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.Two numbers have a sum of 45. The larger number is 9 more than the smaller number. Form an equation using n for the smaller number, and solve it to find the larger number.
- 2.The graph of W = 840 ÷ L shows how the width, W metres, of a rectangular field of area 840 m² depends on its length, L metres. Use the relationship to work out the width when the length is 35 m, and hence the perimeter of the field.
- 3.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.
- 4.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.
- 5.The graph of y = 20/x is drawn for x > 0. As the value of x increases, what happens to the value of y?
- 6.A cyclist's distance-time graph shows the following: she travels 12 km in the first 30 minutes at a steady speed, then rests for 30 minutes, then travels a further 8 km in the next 15 minutes. Work out her average speed, in km/h, for the whole journey.
- 7.The formula for converting a temperature in Fahrenheit, F, to Celsius, C, is C = 5(F − 32) ÷ 9. Work out C, to 1 decimal place, when F = 98.6.
- 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.Work out the value of 5m² + 3m when m = −2.4
- 10.A model rocket's height is given by h = −5t² + 20t, where h is in metres and t is in seconds. A student says the graph of h against t is n-shaped. Which statement gives the correct verdict on the SHAPE and the reason that settles it from the equation?
- 11.Which of these is an equation, rather than an expression, a formula or an identity?
- 12.A person's body mass index is given by the formula B = m ÷ h², where m is the mass in kg and h is the height in metres. Work out B, to 1 decimal place, when m = 68 and h = 1.6.
- 13.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.
- 14.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.
- 15.A ferry's distance-time graph is a straight line from (0 minutes, 0 km) to (25 minutes, 15 km). Work out the ferry's speed in kilometres per hour.
Answer key
- (a) 27 — Method: let the smaller number be n, so the larger number is n + 9. Form the equation n + (n + 9) = 45. Working: simplify: 2n + 9 = 45, so 2n = 36, giving n = 18. The larger number is 18 + 9 = 27. Answer: 27. 18 is the smaller number, found correctly, but given as the answer to the question asking for the larger number. 22.5 comes from ignoring the "9 more" part and simply halving the total, 45 ÷ 2. 36 comes from a sign error when isolating the constant, treating the equation as 2n = 45 + 9 instead of 2n = 45 − 9.
- (b) 118 m — Width = 840 ÷ 35 = 24 m. Perimeter = 2 × (length + width) = 2 × (35 + 24) = 2 × 59 = 118 m. The option 59 m gives the sum of the length and width but forgets to double it for the perimeter. The option 70 m doubles only the length (2 × 35 = 70) and leaves out the width entirely. The option 48 m doubles only the width (2 × 24 = 48) and leaves out the length entirely.
- (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.
- (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) y decreases towards zero but never reaches it — Method: as x gets larger, dividing 20 by a bigger number gives a smaller result, so y decreases; because 20/x can never be exactly zero for any positive x, the curve gets closer to zero without ever reaching it, so y decreases towards zero but never reaches it. Distractor origins: 'y increases towards a limit but never reaches it' has the relationship backwards, treating y as increasing when it is actually decreasing; 'y decreases at a steady rate and reaches zero' wrongly assumes the graph behaves like a straight line that eventually hits zero; 'y stays the same however large x becomes' wrongly assumes there is no change in y at all.
- (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.
- (d) 37.0 — C = 5(98.6 − 32) ÷ 9 = 5 × 66.6 ÷ 9 = 333 ÷ 9 = 37.0. A candidate who forgets to subtract 32 first gets 5 × 98.6 ÷ 9 = 54.8 (1 d.p.). A candidate who forgets the 5 ÷ 9 factor entirely and just works out F − 32 gets 66.6. A candidate who multiplies by 9 ÷ 5 instead of 5 ÷ 9 gets 66.6 × 9 ÷ 5 = 119.9 (1 d.p.).
- (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.
- (c) 21.6 — m² = (−2.4)² = 5.76, since a negative number squared is positive, so 5m² = 28.8. Then 3m = 3 × (−2.4) = −7.2. Total: 28.8 + (−7.2) = 21.6. −36 comes from taking m² as negative, 5 × (−5.76) − 7.2 = −36, forgetting that squaring removes the negative sign. 36 comes from ignoring the negative sign on m altogether and using m = 2.4 throughout. 29.4 comes from treating 3m as 3 + m instead of 3 × m, giving 28.8 + (3 + (−2.4)).
- (a) Yes — the t² coefficient is negative, giving an n-shape. — The coefficient of t² in h = −5t² + 20t is −5, which is negative, so the graph is n-shaped with a maximum point — this matches the physical story of the rocket rising then falling, but the shape itself is decided by the negative coefficient of t², not by the story alone. Saying the shape comes from the story rather than the coefficient gets the reasoning backwards — the algebra determines the shape, and the story happens to agree with it. Saying it is U-shaped because height starts by increasing confuses the early part of the curve with its overall shape; a U-shaped curve would mean the rocket's height eventually increases again forever, which does not happen here. Saying it is n-shaped only because the rocket lands treats a consequence of the shape as if it were the cause.
- (c) 3x + 5 = 17 — An equation contains an equals sign and is true only for particular value(s) of the unknown — solving 3x + 5 = 17 gives the single value x = 4. 3x + 5 is an expression: it has no equals sign, so it cannot be solved, only simplified or evaluated. A = πr² is a formula: it shows the general relationship between different quantities (area and radius), rather than asking for one unknown value. 3(x + 5) ≡ 3x + 15 is an identity: the ≡ sign shows it is true for every value of x, not just one. The equation is 3x + 5 = 17.
- (c) 26.6 — h² = 1.6² = 2.56, so B = 68 ÷ 2.56 = 26.5625, which rounds to 26.6 (1 d.p.). A candidate who forgets to square the height gets 68 ÷ 1.6 = 42.5. A candidate who squares the mass instead of the height gets 68² ÷ 1.6 = 2890.0. A candidate who truncates 26.5625 instead of rounding it gets 26.5.
- (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.
- (a) 23 — The number of logs increases by 4 for each layer down, starting from 3 in the top layer, so the nth layer has 3+(n−1)×4 logs. For the 6th layer: 3+5×4=3+20=23. A candidate who adds the difference of 4 six times instead of five, treating the top layer as needing an addition too, would compute 3+6×4=27. A candidate who uses 4n instead of 4n−1, omitting the adjustment for the first layer, would compute 4×6=24. A candidate who subtracts the common difference, 4, instead of 1 when adjusting the multiplier would compute 4×6−4=20.
- (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.
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