Algebra worksheet — GCSE Foundation

1.The graph of y = x² − 2x − 8 takes these values: when x = −2, y = 0; when x = −1, y = −5; when x = 0, y = −8; when x = 3, y = −5; when x = 4, y = 0. Use these values to write down the two solutions of x² − 2x − 8 = 0.

2.A stack of firewood has 3 logs in the top layer. Each layer below has 4 more logs than the layer above it. Work out the number of logs in the 6th layer from the top.

3.The formula for the energy stored in a stretched spring is E = (1/2)kx², where E is in joules, k is the spring constant in N/m and x is the extension in m. A spring with a spring constant of 40 N/m is stretched so that its extension is 0.3 m. Work out the energy stored in the spring.

4.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.

5.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.

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.Which expression is equivalent to 0.5(4x + 6) − x?

8.A square tile has an area of 144 cm². Work out the side length of the tile.

9.Solve 5x² − 15x = 0.

10.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.

11.A photo printing service has two adverts for its price. Advert A: cost in pounds = 3(2n + 4) for n photos. Advert B: cost in pounds = 6n + 12. A customer says the two adverts always charge the same amount. Is the customer correct?

12.Look at the statement 2x + 5 = 17. Considering whether it is an expression, an equation, a formula, or an identity, which classification is correct?

13.A ball's height, h metres, t seconds after being thrown follows h = (t − 1)(9 − t). Given that the ball is at ground level at t = 1 and t = 9, work out at what time t the ball reaches its maximum height, using symmetry.

14.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.

15.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.