Printable · GCSE Foundation · ages 14-16
Forming and solving equations from situations worksheet — GCSE Foundation
Fifteen questions on "forming and solving equations from situations" — DfE statement A21. Print it, or print three versions so neighbours cannot copy by letter; the key gives the letter for each version.
Answer key: Forming and solving equations from situations worksheet — GCSE Foundation
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- (d) 110 — Method: form the equation 15 + 0.08m = 23.80, where m is the number of minutes, then solve for m. Working: subtract the fixed fee: 0.08m = 23.80 − 15 = 8.80. Divide by the cost per minute: m = 8.80 ÷ 0.08 = 110. Answer: 110 minutes. 1.1 comes from using 8 instead of 0.08 as the cost per minute, forgetting to convert pence to pounds. 297.5 comes from dividing the whole bill by the cost per minute without subtracting the fixed fee first, 23.80 ÷ 0.08. 485 comes from adding the fixed fee to the bill instead of subtracting it, before dividing by the cost per minute, (23.80 + 15) ÷ 0.08.
- (d) 4y — Method: the perimeter is the total distance round the outside of the shape, and a square has four sides all of the same length. Working: the four sides are y, y, y and y, so the perimeter is y + y + y + y, which is written as 4y. Answer: 4y. The distractors: 2y comes from adding only two of the four sides, as if the square were being treated like a rectangle with one length and one width; y² comes from working out the area of the square, y × y, instead of its perimeter; y + 4 comes from adding 4 to the side length instead of multiplying the side length by 4.
- (c) £2.80 — 3n + 1.50 = 9.90, so 3n = 8.40 and n = 2.80, so each notebook costs £2.80. A candidate who forgets to subtract the cost of the pen and divides the total by 3 gets n = 9.90 ÷ 3 = £3.30. A candidate who adds the cost of the pen instead of subtracting it gets 3n = 11.40 and n = £3.80. A candidate who treats the pen as a fourth notebook and divides the total by 4 gets n = 9.90 ÷ 4 = £2.48 (2 d.p.).
- (a) y + 1 — The brother's age now is y − 4. In 5 years' time this becomes y − 4 + 5 = y + 1. A candidate who adds 4 instead of subtracting it, then adds 5, gets y + 4 + 5 = y + 9. A candidate who subtracts 5 instead of adding it gets y − 4 − 5 = y − 9. A candidate who works out the brother's current age but forgets to add on the 5 years gets y − 4.
- (b) 12 years — Method: call the number of years t, add t to both ages, and form an equation from the comparison at that future time. Working: in t years the leader will be 44 + t and the scout will be 16 + t, so 44 + t = 2(16 + t); expanding gives 44 + t = 32 + 2t, and subtracting t and 32 from both sides gives t = 12. Checking: in 12 years the leader will be 56 and the scout 28, and 56 = 2 × 28. Answer: 12 years. The distractors: 6 years comes from halving the leader's present age instead, so that the scout has to reach 22, which takes 6 years; 14 years comes from halving the 28-year gap between the two ages; 28 years comes from giving the age gap itself as the number of years.
- (b) 8 — The perimeter is 2(x + (x + 3)) = 4x + 6, so 4x + 6 = 38, which gives 4x = 32 and x = 8. A candidate who forgets the '+3' and treats the rectangle as a square, solving 2(2x) = 38, gets x = 9.5. A candidate who forgets to double the sum of the sides, solving 2x + 3 = 38, gets x = 17.5. A candidate who uses 3x instead of x + 3 for the length, solving 2(x + 3x) = 38, gets x = 4.75.
- (a) 2b — Method: a total cost is the price of one item multiplied by the number of items bought. Working: each pen costs £2 and there are b pens, so the total cost is 2 × b, which is written as 2b. Answer: 2b. The distractors: b + 2 comes from adding the price to the number of pens instead of multiplying, treating £2 as a one-off charge; 2b + 2 comes from multiplying correctly and then adding the £2 a second time as though there were also a fixed charge; b/2 comes from dividing by the price instead of multiplying by it.
- (d) 8 — Method: write both present ages in terms of one letter, take four years off each of them, and form an equation from the comparison four years ago. Working: let Amelia be y, so her mother is 3y; four years ago they were y − 4 and 3y − 4, giving 3y − 4 = 5(y − 4); expanding gives 3y − 4 = 5y − 20; adding 20 and subtracting 3y from both sides gives 16 = 2y, and dividing by 2 gives y = 8. Checking: Amelia is 8 and her mother 24; four years ago they were 4 and 20, and 20 = 5 × 4. Answer: 8. The distractors: 4 comes from solving correctly but giving Amelia's age four years ago instead of her age now; 16 comes from stopping at 2y = 16 and giving 16 as her age; 2 comes from collecting the y terms by adding the 5y instead of subtracting it, giving 3y + 5y = 20 − 4 and so 8y = 16.
- (a) Subtracting 4x gives 3 = 10, which is never true. — Method: try to solve the equation as normal and see what happens. Working: subtract 4x from both sides: 4x + 3 − 4x = 4x + 10 − 4x, giving 3 = 10. This statement is false for every value of x, so the equation has no solution. Answer: subtracting 4x gives 3 = 10, which is never true. "x would have to be negative" invents a constraint on x that the equation never states. "It's true for every x" confuses this equation with an identity, where both sides would simplify to the same expression. "x = 7" misreads the false statement 3 = 10 as something to solve for x, rather than recognising it means no solution exists.
- (a) 6x + 12 — A regular hexagon has 6 equal sides, so the perimeter is 6(x + 2) = 6x + 12. A candidate who multiplies only the x-term by 6 and forgets to multiply the 2 gets 6x + 2. A candidate who multiplies only the number term by 6 and forgets to multiply the x gets x + 12. A candidate who adds 6 and 2 to make a single coefficient of x instead of expanding the brackets gets 8x.
- (a) 3x — Method: "three times as old" is a multiplication by 3, applied to the age that is known in terms of the letter. Working: the brother's age is x, and Leo's age is 3 × x, which is written as 3x. Answer: 3x. The distractors: x + 3 comes from reading "three times as old" as "three years older" and adding instead of multiplying; x − 3 comes from reading it as "three years younger" and subtracting; x/3 comes from dividing by 3 instead of multiplying, which describes the brother in terms of Leo rather than Leo in terms of the brother.
- (a) 11 — 2n − 7 = 15, so adding 7 to both sides gives 2n = 22, and dividing by 2 gives n = 11. A candidate who makes a sign error and forms the equation 2n + 7 = 15 gets 2n = 8 and n = 4. A candidate who correctly adds 7 to get 2n = 22 but forgets to divide by 2 gets n = 22. A candidate who ignores the −7 altogether and solves 2n = 15 gets n = 7.5.
- (d) 12 — Method: call the younger brother's age x, write the elder brother's age in terms of x, and form an equation from the total. Working: the elder brother is x + 6, so x + (x + 6) = 30; simplifying gives 2x + 6 = 30, subtracting 6 from both sides gives 2x = 24, and dividing by 2 gives x = 12. Checking: 12 and 18 add up to 30 and differ by 6. Answer: 12. The distractors: 18 comes from solving correctly and then giving the elder brother's age, which is not the age asked for; 15 comes from halving 30 and ignoring the 6-year difference altogether; 24 comes from taking 6 off the total, 30 − 6 = 24, and giving that as an age.
- (c) 5 — Method: call the number of years ago t, take t off both ages, and form an equation from the comparison at that time. Working: t years ago the father was 40 − t and Ethan was 10 − t, so 40 − t = 7(10 − t); expanding gives 40 − t = 70 − 7t, adding 7t to both sides gives 40 + 6t = 70, subtracting 40 gives 6t = 30, and dividing by 6 gives t = 5. Checking: five years ago the father was 35 and Ethan was 5, and 35 = 7 × 5. Answer: 5. The distractors: 35 comes from finding the right moment but giving the father's age at that time instead of the number of years; 30 comes from using Ethan's present age on the right-hand side, 40 − t = 7 × 10, which gives t = −30 and is then written as 30; 24 comes from reaching 6t = 30 correctly and subtracting 6 instead of dividing by 6.
- (b) k − 5 — Method: the width is 5 less than the length, so subtract 5 from the length. Working: length − 5 = k − 5. Answer: k − 5. 5 − k comes from subtracting in the wrong order, taking the length away from 5 instead of the other way round. k + 5 comes from adding instead of subtracting, missing that the width is smaller than the length. 5k comes from multiplying the length by 5 instead of subtracting 5 from it.
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