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
MathsUKwww.geekhero.co.uk
- 1.The simple interest, £I, earned on a savings account is given by the formula I = PRT/100, where P is the amount invested in pounds, R is the interest rate per year and T is the time in years. Rearrange the formula to make P the subject.
- 2.Solve the simultaneous equations 3x + y = 14 and x + y = 6. Work out the value of x.
- 3.A straight line has gradient 3 and passes through the point (1, 4). Work out the equation of the line.
- 4.Solve the simultaneous equations 3x + 2y = 16 and x + y = 7. Work out the value of y.
- 5.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.
- 6.A cyclist's speed-time graph is described as follows: speed increases steadily from 0 m/s to 8 m/s over the first 4 seconds, stays constant at 8 m/s for the next 6 seconds, then decreases steadily to 0 m/s over the final 2 seconds. Work out the total distance travelled.
- 7.The first four terms of a sequence are 11, 18, 25, 32. Ravi thinks the nth term is 7n. Work out the correct expression for the nth term.
- 8.A circle has centre (0, 0) and equation x² + y² = 8. Work out the radius of the circle, giving your answer as a surd in its simplest form.
- 9.A company's profit was £2000 in its first year. Each following year, the profit increases by £800. Work out the first year in which the profit is more than £7000.
- 10.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.
- 11.Expand and simplify (2x − 1)(x + 5)(x − 2). Write down the coefficient of x in your answer.
- 12.Solve x² − 8x + 3 = 0 by completing the square. Give your answers in surd form.
- 13.Solve the simultaneous equations y = 2x − 1 and y = x² − x − 1, giving both pairs of solutions.y = 2x − 1y = x²
- 14.The iterative formula xₙ₊₁ = 5 − 3/xₙ is used with starting value x₀ = 2.5, so that x₁ is the value after the formula has been used once. Work out x₄ correct to 3 significant figures.
- 15.The point (3, 4) lies on the circle x² + y² = 25, which has centre (0, 0). Work out the gradient of the tangent to the circle at (3, 4).
Answer key
- (a) P = 100I / (RT) — Method: undo the operations done to P in reverse order — multiply by 100, then divide by R and by T. Working: I = PRT/100, so 100I = PRT, so P = 100I / (RT). Answer: P = 100I / (RT). P = IRT/100 comes from leaving R and T in the numerator instead of moving them to the denominator. P = 100RT/I comes from swapping P and I when rearranging. P = 100I/R comes from dividing by R only and forgetting to also divide by T.
- (d) 4 — Subtracting the second equation from the first: the y-terms cancel, and the x-terms combine as 3x − x = 2x; the right-hand sides give 14 − 6 = 8. This gives 2x = 8, so x = 4. A candidate who subtracts in the wrong order would get 2x = 6 − 14 = −8, so x = −4. A candidate who uses only the first equation, dividing 14 by 3 as if y were 0, would get x ≈ 4.67. A candidate who reports the value of y instead of x would get y = 6 − 4 = 2.
- (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.
- (b) 5 — From x + y = 7, x = 7 − y. Substituting into 3x + 2y = 16: 3(7 − y) + 2y = 16, so 21 − 3y + 2y = 16, giving 21 − y = 16, so y = 5 (then x = 2). A candidate who forgets to multiply the y-term inside the bracket by 3 would write 21 − y + 2y = 16, giving 21 + y = 16, so y = −5. A candidate who subtracts the two equations directly, (3x + 2y) − (x + y) = 16 − 7, gets 2x + y = 9, and if they wrongly treat this as giving y alone, ignoring the x term, they would answer 9. A candidate who reports the value of x instead of y would answer 2.
- (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.
- (c) 72 m — The first section is a triangle with base 4 s and height 8 m/s, area = 1/2 × 4 × 8 = 16 m. The middle section is a rectangle with base 6 s and height 8 m/s, area = 6 × 8 = 48 m. The final section is a triangle with base 2 s and height 8 m/s, area = 1/2 × 2 × 8 = 8 m. Total distance = 16 + 48 + 8 = 72 m. Treating both sloping sections as full rectangles instead of triangles — forgetting the factor of a half — gives 4 × 8 + 6 × 8 + 2 × 8 = 96 m, which is wrong because a sloping speed-time section is a triangle, not a rectangle. Leaving out the final, decreasing section entirely gives 16 + 48 = 64 m, which is wrong because the cyclist is still moving and covering distance while slowing down. Leaving out the middle, constant-speed section gives 16 + 8 = 24 m, which is wrong because the cyclist covers the most distance of all during the 6 seconds of constant speed.
- (a) 7n + 4 — Method: find the common difference, then find the constant that fits the first term. Working: 18 − 11 = 7, 25 − 18 = 7, 32 − 25 = 7, so the terms increase by 7 each time and the nth term has the form 7n + c. Substituting n = 1: 7(1) + c = 11, so c = 4. Answer: the correct nth term is 7n + 4. The value 7n is Ravi's value, which comes from using only the common difference and leaving out the constant. The value 7n + 11 comes from using the first term as the constant directly, without subtracting the common difference first. The value 11n + 7 comes from swapping the roles of the first term and the common difference — using the first term, 11, as the coefficient of n and the difference, 7, as the constant.
- (d) 2√2 — The radius satisfies r² = 8, so r = √8 = √(4 × 2) = √4 × √2 = 2√2. Choosing 8 forgets to take the square root of r² at all. Choosing 4 comes from halving 8 instead of finding its square root. Choosing √2 splits off the factor of 4 correctly but forgets to multiply the 2 back in front of the root.
- (d) 8 — Method: write the nth term of the sequence, 2000 + 800(n − 1), and find the smallest whole n for which it is greater than 7000. Working: 2000 + 800(n − 1) > 7000, so 800(n − 1) > 5000, giving n − 1 > 6.25. Since n − 1 must be a whole number, the smallest value is 7, so n = 8. Check: year 8's total is 2000 + 800 × 7 = 7600, which is more than £7000, while year 7's total is 2000 + 800 × 6 = 6800, which is not. Answer: year 8. 7 comes from rounding 6.25 to the nearest whole number, 6, and then adding 1, instead of rounding up to the next whole number before adding 1. 6 comes from using 6.25 rounded down to 6 as the year number directly, without adding the 1 needed to convert from the number of increases to the year number. 9 comes from adding one extra year beyond the year that already satisfies the condition.
- (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) −23 — Expand two of the three brackets first: (x + 5)(x − 2) = x² + 3x − 10. Then multiply this by the remaining bracket: (2x − 1)(x² + 3x − 10) = 2x³ + 6x² − 20x − x² − 3x + 10, which simplifies to 2x³ + 5x² − 23x + 10, so the coefficient of x is −23. Writing −13 comes from a sign slip in the first expansion, combining 5x − 2x as −5x − 2x = −7x instead of +3x, which carries through to a wrong final coefficient. Writing −20 comes from forgetting to distribute the −1 across every term of x² + 3x − 10, dropping the −1 × 3x = −3x contribution. Writing 10 comes from reading off the constant term of the expansion instead of the coefficient of x.
- (a) x = 4 ± √13 — Method: halve the coefficient of x to form the bracket, subtract the square of that number to keep the expression equal to the original, then rearrange and take the square root of both sides. Working: half of −8 is −4, so x² − 8x + 3 = (x − 4)² − 16 + 3 = (x − 4)² − 13; setting this equal to zero gives (x − 4)² = 13, so x − 4 = ±√13 and x = 4 ± √13. Answer: x = 4 ± √13. The distractors: x = 8 ± √13 comes from putting the whole coefficient 8 inside the bracket instead of half of it; x = 4 ± √19 comes from adding 16 and 3 rather than subtracting 3 from 16; x = −4 ± √13 comes from writing the bracket as (x + 4)², which reverses the sign of the number that comes out of it.
- (d) x = 0, y = −1 and x = 3, y = 5 — Set the two expressions for y equal: 2x − 1 = x² − x − 1. Rearranging, subtracting 2x and adding 1 to both sides: 0 = x² − x − 1 − 2x + 1 = x² − 3x, so x² − 3x = 0. Factorise: x(x − 3) = 0, giving x = 0 or x = 3. Using y = 2x − 1: x = 0 gives y = −1; x = 3 gives y = 5. Distractor routes: x = 0, y = −1 alone stops after the factor x = 0 and never checks the second factor, x − 3 = 0. x = 0, y = −1 and x = −3, y = −7 comes from mis-factorising x² − 3x as x(x + 3), a sign error that gives a second root of −3 instead of 3. x = −2, y = −5 and x = 1, y = 1 comes from adding 2x to both sides instead of subtracting it when rearranging, giving x² + x − 2 = 0 instead of x² − 3x = 0.
- (c) 4.30 — Method: substitute the starting value into the right-hand side to get x₁, then feed each value back in, keeping the whole display and respecting the order of operations, which divides before it subtracts. Working: x₁ = 5 − 3 ÷ 2.5 = 5 − 1.2 = 3.8; x₂ = 5 − 3 ÷ 3.8 = 5 − 0.78947… = 4.21052…; x₃ = 5 − 3 ÷ 4.21052… = 5 − 0.7125 = 4.2875; x₄ = 5 − 3 ÷ 4.2875 = 5 − 0.69970… = 4.30029…, which is 4.30 correct to 3 significant figures. Answer: 4.30. The distractors: 4.29 is x₃ = 4.2875 rounded, reached by counting the starting value itself as the first iterate and so stopping one use of the formula early; 3.80 is x₁, the value after a single use of the formula; 2.50 comes from working out (5 − 3) ÷ xₙ instead of 5 − (3 ÷ xₙ), subtracting before dividing, which produces the sequence 0.8, 2.5, 0.8, 2.5 and lands on 2.5 at the fourth step.
- (b) −3/4 — The tangent to a circle at a point is always perpendicular to the radius drawn to that point. The radius from (0, 0) to (3, 4) has gradient 4/3. The gradient of a line perpendicular to a line with gradient m is the negative reciprocal, −1/m, so the tangent's gradient here is −3/4. Using the radius's own gradient, forgetting that the tangent is perpendicular to it, gives 4/3. Negating the radius's gradient without also taking the reciprocal gives −4/3. Taking the reciprocal of the radius's gradient without negating it gives 3/4.
Build your own mix at the worksheet builder.