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.
Calculator
Algebra worksheet — GCSE Foundation
MathsUKwww.geekhero.co.uk
- 1.Priya buys b balloons for a party. Each balloon costs 15p. Write down an expression for the total cost, in pence.
- 2.A square tile has an area of 144 cm². Work out the side length of the tile.
- 3.On Monday, a runner covers 15 km in 2.5 hours. On Tuesday, she covers 12 km in 1.5 hours. Using the formula speed = distance ÷ time, work out on which day she ran faster, and by how much.
- 4.A hiker's distance-time graph shows the following: he walks 5 km in the first 1 hour at a steady speed, rests for 1 hour, then walks a further 9 km in the next 2 hours, also at a steady speed. Work out his average speed for the whole journey, in km/h.
- 5.A taxi fare in pounds is given by F = 2.50 + 1.20m, where m is the number of miles travelled. Sam takes a taxi for a journey of 6 miles. Work out the fare.
- 6.The graph of P = 480 ÷ V shows how the pressure, P pascals, of a gas depends on its volume, V cm³, at constant temperature. Use the relationship to work out the volume when the pressure is 32 pascals, and then work out by how much the volume increases when the pressure drops to 24 pascals.
- 7.A car park charges by the hour. Parking for 1 hour costs £5, 2 hours costs £9, 3 hours costs £13 and 4 hours costs £17, with the cost increasing by the same amount for each extra hour. Work out an expression, in terms of n, for the cost, in pounds, of parking for n hours.
- 8.A parcel delivery company's cost graph shows the following: the cost is a flat £5 for parcels weighing up to 2 kg, and then rises by £2 for each additional kg above 2 kg. Use the graph to work out the weight of a parcel that costs £17.
- 9.A gym charges a joining fee plus a monthly fee. Anna paid £100 in total after 3 months of membership. Ben paid £160 in total after 6 months of membership (same joining fee and monthly fee as Anna). Work out the monthly fee.
- 10.Work out the value of 3b − 4 when b = 6
- 11.A regular hexagon has sides of length (x + 2) cm. Write down an expression, in terms of x, for the perimeter of the hexagon.
- 12.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.
- 13.The formula for the perimeter of a square is P = 4s, where s is the side length. Make s the subject of the formula.
- 14.A number, n, is tripled and then 4.5 is added. The result is 22.5. Form an equation and solve it to find n.
- 15.A car's speed-time graph shows the following: its speed increases steadily from 0 m/s to 20 m/s over the first 10 seconds, then stays constant at 20 m/s for the next 15 seconds. Work out the total distance travelled in the first 25 seconds.
Answer key
- (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.
- (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.
- (a) Tuesday, by 2 km/h — Monday's speed is 15 ÷ 2.5 = 6 km/h and Tuesday's speed is 12 ÷ 1.5 = 8 km/h, so Tuesday was faster, by 8 − 6 = 2 km/h. A candidate who works out the correct speeds but mislabels which day is faster gets Monday, by 2 km/h. A candidate who divides 15 ÷ 2.5 incorrectly as 5 instead of 6 gets a difference of 8 − 5 = 3 km/h, still crediting Tuesday. A candidate who forgets to find Monday's speed and gives Tuesday's speed itself as the difference states Tuesday, by 8 km/h.
- (a) 3.5 km/h — Average speed is total distance ÷ total time. Total distance = 5 + 9 = 14 km. Total time, including the rest, = 1 + 1 + 2 = 4 hours. So average speed = 14 ÷ 4 = 3.5 km/h. Leaving out the 1 hour rest and dividing by only the 3 hours of walking gives 14 ÷ 3 = 4.67 km/h. Averaging the two separate speeds, 5 km/h and 4.5 km/h, instead of using total distance over total time, gives (5 + 4.5) ÷ 2 = 4.75 km/h. Dividing only the second leg's distance by the total time, 9 ÷ 4 = 2.25 km/h, ignores the first leg's distance.
- (c) £9.70 — F = 2.50 + 1.20 × 6 = 2.50 + 7.20 = £9.70. £7.20 comes from forgetting to add the £2.50 fixed charge. £10.20 comes from adding the 2.50 to the number of miles before multiplying by 1.20, 1.20 × (6 + 2.50). £22.20 comes from adding the fixed charge and the rate together first and then multiplying by the miles, (2.50 + 1.20) × 6.
- (b) 5 cm³ — When P = 32, V = 480 ÷ 32 = 15 cm³. When P = 24, V = 480 ÷ 24 = 20 cm³. The increase in volume is 20 − 15 = 5 cm³. Giving only the second volume, 20 cm³, forgets to subtract the first volume. Giving only the first volume, 15 cm³, answers the wrong part of the question. Subtracting the two pressures instead of the two volumes, 32 − 24 = 8, mixes up which quantity the question asks for.
- (a) 4n + 1 — Method: find the increase in cost per hour, then find the constant by adjusting the 1-hour cost. Working: the cost goes up by £4 for each extra hour (9 − 5 = 4), so the coefficient of n is 4. The constant is the 1-hour cost minus the common difference: 5 − 4 = 1. Answer: the nth term is 4n + 1. 4n + 5 comes from using the 1-hour cost, 5, as the constant without subtracting the common difference. 4n − 3 comes from a slip in working out the constant, subtracting the common difference twice (5 − 4 − 4 = −3) instead of once. n + 4 comes from swapping the hourly increase and the constant.
- (b) 8 kg — The cost above the flat £5 charge is 17 − 5 = £12. At £2 per kg, this covers 12 ÷ 2 = 6 kg above the first 2 kg, so the total weight is 2 + 6 = 8 kg. Dividing the full £17 by £2 per kg without first taking off the £5 flat charge gives 17 ÷ 2 = 8.5 kg. Taking off the £5 flat charge and dividing by £2 per kg, but forgetting to add back the 2 kg that the flat charge covers, gives 12 ÷ 2 = 6 kg. Taking off £2 instead of £5 as the flat charge, (17 − 2) ÷ 2 = 7.5 kg, swaps which number is the fixed fee.
- (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.
- (c) 14 — Substitute b = 6 into 3b − 4: 3 × 6 − 4 = 18 − 4 = 14. 5 comes from treating 3b as 3 + b instead of 3 × b, giving 3 + 6 − 4. 6 comes from subtracting 4 from b before multiplying by 3, 3 × (6 − 4). 18 comes from working out 3 × 6 correctly but forgetting to subtract the 4.
- (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) 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.
- (c) s = P / 4 — Method: undo the multiplication by 4 by dividing both sides by 4. Working: P = 4s, so dividing both sides by 4 gives s = P / 4. The value s = 4P comes from multiplying by 4 instead of dividing — using the wrong inverse operation. The value s = P + 4 comes from adding 4 instead of dividing. The value s = P − 4 comes from subtracting 4 instead of dividing.
- (b) 6 — 3n + 4.5 = 22.5, so 3n = 18 and n = 6. A candidate who divides 22.5 by 3 first and ignores the 4.5 gets n = 7.5. A candidate who makes a sign error and forms the equation 3n − 4.5 = 22.5 gets 3n = 27 and n = 9. A candidate who divides by 3 before subtracting the 4.5, working out 22.5 ÷ 3 + 4.5, gets n = 12.
- (d) 400 m — The distance travelled equals the area under the speed-time graph. The first 10 seconds form a triangle with base 10 and height 20, giving an area of 0.5 × 10 × 20 = 100 m. The next 15 seconds form a rectangle with base 15 and height 20, giving an area of 15 × 20 = 300 m. The total distance is 100 + 300 = 400 m. The option 500 m treats the whole 25 seconds as travelled at the constant 20 m/s, ignoring that the speed was building up during the first 10 seconds: 25 × 20 = 500. The option 300 m only counts the constant-speed section and forgets the triangle section entirely. The option 200 m comes from working out the triangle's area without halving it (10 × 20 = 200) and forgetting the rectangle altogether.
Build your own mix at the worksheet builder.