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.
Forming and solving equations from situations worksheet — GCSE Foundation
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- 1.A rectangle has width x cm. Its length is 3 cm more than its width. The perimeter of the rectangle is 38 cm. Form an equation and solve it to find x.
- 2.Oliver buys b pens. Each pen costs £2. Write down an expression for the total cost, in pounds.
- 3.A square has sides of length y cm. Write down an expression for the perimeter of the square.
- 4.Harry is 14 years old and Mia is 8 years old. Will Harry ever be exactly twice as old as Mia? Give a reason for your answer.
- 5.A scout leader is 44 years old and one of the scouts is 16 years old. Work out how many years it will be until the leader is exactly twice as old as the scout.
- 6.A cinema sells adult tickets for £10 and child tickets for £6. A family group buys 9 tickets in total, spending £74. Work out how many adult tickets were bought.
- 7.Priya buys b balloons for a party. Each balloon costs 15p. Write down an expression for the total cost, in pence.
- 8.The mean of four numbers is 12.5. Three of the numbers are 8, 15 and 19. Form an equation using n for the fourth number, and solve it to find n.
- 9.A regular hexagon has sides of length (x + 2) cm. Write down an expression, in terms of x, for the perimeter of the hexagon.
- 10.Three consecutive integers add up to 72. Using n for the smallest integer, form an equation and solve it to find the smallest of the three integers.
- 11.A rectangle has length k cm and width 5 cm less than its length. Write down an expression for the width of the rectangle, in cm.
- 12.Sophie's mother is 34 years old. In x years' time she will be 50 years old. Work out the value of x.
- 13.Yusuf has £40 saved and adds £6 each week. His sister Aisha has £10 saved and adds £11 each week. After how many weeks will they have saved the same amount?
- 14.A number, n, is doubled and then 7 is subtracted. The result is 15. Form an equation and solve it to find n.
- 15.A taxi charges a fixed fee of £3 plus £2 for every mile travelled. The journey is m miles. Write down an expression for the total cost, in pounds.
Answer key
- (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) 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) No — that moment has already passed — Method: call the number of years from now x, add x to both ages, form the equation from the comparison and then interpret the value of x that comes out. Working: in x years Harry will be 14 + x and Mia will be 8 + x, so 14 + x = 2(8 + x); expanding gives 14 + x = 16 + 2x, and subtracting x and 16 from both sides gives x = −2. A negative value of x places the moment two years in the past, when Harry was 12 and Mia was 6 and 12 = 2 × 6, so it is not something still to come. Answer: no — that moment has already passed. The distractors: the claim that it has never happened and never will comes from reaching x = −2 and reading a negative number of years as no solution at all, when x = −2 does not say that no such moment exists but says where it is — two years before now; the claim that it happens when Harry is 16 comes from doubling Mia's present age, 2 × 8 = 16, and reading that as the age Harry has to reach; the claim that it happens when Harry is 20 comes from expanding 2(8 + x) as 8 + 2x, doubling only the x, which gives x = 6.
- (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.
- (c) 5 — Method: let a be the number of adult tickets, so the number of child tickets is 9 − a. Form the equation 10a + 6(9 − a) = 74. Working: expand the bracket: 10a + 54 − 6a = 74, so 4a = 20, giving a = 5. Answer: 5 adult tickets. 4 comes from correctly finding the number of child tickets but reporting it instead of the number of adult tickets asked for. 1.25 comes from a sign error expanding the bracket, writing 10a + 54 + 6a = 74 instead of subtracting, which gives 16a = 20. 7.4 comes from dividing the total takings by the adult ticket price only, ignoring the children entirely, 74 ÷ 10.
- (d) 15b — Method: each balloon costs 15p, so b balloons cost 15 × b pence. Working: 15 × b = 15b. Answer: 15b. b + 15 comes from treating the 15p as a one-off addition rather than a rate applied to every balloon. b − 15 comes from misreading the situation as a discount rather than a cost per balloon. b/15 comes from dividing instead of multiplying, swapping the two operations.
- (d) 8 — Since the mean of the four numbers is 12.5, their total is 4 × 12.5 = 50. The three known numbers add up to 8 + 15 + 19 = 42, so n = 50 − 42 = 8. A candidate who multiplies the mean by 3 instead of 4 gets a total of 37.5, giving n = 37.5 − 42 = −4.5. A candidate who forgets to subtract the three known numbers and gives the total itself as n states n = 50. A candidate who subtracts the mean from the total of the three known numbers instead of the other way round gets n = 42 − 12.5 = 29.5.
- (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.
- (d) 23 — Method: let the smallest integer be n, so the three consecutive integers are n, n + 1 and n + 2. Form the equation n + (n + 1) + (n + 2) = 72. Working: simplify the left side: 3n + 3 = 72, so 3n = 69, giving n = 23. Answer: 23. 24 comes from dividing 72 by 3 directly, 72 ÷ 3 = 24, which finds the middle integer rather than realising the three numbers differ. 25 comes from correctly finding n = 23 but reading off the largest integer, n + 2, instead of the smallest as asked. 21 comes from dividing first and subtracting after, (72 ÷ 3) − 3, instead of subtracting 3 before dividing by 3.
- (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.
- (a) 16 — Method: the future age is the present age plus the number of years, so form that equation and solve it. Working: 34 + x = 50, and subtracting 34 from both sides gives x = 50 − 34 = 16. Answer: 16. The distractors: 84 comes from adding the two ages instead of subtracting, 34 + 50; 50 comes from giving her future age rather than the number of years that pass; 24 comes from a column subtraction in which the smaller digit is taken from the larger in each column, 4 − 0 and 5 − 3, instead of exchanging.
- (d) 6 — Method: write an expression for each person's savings after w weeks, and set them equal. Working: 40 + 6w = 10 + 11w. Subtract 6w from both sides: 40 = 10 + 5w. Subtract 10: 30 = 5w, so w = 6. Answer: 6 weeks. 0 comes from setting only the weekly amounts equal, 6w = 11w, and ignoring the different starting amounts entirely. 10 comes from adding the two starting amounts and dividing by the difference in weekly amounts, (40 + 10) ÷ (11 − 6), instead of forming and solving the correct equation. 1.76 comes from adding the two weekly amounts instead of subtracting them when rearranging, (40 − 10) ÷ (11 + 6).
- (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.
- (b) 3 + 2m — Method: add the fixed fee to the cost of the miles travelled. Working: the miles cost 2 × m = 2m pounds, and the fixed fee is £3, so the total is 3 + 2m. Answer: 3 + 2m. 2 + 3m comes from swapping which number is the fixed fee and which is the per-mile rate. 5 + m comes from adding the fixed fee and the rate together into one number, 3 + 2 = 5, before adding the miles, instead of multiplying the rate by the number of miles. 5m comes from multiplying that combined total by the miles, treating the whole £5 as if it were a per-mile rate.
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