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
MathsUKwww.geekhero.co.uk
- 1.Amelia's mother is three times as old as Amelia. Four years ago her mother was five times as old as Amelia was then. Work out Amelia's age now.
- 2.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.
- 3.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.
- 4.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.
- 5.Oliver buys b pens. Each pen costs £2. Write down an expression for the total cost, in pounds.
- 6.A plumber charges a call-out fee of £30 plus £25 for each hour worked. The total bill for a job was £130. Work out how many hours the plumber worked.
- 7.Explain why the equation 4x + 3 = 4x + 10 has no solution for x.
- 8.Mia buys 3 identical notebooks and a pen costing £1.50. She pays a total of £9.90. Form an equation using n for the cost of one notebook, and solve it to find n.
- 9.Leo is three times as old as his brother. His brother is x years old. Write down an expression for Leo's age.
- 10.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.
- 11.Sophie's mother is 34 years old. In x years' time she will be 50 years old. Work out the value of x.
- 12.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.
- 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.Priya buys b balloons for a party. Each balloon costs 15p. Write down an expression for the total cost, in pence.
- 15.A regular hexagon has sides of length (x + 2) cm. Write down an expression, in terms of x, for the perimeter of the hexagon.
Answer key
- (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.
- (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.
- (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.
- (b) 4 — Method: form the equation 30 + 25h = 130, where h is the number of hours, then solve for h. Working: subtract the call-out fee from the total bill: 25h = 130 − 30 = 100. Divide by the hourly rate: h = 100 ÷ 25 = 4. Answer: 4 hours. 5.2 comes from dividing the whole bill by the hourly rate without subtracting the fixed fee first, 130 ÷ 25. 3.5 comes from swapping the fee and the rate, subtracting the rate from the bill and dividing by the fee, (130 − 25) ÷ 30. 6.4 comes from adding the call-out fee to the bill instead of subtracting it, before dividing by the rate, (130 + 30) ÷ 25.
- (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.
- (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) 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.
- (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.
- (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) 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.
- (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).
- (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.
- (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.
Build your own mix at the worksheet builder.