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.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.
- 2.A taxi fare, in pounds, for a journey of m miles is given by the formula F = 3 + 2.5m. A journey costs £15.50. Work out the distance travelled, m, by first rearranging the formula to make m the subject, then substituting F = 15.50.
- 3.A circle has centre (0, 0) and passes through the point (12, 35). Work out the equation of the circle.
- 4.A cyclist accelerates uniformly from rest to 6 m/s in 4 seconds, travels at a constant 6 m/s for 10 seconds, then decelerates uniformly to rest in 3 seconds. Work out the total distance the cyclist travels.
- 5.A car accelerates uniformly from rest at 2.5 m/s² until it reaches a speed of 20 m/s, then travels at this constant speed for a further 30 seconds. Work out the total distance travelled.
- 6.Work out the value of 5m² + 3m when m = −2.4
- 7.f(x) = x³ − 5x − 6. Given that f(2.6) = −1.424 and f(2.7) = 0.183, work out what this shows about the equation x³ − 5x − 6 = 0.y = x
- 8.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.
- 9.A Fibonacci-type sequence is defined by aₙ = aₙ₋₁ + aₙ₋₂ for n ≥ 3. Given that a₁ = 5 and a₃ = 17, work out a₂.
- 10.An exponential graph y = A × rˣ passes through the points (0, 8) and (2, 18). Work out the value of r, correct to 2 decimal places.
- 11.The equation x² + 2x − 5 = 0 can be solved using the iterative formula xₙ₊₁ = 5/(xₙ + 2). The starting value is x₀ = 1, so x₁ is the value after the formula has been used once. Work out x₂ correct to 2 decimal places.
- 12.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.
- 13.Points A(−1, 2) and B(5, 8) are the endpoints of a line segment. Work out the equation of the perpendicular bisector of AB.
- 14.A cinema sells adult tickets for £10 and child tickets for £6. A family group buys 9 tickets in total, spending £74. Work out how many adult tickets were bought.
- 15.Solve the inequality 2(3x − 1) ≥ 4x + 8.
Answer key
- (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) 5 — Method: rearrange the formula to make m the subject, then substitute F = 15.50. Working: F = 3 + 2.5m, so subtracting 3 from both sides gives F − 3 = 2.5m, then dividing by 2.5 gives m = (F − 3) / 2.5. Substituting F = 15.50: m = (15.50 − 3) / 2.5 = 12.50 / 2.5 = 5. The value 6.2 comes from dividing 15.50 by 2.5 without subtracting the fixed £3 first. The value 3.2 comes from dividing first and subtracting afterwards, in the wrong order: (15.50 / 2.5) − 3 = 3.2. The value 7.4 comes from adding £3 instead of subtracting it before dividing: (15.50 + 3) / 2.5 = 7.4.
- (b) x² + y² = 1369 — For a circle centred at the origin, the radius squared equals the sum of the squares of the coordinates of any point on it: r² = 12² + 35² = 144 + 1225 = 1369. The equation is x² + y² = 1369. x² + y² = 2209 comes from adding the coordinates first and then squaring the sum: (12 + 35)² = 47² = 2209, instead of squaring each coordinate separately. x² + y² = 1225 comes from using only 35² and leaving out the 12² term. x² + y² = 144 comes from using only 12² and leaving out the 35² term.
- (c) 81 m — Split the velocity-time graph into its three phases. The accelerating phase (0 to 4 s) is a triangle: 1/2 × 4 × 6 = 12. The constant phase (4 s at 6 m/s, for 10 s) is a rectangle: 10 × 6 = 60. The decelerating phase (3 s) is a triangle: 1/2 × 3 × 6 = 9. Total distance: 12 + 60 + 9 = 81 m. Forgetting the final decelerating phase entirely and adding only the first two areas gives 12 + 60 = 72 m. Treating all three phases as rectangles, forgetting to halve either triangular phase, gives 4 × 6 + 10 × 6 + 3 × 6 = 102 m. Halving only the deceleration triangle correctly but treating the acceleration phase as a rectangle too gives 4 × 6 + 60 + 9 = 93 m.
- (d) 680 m — First find how long the acceleration takes: acceleration = change in speed ÷ time, so 2.5 = 20 ÷ t, giving t = 20 ÷ 2.5 = 8 seconds. The distance during this phase is the area of a triangle with base 8 and height 20: 1/2 × 8 × 20 = 80 m. The distance during the constant-speed phase is 20 × 30 = 600 m, since distance = speed × time at a constant speed. The total distance is 80 + 600 = 680 m. Using the given 30 seconds for the acceleration phase as well, 1/2 × 30 × 20 = 300, plus the correct 600, gives 900 m — but 30 seconds is only stated for the constant-speed phase. Leaving out the 1/2 and using the full rectangle for the acceleration phase, 8 × 20 = 160, plus the correct 600, gives 760 m — the speed is not constant during acceleration, so this area is a triangle, not a rectangle. Swapping the two times round, and using 8 seconds for the constant-speed distance instead of 30, 20 × 8 = 160, plus the correct triangle area of 80, gives 240 m.
- (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) It has a solution between x = 2.6 and x = 2.7 — f(2.6) is negative and f(2.7) is positive, so the graph of f crosses the x-axis between x = 2.6 and x = 2.7, meaning the equation has a solution there. Choosing 'x = 2.6 is a solution' reads an end of the interval as the root itself, but f(2.6) = −1.424, which is not zero — the change of sign locates a root between the two values, it does not land on either of them. Choosing 'between x = −2.6 and x = −2.7' confuses the negative f-VALUE at 2.6 with a negative x-value. Choosing 'no root in this interval' misapplies the rule, which needs a CHANGE of sign — and a change of sign is exactly what these two values show.
- (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.
- (d) 12 — Since a₃ = a₂ + a₁, rearranging gives a₂ = a₃ − a₁ = 17 − 5 = 12. 22 comes from adding instead of subtracting: a₃ + a₁ = 17 + 5 = 22. 11 comes from treating the sequence as if it were arithmetic and averaging the two given terms: (5 + 17) ÷ 2 = 11 — but a Fibonacci-type sequence isn't arithmetic, so this doesn't apply. −12 comes from subtracting in the wrong order: a₁ − a₃ = 5 − 17 = −12.
- (b) 1.50 — Substituting (0, 8) gives A = 8, since r⁰ = 1. Substituting (2, 18) gives 8 × r² = 18, so r² = 18 ÷ 8 = 2.25. Taking the square root of 2.25 gives r = 1.50 (2 d.p.). Stopping after finding r² and giving 2.25 as the final answer, without taking the square root, is wrong because r² is not the same as r. Dividing r² by the exponent 2 instead of taking its square root — treating the power as something you divide by rather than root — gives 2.25 ÷ 2 = 1.13 (2 d.p.), which is wrong. Inverting the ratio, working out 8 ÷ 18 instead of 18 ÷ 8, gives 0.44 (2 d.p.), which is wrong because the LATER value must be divided by the EARLIER one to find the growth multiplier, not the other way round.
- (a) 1.36 — Method: put the starting value into the right-hand side to get x₁, feed that value back in to get x₂, and round only once the second value has been found. Working: x₁ = 5 ÷ (1 + 2) = 5 ÷ 3 = 1.66666…; x₂ = 5 ÷ (1.66666… + 2) = 5 ÷ 3.66666… = 1.36363…. The digit in the third decimal place is 3, so x₂ = 1.36 correct to 2 decimal places. Answer: 1.36. The distractors: 1.67 is x₁, the value after a single use of the formula, given by a candidate who counts the starting value itself as x₁; 1.49 is x₃ = 1.48648…, one use of the formula too many; 1.37 comes from writing x₁ down as 1.66, truncating the display instead of keeping it in full, and then working out 5 ÷ 3.66 = 1.36612…, which rounds up to 1.37.
- (b) 4 — Method: form the equation 30 + 25h = 130, where h is the number of hours, then solve for h. Working: subtract the call-out fee from the total bill: 25h = 130 − 30 = 100. Divide by the hourly rate: h = 100 ÷ 25 = 4. Answer: 4 hours. 5.2 comes from dividing the whole bill by the hourly rate without subtracting the fixed fee first, 130 ÷ 25. 3.5 comes from swapping the fee and the rate, subtracting the rate from the bill and dividing by the fee, (130 − 25) ÷ 30. 6.4 comes from adding the call-out fee to the bill instead of subtracting it, before dividing by the rate, (130 + 30) ÷ 25.
- (d) x + y = 7 — The midpoint of AB is (−1 + 5)/2, (2 + 8)/2, which is (2, 5). The gradient of AB is (8 − 2)/(5 − (−1)) = 6/6 = 1. The perpendicular bisector has gradient −1 and passes through (2, 5): y − 5 = −(x − 2), which rearranges to x + y = 7. Distractor routes: x − y = −3 uses the gradient of AB itself, 1, rather than its negative reciprocal, giving a line PARALLEL to AB through the midpoint instead of perpendicular to it. x + y = 12 comes from adding the y-coordinates, 2 + 8 = 10, but forgetting to divide by 2, using the midpoint (2, 10) instead of (2, 5). x + y = 4 comes from using A's x-coordinate, −1, directly instead of averaging it with B's, giving the point (−1, 5) instead of the true midpoint (2, 5).
- (c) 5 — Method: let a be the number of adult tickets, so the number of child tickets is 9 − a. Form the equation 10a + 6(9 − a) = 74. Working: expand the bracket: 10a + 54 − 6a = 74, so 4a = 20, giving a = 5. Answer: 5 adult tickets. 4 comes from correctly finding the number of child tickets but reporting it instead of the number of adult tickets asked for. 1.25 comes from a sign error expanding the bracket, writing 10a + 54 + 6a = 74 instead of subtracting, which gives 16a = 20. 7.4 comes from dividing the total takings by the adult ticket price only, ignoring the children entirely, 74 ÷ 10.
- (b) x ≥ 5 — Expand the bracket: 2(3x − 1) = 6x − 2, so the inequality is 6x − 2 ≥ 4x + 8. Subtract 4x from both sides and add 2 to both sides: 2x ≥ 10. Divide both sides by 2: x ≥ 5. A candidate who only multiplies the 3x by 2 and forgets to multiply the −1 gets 6x − 1 ≥ 4x + 8, leading to x ≥ 4.5. A candidate who adds 4x instead of subtracting it gets 10x ≥ 10, leading to x ≥ 1. A candidate who multiplies by 2 instead of dividing gets x ≥ 20.
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