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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Answer key: Algebra worksheet — GCSE Foundation
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- (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.
- (a) 5.8 — The horizontal distance is 3 and the vertical distance is 5, so using Pythagoras' theorem the distance is √(3² + 5²) = √34 = 5.8 (1 d.p.). A candidate who adds the two differences instead of using Pythagoras gets 3 + 5 = 8.0. A candidate who works out 3² + 5² = 34 but forgets to take the square root gets 34.0. A candidate who subtracts the squares instead of adding them gets √(5² − 3²) = √16 = 4.0.
- (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.
- (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.
- (c) x + 3 — Expand the bracket: 0.5(4x + 6) = 2x + 3. Then subtract the x: 2x + 3 − x = x + 3. The option 2x + 3 comes from expanding the bracket correctly but then forgetting to subtract the x at all. The option x + 6 comes from forgetting to multiply the 6 inside the bracket by 0.5 (treating it as 2x + 6), then subtracting x. The option 3x + 3 comes from adding the x instead of subtracting it: 2x + 3 + x = 3x + 3.
- (c) 12 cm — For a square, area = side². So side² = 144, giving side = ±12. Since a length must be positive, the side length is 12 cm. A candidate who gives both square roots without rejecting the negative one, which cannot be a length, answers 12 cm or −12 cm. A candidate who halves 144 instead of taking its square root gets 72 cm. A candidate who divides 144 by 4, confusing the area formula with a perimeter calculation, gets 36 cm.
- (c) 16 km/h — Method: total distance = 12 + 8 = 20 km. Total time = 30 + 30 + 15 = 75 minutes = 1.25 hours. Average speed = total distance ÷ total time = 20 ÷ 1.25 = 16 km/h. Distractor origins: 20 km/h gives the total distance without ever dividing by the total time; 24 km/h uses only the speed of the first stage (12 km in 30 minutes), ignoring the second stage and the rest; 28 km/h averages the two separate stage speeds, 24 km/h and 32 km/h, instead of using total distance over total time.
- (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.
- (a) 36 km/h — Speed is distance ÷ time. Convert 25 minutes to hours: 25 ÷ 60 = 5/12 hours. Then 15 ÷ (5/12) = 15 × 12/5 = 36 km/h. Dividing 15 by 25 without converting minutes to hours gives 0.6 km/h. Multiplying 15 by 25/60 instead of dividing by it gives 6.25 km/h. Working out the speed in km per minute, 0.6, and then multiplying by 24 instead of 60 gives 14.4 km/h.
- (d) £20 — Let f be the joining fee and m the monthly fee: f + 3m = 100 and f + 6m = 160. Subtracting the first equation from the second eliminates f: 3m = 60, so m = 20. A candidate who finds the joining fee instead of the monthly fee would get f = 100 − 3(20) = £40. A candidate who divides Ben's total by his number of months, ignoring that part of the cost is a fixed joining fee, would get 160 ÷ 6 ≈ £26.67. A candidate who divides the difference in cost by the total number of months instead of the difference in months would get (160 − 100) ÷ 9 ≈ £6.67.
- (b) £47, and T=15+8h is a formula (relates T and h). — The charge is £15 fixed plus £8 for each hour: T=15+8h, so for h=4, T=15+8×4=15+32=47. Because T=15+8h relates two different quantities, the total charge T and the number of hours h, it is a formula, not an equation — an equation is solved for one particular value of an unknown, but this relationship holds for every value of h a job might last. A candidate who forgets to include the fixed £15 fee would compute only 8×4=32. A candidate who adds the three numbers in the formula together instead of multiplying the hourly rate by the number of hours would compute 15+8+4=27. A candidate who correctly finds the charge but mistakes the formula for an equation, treating it as something to be solved for one specific value of h rather than a general relationship between T and h, would pick the correct charge with the wrong classification.
- (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.
- (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) 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.
- (a) £14.50 — Method: the first 200 minutes are already covered by the £12, so only 250 − 200 = 50 minutes are charged extra. Extra charge = 50 × £0.05 = £2.50. Total cost = £12 + £2.50 = £14.50. Distractor origins: £24.50 charges 5p for all 250 minutes instead of only the 50 minutes over 200 (250 × £0.05 = £12.50, plus £12 = £24.50); £12.50 makes that same slip of charging all 250 minutes but then forgets to add the £12 monthly fee; £13.50 works out 250 − 200 wrongly as 30 extra minutes instead of 50 (30 × £0.05 = £1.50, plus £12 = £13.50).
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