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.
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Algebra worksheet — GCSE Higher
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- 1.The graph of y = 20/x is drawn for x > 0. As the value of x increases, what happens to the value of y?
- 2.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
- 3.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.
- 4.The equation x² − 5x − 2 = 0 can be solved using the iterative formula xₙ₊₁ = √(5xₙ + 2). The starting value is x₀ = 2, so x₁ is the value after the formula has been used once. Work out x₃ correct to 3 decimal places.
- 5.A tram sets off from a stop. Its velocity-time graph rises in a straight line from 0 m/s to 12 m/s over the first 8 seconds, then stays constant at 12 m/s for a further 10 seconds. Work out the total distance the tram travels in these 18 seconds, using the area under the graph.
- 6.The equation x² − 7 = 0 has a positive root. Let f(x) = x² − 7. Given that x₁ = 2.6 and x₂ = 2.65, work out which of these is correct.y = x² − 7
- 7.A distance-time graph is a straight line from (0, 0) to (4, 100), where time is in hours and distance is in kilometres. Work out the gradient of the line.
- 8.The equation x³ − 3x − 4 = 0 has a root near x = 2. Four students each try a different iterative formula, all starting from x₀ = 2: xₙ₊₁ = ∛(3xₙ + 4); xₙ₊₁ = (xₙ³ − 4) ÷ 3; xₙ₊₁ = 4 ÷ (xₙ² − 3); xₙ₊₁ = xₙ³ − 2xₙ − 4. Only one of these formulas keeps producing values that settle near the root when it is repeated. Work out x₁, correct to 3 decimal places, for the formula that does this.
- 9.f(x) = x³ − 3x − 5. Given that f(2.2) = −0.952 and f(2.3) = 0.267, work out what this shows about the equation x³ − 3x − 5 = 0.y = x
- 10.A car's journey is represented by a velocity-time graph. The total distance travelled, found from the area under the graph, is 350 metres, and the total time for the journey is 25 seconds. Work out the car's average speed for the whole journey, in m/s.
- 11.The point (20, 21) lies on the circle x² + y² = 841, which has centre O(0, 0). The tangent to the circle at (20, 21) crosses the x-axis at P and the y-axis at Q. Work out the area of triangle OPQ, correct to 1 decimal place.
- 12.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.
- 13.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.
- 14.The iterative formula xₙ₊₁ = xₙ³ − 2 is used repeatedly, starting from x₀ = 2. Which of these describes what happens to the sequence of values as n increases?
- 15.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.
Answer key
- (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.
- (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.
- (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) 4.897 — Method: substitute the starting value into the right-hand side of the formula to get x₁, then feed each new value back in, keeping the whole calculator display every time and rounding only at the very end. Working: x₁ = √(5 × 2 + 2) = √12 = 3.46410…; x₂ = √(5 × 3.46410… + 2) = √19.32050… = 4.39551…; x₃ = √(5 × 4.39551… + 2) = √23.97755… = 4.89668…, which is 4.897 correct to 3 decimal places. Answer: 4.897. The distractors: 4.396 is x₂, written down by a candidate who counts the starting value x₀ as the first iterate and so stops one use of the formula early; 3.464 is x₁, the value after using the formula only once; 5.146 is x₄, one use of the formula too many — the mirror image of the first slip, made by a candidate who labels the first value worked out as x₀ rather than as x₁ and so runs the count a step long.
- (d) 168 m — Split the area under the graph into two parts. The rising section (0 to 8 s) is a triangle: area = 0.5 × 8 × 12 = 48. The constant section, from 8 s to 18 s (10 s), is a rectangle: area = 12 × 10 = 120. Total distance = 48 + 120 = 168 m. Leaving out the 0.5 doubles the triangle, giving 8 × 12 + 120 = 216 m; ignoring the triangle section altogether gives only 120 m; averaging the start and final speeds over the whole 18 seconds, (0 + 12) ÷ 2 = 6, then 6 × 18 = 108 m, wrongly treats the tram as accelerating the whole time, when the graph is flat for the last 10 seconds.
- (d) x₂ closer: f(x₂) = 0.0225, nearer to 0 — f(2.6) = 2.6² − 7 = 6.76 − 7 = −0.24, and f(2.65) = 2.65² − 7 = 7.0225 − 7 = 0.0225. The closer a value of x is to the root, the closer f(x) is to zero — regardless of sign. Since |0.0225| = 0.0225 is much smaller than |−0.24| = 0.24, x₂ = 2.65 is closer to the root. 'x₁ closer: −0.24 is the smaller value' comes from comparing the SIGNED values of f(x) rather than their distances from zero — −0.24 is indeed less than 0.0225 as a number, but that does not mean x₁ is closer to the root. 'x₁ closer: f(x₁) negative ⇒ nearer root' invents a rule that a negative f(x) means x is closer to the root; the sign of f(x) only tells you which side of the root x is on, not how close it is. 'x₂ is the exact root, since f(x₂) ≈ 0' misreads f(x₂) = 0.0225 as zero; the true root is √7 ≈ 2.6458, so f(2.65) is close to zero but not equal to it, and x₂ is an approximation, not the exact root.
- (a) 25 — Gradient = change in y ÷ change in x = (100 − 0) ÷ (4 − 0) = 100 ÷ 4 = 25. Dividing time by distance instead of distance by time gives 0.04; multiplying the two values instead of dividing gives 400; stopping at the change in distance, 100, forgets to divide by the change in time.
- (a) 2.154 — Continuing xₙ₊₁ = ∛(3xₙ + 4) from x₁ = 2.154 gives x₂ = 2.187, x₃ = 2.195, settling towards the root near 2.196 — this is the formula that converges. Continuing xₙ₊₁ = (xₙ³ − 4) ÷ 3 from x₁ = 1.333 gives x₂ = −0.543 and then x₃ = −1.387, moving further from the root each time. Continuing xₙ₊₁ = 4 ÷ (xₙ² − 3) from x₁ = 4 gives x₂ = 0.308 and then x₃ = −1.377, swinging wildly rather than settling. Continuing xₙ₊₁ = xₙ³ − 2xₙ − 4 from x₁ = 0 gives x₂ = −4 and then x₃ = −60, running away from the root entirely.
- (d) It has a solution between x = 2.2 and x = 2.3 — Since f(2.2) is negative and f(2.3) is positive, the graph of f crosses the x-axis somewhere between x = 2.2 and x = 2.3, so the equation has a solution in that interval. Choosing 'between x = −2.2 and x = −2.3' confuses the negative f-VALUE at 2.2 with a negative x-value. Choosing 'no solution' misreads a change of sign as meaning the opposite of what it shows. Choosing 'exactly two solutions' assumes a single change of sign must give two roots, which is not what the rule guarantees.
- (b) 14 m/s — Average speed for a whole journey is total distance divided by total time: 350 ÷ 25 = 14 m/s. Multiplying the distance and time instead of dividing gives 350 × 25 = 8750 m/s. Adding the distance and time instead of dividing gives 350 + 25 = 375 m/s. Inverting the division, working out time divided by distance, gives 25 ÷ 350, which rounds to 0.07 m/s.
- (a) 842.0 — The radius to (20, 21) has gradient 21/20, so the tangent's gradient is −20/21. The tangent line is y − 21 = −20/21(x − 20), i.e. y = −20/21x + 841/21. Setting y = 0 gives the x-intercept x = 841/20 = 42.05; setting x = 0 gives the y-intercept y = 841/21 ≈ 40.048. The area of triangle OPQ is 1/2 × 42.05 × 40.048 ≈ 842.0. 1684.0 comes from multiplying the two intercepts without the 1/2 that a triangle's area needs — twice the correct area. 580.7 comes from using the circle's radius, 29, as a side of the triangle instead of the x-intercept, 42.05: 1/2 × 29 × 40.048 ≈ 580.7. 2.0 comes from a sign error in the tangent's gradient — using 20/21 instead of −20/21 — which gives a different line, with intercepts x ≈ −2.05 and y ≈ 1.952, and area 1/2 × 2.05 × 1.952 ≈ 2.0.
- (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.
- (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) It diverges, growing rapidly without limit. — x₁ = 2³ − 2 = 8 − 2 = 6. x₂ = 6³ − 2 = 216 − 2 = 214. x₃ = 214³ − 2 = 9800344 − 2 = 9800342. The values 6, 214, 9800342, … grow far larger at every step, so the sequence diverges rather than settling anywhere. Checking whether the sequence converges to a fixed value near 2 fails, since the terms grow enormously instead of levelling off. Checking for a repeating pair of values also fails, since 6, 214 and 9800342 are all different, with no sign of a return to 6. x₀ = 2 is a fixed point only if 2³ − 2 = 2, but 2³ − 2 = 6, not 2, so the sequence does not stay constant.
- (d) 8 — Since the mean of the four numbers is 12.5, their total is 4 × 12.5 = 50. The three known numbers add up to 8 + 15 + 19 = 42, so n = 50 − 42 = 8. A candidate who multiplies the mean by 3 instead of 4 gets a total of 37.5, giving n = 37.5 − 42 = −4.5. A candidate who forgets to subtract the three known numbers and gives the total itself as n states n = 50. A candidate who subtracts the mean from the total of the three known numbers instead of the other way round gets n = 42 − 12.5 = 29.5.
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