Printable · GCSE Higher · ages 14-16
Algebra worksheet — GCSE Higher
Fifteen questions across the algebra statements at Higher tier. Choose the non-calculator filter to rehearse Paper 1, which counts for a third of the marks.
Algebra worksheet — GCSE Higher
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- 1.Two students each use the trapezium rule to estimate the area under the same curve between x = 0 and x = 8. Student A uses 4 strips of width 2. Student B uses 8 strips of width 1. Which estimate is more likely to be closer to the true area, and why?
- 2.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.
- 3.f(x) = x² − 1 and g(x) = 3x. Work out fg(4).y = x² − 1
- 4.A straight line has gradient 3 and passes through the point (1, 4). Work out the equation of the line.
- 5.The graph of y = cos x is transformed onto the graph of y = cos(x − 90°). State the direction of the translation and which standard graph the image is.y = cos(x)
- 6.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.
- 7.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
- 8.The diagram shows a distance–time graph for a cyclist travelling at a constant speed, for the first 6 minutes of a journey. Distance is in kilometres and time is in minutes. Work out how far the cyclist would travel in 20 minutes at the same speed.
- 9.A straight line passes through the points (1, 3) and (2, 5). Work out the equation of the line.
- 10.Priya has a budget of £50 for a school trip. The coach costs £14 and each student ticket costs £4. Using the inequality 14 + 4s ≤ 50, work out the greatest number of student tickets, s, she can buy.
- 11.The equation x³ = 6x + 20 can be solved using the iterative formula xₙ₊₁ = ∛(6xₙ + 20). Taking x₀ = 3, x₁ = 3.3620 correct to 4 decimal places. Using the full unrounded value of x₁, work out x₂ correct to 3 decimal places.
- 12.A cycle route is 84 km long. Freya sets off along it at a steady 14 km/h. Write down the function for the distance y, in kilometres, that is still to be cycled after x hours.
- 13.n = 3. Work out the value of (2n)².
- 14.Work out the value of 4p² − 3 when p = 2.5
- 15.Aisha saves money each week. In week 1 she has saved £15 in total, in week 2 she has saved £23 in total, in week 3 she has saved £31 in total and in week 4 she has saved £39 in total, with the total increasing by the same amount each week. Work out an expression, in terms of n, for the total amount she has saved, in pounds, after week n.
Answer key
- (b) Student B's — narrower strips fit the curve more closely. — The trapezium rule replaces the curve with straight-line segments; the narrower each strip, the more closely its straight edge follows the curve, so Student B's estimate with 8 narrower strips is more likely to be closer to the true area. Reasoning from the number of arithmetic steps rather than from how well the straight lines fit the curve gives the wrong claim that Student A makes fewer rounding errors. Believing the trapezium rule is exact, rather than an estimate that improves with narrower strips, gives the wrong claim that the two are always the same. Believing wider strips smooth out the curve better, rather than following it less closely, gives the wrong claim in favour of Student A's wider strips.
- (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.
- (b) 143 — fg(4) means f(g(4)): work out g(4) first, then substitute the result into f. g(4) = 3 × 4 = 12, then f(12) = 12² − 1 = 144 − 1 = 143. Working out gf(4) instead swaps the order: f(4) = 4² − 1 = 15, then g(15) = 3 × 15 = 45 — that is the wrong composition. Treating f(x) as x − 1 (forgetting to square the input) gives f(12) = 12 − 1 = 11. Applying g twice instead of applying g then f gives g(g(4)) = g(12) = 3 × 12 = 36, which mixes up which function should be applied second.
- (c) y = 3x + 1 — Method: a line of known gradient m has equation y = mx + c, and c is found by substituting the coordinates of a point known to lie on it. Working: the gradient is 3, so the line is y = 3x + c; substituting x = 1 and y = 4 gives 4 = 3 × 1 + c, so c = 4 − 3 = 1. Answer: y = 3x + 1. The distractors: y = 3x − 1 comes from working out the constant as mx − y, 3 − 4 = −1, instead of y − mx; y = x + 3 comes from swapping the two numbers over, putting the gradient 3 in the constant position and the x-coordinate 1 in front of x; y = 3x + 4 comes from using the y-coordinate 4 as the constant without substituting.
- (a) Positive x-direction, 90°; image is y = sin x. — Writing cos(x − 90°) as cos(x − a) with a = 90 shows this is a horizontal translation, y = f(x − a), which moves the graph 90° in the positive x-direction; the identity cos(x − 90°) = sin x confirms the image is y = sin x. Choosing the negative x-direction reverses the sign inside the bracket — subtracting inside the bracket always translates in the positive x-direction, not the negative one, so that statement is wrong on direction. Getting the direction right but conflating the subtraction inside the bracket with an extra reflection of the output flips the sign of the resulting graph, wrongly giving y = −sin x. Treating the subtraction as if it changed the output directly, rather than the input, wrongly calls this a vertical translation even while still correctly recalling that the image simplifies to y = sin x.
- (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) £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.
- (b) 10 km — Method: find the constant speed from the graph (distance ÷ time for any point on the line), then multiply that speed by 20 minutes. Working: the line passes through (4 minutes, 2 km), so the speed is 2 ÷ 4 = 0.5 km per minute; in 20 minutes the cyclist travels 0.5 × 20 = 10 km. Answer: 10 km. Distractor refutation: 3 km comes from reading off the distance shown at the end of the plotted section (6 minutes) and stopping there, instead of extending the line to 20 minutes. 20 km comes from misreading the speed as 1 km per minute instead of 0.5 km per minute, doubling the true rate. 40 km comes from dividing 20 by the speed instead of multiplying by it, a reciprocal mix-up.
- (c) y = 2x + 1 — Method: the gradient of the line through two points is the change in y divided by the change in x, and the constant is then found by substituting one of the points into y = mx + c. Working: m = (5 − 3) ÷ (2 − 1) = 2 ÷ 1 = 2, so the line is y = 2x + c; substituting x = 1 and y = 3 gives 3 = 2 × 1 + c, so c = 3 − 2 = 1. Answer: y = 2x + 1. The distractors: y = x + 2 comes from taking the gradient as the change in x, 2 − 1 = 1, and then substituting (1, 3) to reach a constant of 2; y = 2x − 1 comes from working out the constant as mx − y, 2 × 1 − 3 = −1, instead of y − mx; y = 2x + 3 comes from using the y-coordinate of (1, 3) as the constant without substituting at all.
- (b) 9 — Subtract 14 from both sides: 4s ≤ 36. Divide both sides by 4: s ≤ 9, so the greatest number of tickets is 9. A candidate who forgets the £14 coach cost solves 4s ≤ 50, getting s ≤ 12.5, rounded down to 12. A candidate who adds the £14 instead of subtracting it solves 4s ≤ 64, getting s = 16. A candidate who miscalculates 50 − 14 as 32 solves 4s ≤ 32, getting s = 8.
- (d) 3.425 — x₁ = ∛(6 × 3 + 20) = ∛38 = 3.3620 (unrounded, 3.36198...). x₂ = ∛(6 × 3.3620 + 20) = ∛40.172 = 3.425 (3 d.p.). Choosing 3.362 stops at x₁ instead of continuing to x₂. Choosing 2.722 leaves out the '+ 20' inside the root, working out ∛(6 × 3.3620) = ∛20.172 = 2.722. Choosing 0.556 subtracts 20 instead of adding it, working out ∛(6 × 3.3620 − 20) = ∛0.172 = 0.556.
- (b) y = 84 − 14x — Method: the distance still to go is the whole route minus the distance already covered, and at a steady speed the distance covered after x hours is the speed multiplied by x. Working: after x hours Freya has cycled 14x km, so the distance remaining is the 84 km route take away 14x, giving y = 84 − 14x, a straight line with intercept 84 and gradient −14. Answer: y = 84 − 14x. The distractors: y = 84 + 14x comes from adding the distance cycled to the length of the route rather than taking it away, so the ride would grow longer the further she goes; y = 14x − 84 comes from carrying out the subtraction the wrong way round, which is negative for the whole of the ride; y = 14 − 84x comes from swapping the two numbers over, treating 84 km as the hourly speed and 14 km as the length of the route.
- (d) 36 — (2n)² means the whole of 2n is squared, so with n = 3: (2n)² = (2 × 3)² = 6² = 36. Answering 18 instead works out 2n² — squaring only the n and then multiplying by 2 — which is a different expression because the brackets around 2n are missing. Answering 12 squares only the coefficient, treating (2n)² as 2² × n = 4 × 3 = 12, and forgets to square the n as well. Answering 9 ignores the coefficient of 2 altogether and works out n² on its own. The value of (2n)² when n = 3 is 36.
- (c) 22 — p² means p × p. Substitute p = 2.5: p² = 6.25, so 4p² = 4 × 6.25 = 25, and 25 − 3 = 22. 97 comes from squaring 4p together instead of just p, (4 × 2.5)² − 3 = 10² − 3 = 97. 7 comes from using p instead of p², 4 × 2.5 − 3. 1 comes from squaring the 3 instead of the p, 4 × 2.5 − 3².
- (c) 8n + 7 — Method: find the weekly increase, then find the constant by adjusting the week 1 total. Working: the total goes up by £8 each week (23 − 15 = 8), so the coefficient of n is 8. The constant is the week 1 total minus the common difference: 15 − 8 = 7. Answer: the nth term is 8n + 7. 8n + 15 comes from using the week 1 total, 15, as the constant without subtracting the common difference. 8n − 1 comes from a slip in working out the constant, subtracting the common difference twice (15 − 8 − 8 = −1) instead of once. 7n + 8 comes from swapping the weekly increase and the constant.
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