 1/8 comes from working out how much more Oliver eats, 6/8 − 5/8, instead of writing down the greater of the two shares. 7/8 comes from adding 4 to the numerator and 4 to the denominator of 3/4 instead of multiplying both by 2.
- (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.
- (c) 1,000 m² — Method: round each length to 1 significant figure, then use area of a rectangle = length × width on the rounded lengths. Working: 19.6 m rounds to 20 m and 48.3 m rounds to 50 m, so the estimate is 20 × 50 = 1,000 and the area is about 1,000 m². Answer: 1,000 m². The distractors: 800 m² comes from rounding 48.3 down to 40 when the digit after its first significant figure is 8 and sends it up to 50, giving 20 × 40 = 800; 140 m² is the perimeter of the rounded rectangle, 2 × 20 + 2 × 50 = 140, not its area; 70 m² comes from adding the rounded lengths, 20 + 50 = 70, instead of multiplying them.
- (c) 5 × 10⁶ — Method: standard form is written as A × 10ⁿ, where A is at least 1 and less than 10 and n counts the places the decimal point moves. Working: the digits of 5,000,000 give a coefficient of A = 5, and the decimal point travels from the end of 5,000,000 until it sits just after the 5, a move of 6 places, so n = 6. Answer: 5 × 10⁶. The distractors: 50 × 10⁵ comes from stopping before the coefficient has been brought into range, and 50 is not less than 10, so it is not standard form; 5 × 10⁷ comes from counting the seven digits of 5,000,000 instead of the six places the decimal point moves; 5 × 10⁻⁶ comes from making the index negative because the decimal point was carried to the left, when a negative index belongs to a number smaller than 1.
- (d) 1 × 10⁷ seconds — Method: the time for a journey is the distance divided by the speed, so round each number to 1 significant figure and then divide; dividing numbers in standard form means dividing the coefficients and subtracting the indices. Working: 3.1 × 10¹⁵ rounds to 3 × 10¹⁵ and 2.998 × 10⁸ rounds to 3 × 10⁸; 3 ÷ 3 = 1 for the coefficients, and 15 − 8 = 7 for the indices. Answer: about 1 × 10⁷ seconds. The distractors: 3 × 10⁷ seconds comes from subtracting the indices correctly but leaving the coefficient as 3 instead of dividing 3 by 3; 1 × 10⁻⁷ seconds comes from dividing the speed by the distance instead of the distance by the speed; 9 × 10²³ seconds comes from multiplying the two quantities instead of dividing them, since 3 × 3 = 9 and 15 + 8 = 23.
- (b) 24 — List multiples of 8 and of 12: multiples of 8 are 8, 16, 24, 32; multiples of 12 are 12, 24, 36. The lowest number in both lists is 24, so the lighthouses next flash together after 24 minutes. Multiplying the two numbers together, 8 × 12, gives 96, which double-counts the common factor of 4 shared by 8 and 12. Working out the highest common factor instead of the lowest common multiple gives 4, far too soon a time for both lighthouses to line up again. Adding the two numbers, 8 + 12, gives 20, which is not even a multiple of either 8 or 12. So the lighthouses next flash together after 24 minutes.
- (d) 36 — Method: the fraction is acting as an operator on the whole class, so one third of the class equals 12; the operation has to be reversed, and the inverse of dividing by 3 is multiplying by 3. Working: 1/3 × (number of pupils) = 12, so the number of pupils = 12 × 3 = 36. Answer: 36 pupils. The distractors: 4 comes from applying the operator instead of reversing it, working out 12 ÷ 3 = 4; 18 comes from reading the 12 girls as two thirds of the class, giving 12 ÷ 2 × 3 = 18; 24 comes from working out the number of boys, the other two thirds, as 2 × 12 = 24 and giving that instead of the size of the class.
- (d) 27/80 — Method: write the total as a fraction of a litre, then divide by the number of glasses. Working: 1.35 = 27/20, so each glass holds 27/20 ÷ 4 = 27/80 of a litre. Answer: 27/80. 27/20 comes from converting the total correctly to a fraction but forgetting to divide by the number of glasses. 27/5 comes from multiplying the total by 4 instead of dividing. 17/50 comes from rounding 1.35 ÷ 4 to 0.34 before converting to a fraction.
- (a) They cannot both be describing the same path — Jon's measurement means the true length, l, satisfies 11.5 m ≤ l < 12.5 m. Mia's measurement means the true length satisfies 12.55 m ≤ l < 12.65 m. These two ranges do not overlap, so the two measurements cannot both be describing the same path. 'They must both be describing the same path' ignores that the two ranges do not overlap at all. 'Jon's measurement must be wrong' wrongly assumes Jon is the one at fault, when the mismatch does not show which measurement, if either, is wrong. 'Mia's measurement must be wrong' makes the same unjustified assumption in the other direction.
- (d) 9 — Method: split into two cases — the soups with no restriction, and the mushroom soup on its own — then add the totals. Working: the 2 soups other than mushroom can be paired with any of the 4 sandwiches: 2 × 4 = 8. The mushroom soup can only be paired with the cheese sandwich: 1 combination. Total = 8 + 1 = 9. Answer: 9. 12 comes from working out 3 × 4 = 12 without applying the restriction at all. 8 comes from correctly finding the 2 unrestricted soups' 8 combinations, but forgetting to add back the 1 allowed mushroom-and-cheese combination. 11 comes from taking the unrestricted total of 12 and removing only 1 mushroom combination instead of all 3 disallowed ones.
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