Printable · GCSE Foundation · ages 14-16
Algebra worksheet — GCSE Foundation
Fifteen questions across the algebra 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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Algebra worksheet — GCSE Foundation
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- 1.Solve 5x² − 15x = 0.
- 2.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.
- 3.Four sequences are shown below. Sequence P: 4, 8, 16, 32, ... Sequence Q: 4, 8, 12, 16, ... Sequence R: 4, 7, 12, 19, ... Sequence S: 4, 9, 16, 25, ... Work out which of the sequences P, Q, R and S is geometric.
- 4.The graph of t = 240 ÷ s shows how the time, t hours, taken for a car journey of 240 miles depends on its average speed, s mph. Use the relationship to work out the time taken when the average speed is 48 mph.
- 5.Yuki is y years old. Her brother is 4 years younger than Yuki. Write down an expression, in terms of y, for the brother's age in 5 years' time.
- 6.A plumber charges a call-out fee plus an hourly rate. The total cost, y in pounds, of a job lasting x hours is given by y = 45x + 60. Work out the total cost of a job that lasts 3 hours.y = 45x + 60
- 7.The braking distance of a car, in metres, is estimated using the formula d = 0.4v² + 6, where v is the speed in mph. A car travelling at 35 mph brakes. Work out the estimated braking distance.
- 8.A ball is thrown in the air. Its height, h metres, above the ground after t seconds is given in this table: when t = 0, h = 0; when t = 1, h = 15; when t = 2, h = 20; when t = 3, h = 15; when t = 4, h = 0. Use the table to find the two times, in seconds, at which the ball is at ground level.
- 9.A taxi company charges a £2.50 booking fee plus £1.80 per mile. A journey costs £15.10 in total. Work out the number of miles travelled.
- 10.The formula connecting speed, distance and time is v = s ÷ t. Work out v, in km/h, when s = 180 and t = 4.
- 11.A tree is 2 metres tall. Each year its height increases by 10% of its height at the start of that year. Work out the height of the tree after 3 years, giving your answer to 1 decimal place.
- 12.A plumber charges a call-out fee of £30 plus £25 per hour worked. Work out the total charge for a job that takes 3 hours.
- 13.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.
- 14.A colony of bacteria doubles in number every hour. At 9am there are 5 bacteria in the colony. Work out how many bacteria there will be at 12 noon.
- 15.A formula is F = 4s. A pupil says that when s = 5 the value of F is 9, because they added 4 and 5 instead of multiplying. Work out the correct value of F when s = 5.
Answer key
- (c) x = 0 or x = 3 — Method: take out the common factor 5x: 5x(x − 3) = 0, so 5x = 0 or x − 3 = 0, giving x = 0 or x = 3. Distractor origins: x = 0 or x = 15 comes from forgetting to divide the second term by the common factor 5x correctly, leaving 15 instead of 3; x = 3 loses the solution x = 0 by dividing both sides by x; x = 5 or x = 3 mistakes the coefficient 5 itself for one of the solutions.
- (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.
- (c) P — Sequence P has a common ratio of 2 (4 × 2 = 8, 8 × 2 = 16, 16 × 2 = 32), so it is geometric. Sequence Q has a common difference of 4, which is arithmetic, not geometric — a candidate who confuses a constant difference with a constant ratio picks Q. Sequence R has increasing differences of 3, 5, 7, a quadratic sequence, not geometric — a candidate who assumes any fast-growing sequence must be geometric picks R. Sequence S is the square numbers from 2² onwards (2², 3², 4², 5²), which is also quadratic, not geometric — a candidate who thinks squaring always means geometric growth picks S.
- (c) 5 hours — Substitute s = 48 into the relationship: t = 240 ÷ 48 = 5 hours. Inverting the relationship, s ÷ 240 = 48 ÷ 240 = 0.2, gives 0.2 hours. Misplacing the decimal point in the division, 240 ÷ 48 read as 4.8, gives 4.8 hours. Halving the speed to 24 mph by mistake before dividing, 240 ÷ 24 = 10, gives 10 hours.
- (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.
- (d) £195 — Substituting x = 3 into y = 45x + 60 gives y = 45 × 3 + 60 = 135 + 60 = 195. A candidate who forgets to add the call-out fee would get only 45 × 3 = £135. A candidate who adds the hours to the fee and the rate instead of multiplying would get 45 + 60 + 3 = £108. A candidate who multiplies both the hourly rate and the call-out fee by the number of hours would get 45 × 3 + 60 × 3 = £315.
- (c) 496 — d = 0.4v² + 6. First v² = 35 × 35 = 1225. Then 0.4 × 1225 = 490. Add 6: 490 + 6 = 496 m. 490 comes from forgetting to add the 6 at the end. 202 comes from squaring 0.4v together instead of squaring only v, (0.4 × 35)² + 6 = 14² + 6 = 202. 20 comes from using v instead of v², 0.4 × 35 + 6.
- (a) t = 0 or t = 4 — The ball is at ground level exactly when h = 0. From the table, h = 0 at t = 0 and at t = 4, so those are the two times. Distractor origins: t = 1 or t = 3 picks the times with equal (but non-zero) height instead of ground level; t = 2 picks the time of maximum height instead of ground level; t = 0 finds only the starting time and misses the second one.
- (d) 7 — Method: set up the equation 2.50 + 1.80m = 15.10, then subtract the booking fee and divide by the cost per mile. Working: 1.80m = 15.10 − 2.50 = 12.60; m = 12.60 ÷ 1.80 = 7. Answer: 7 miles. 8.39 comes from dividing the whole £15.10 by £1.80 without first subtracting the booking fee: 15.10 ÷ 1.80 ≈ 8.39. 5.32 comes from swapping the two amounts round, subtracting £1.80 and dividing by £2.50: (15.10 − 1.80) ÷ 2.50 ≈ 5.32. 9.78 comes from adding the booking fee instead of subtracting it: (15.10 + 2.50) ÷ 1.80 ≈ 9.78.
- (d) 45 — v = s ÷ t = 180 ÷ 4 = 45 km/h. 720 comes from multiplying s and t instead of dividing, 180 × 4. 184 comes from adding s and t instead of dividing, 180 + 4. 176 comes from subtracting t from s instead of dividing, 180 − 4.
- (d) 2.7 — The height after 3 years of 10% compound growth is 2 × 1.1³ = 2.662 m, which rounds to 2.7 m. A candidate who adds 10% of the original height (0.2 m) in each of the 3 years, instead of compounding on the new height each time, would reach 2 + 3×0.2 = 2.6 m. A candidate who compounds for only 2 years would reach 2 × 1.1² = 2.42 m, rounding to 2.4 m. A candidate who compounds for 4 years instead of 3 would reach 2 × 1.1⁴ = 2.928 m, rounding to 2.9 m.
- (b) £105 — The hourly charge is 25 × 3 = £75. Adding the call-out fee: £75 + £30 = £105. A candidate who forgets the call-out fee gives just the hourly charge, £75. A candidate who adds the call-out fee to the hourly rate before multiplying by the hours, (30 + 25) × 3, gets £165. A candidate who multiplies the call-out fee by the number of hours instead of the hourly rate, 30 × 3, gets £90.
- (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.
- (a) 40 — From 9am to 12 noon is 3 hours, so the population doubles three times: 5 × 2³ = 40. A candidate who counts the elapsed time as 4 hours (an off-by-one counting error) would reach 5 × 2⁴ = 80. A candidate who counts it as only 2 hours would reach 5 × 2² = 20. A candidate who misreads 'doubles' as 'increases by 2' each hour would compute 5 + 3 × 2 = 11.
- (b) 20 — F = 4s, so when s = 5, F = 4 × 5 = 20. 9 comes from adding 4 and 5 instead of multiplying, the mistake described in the question. 25 comes from squaring s (5²) instead of multiplying by 4. 45 comes from writing the digits 4 and 5 next to each other instead of carrying out the multiplication.
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