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 caterer uses the formula C = 6p + 20 to work out the total cost, £C, of a buffet for p people, where £20 covers fixed costs. A customer is charged £92. Work out how many people, p, the buffet was for.
- 2.A bath is filled with water; the graph of volume (litres) against time (minutes) is a curve, since the flow rate changes. The tangent to the curve at t = 10 minutes passes through the points (8, 130) and (12, 190), and the volume in the bath at t = 10 minutes is 160 litres. Use the gradient of this tangent to estimate the volume in the bath 5 minutes after t = 10 minutes.
- 3.Make x the subject of the formula y = 2x − 9.y = 2x − 9
- 4.Solve 6x − 5 = 3x + 13.
- 5.Ben is asked to find the inverse of f(x) = 4 − 3x. He writes f⁻¹(x) = (4 − x)/3. Which statement about Ben's answer is correct?
- 6.A car's velocity–time graph is a straight line from (0 s, 4 m/s) rising to (6 s, V m/s), followed by a straight line falling from (6 s, V m/s) to (9 s, 0 m/s). The gradient of the second line is −8 m/s². Work out the total distance travelled between t = 0 and t = 9 seconds.
- 7.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.
- 8.A car's speed increases steadily while it accelerates. Its speed v m/s after t seconds is given by v = u + at, where u is the starting speed in m/s and a is the acceleration in m/s². A car starts at u = 4 m/s and accelerates at a = 2 m/s² until it reaches v = 20 m/s. Work out t, the time taken in seconds.
- 9.A gardener charges a fixed fee of £15 plus £8 for each hour worked. The total charge, £T, for a job lasting h hours is given by T = 15 + 8h. Work out the charge for a job lasting 4 hours, and identify what kind of statement T = 15 + 8h is.
- 10.Four sequences are shown below. Sequence P: 4, 8, 16, 32, ... Sequence Q: 4, 8, 12, 16, ... Sequence R: 4, 7, 12, 19, ... Sequence S: 4, 9, 16, 25, ... Work out which of the sequences P, Q, R and S is geometric.
- 11.A student says that (x + 4)² is equivalent to x² + 16. For which value of x do the two expressions give the SAME result, making it look (misleadingly) like the student could be right?
- 12.The point (4, 2) lies on the circle x² + y² = 20. Work out the equation of the tangent to the circle at (4, 2).
- 13.A company's weekly profit, P thousand pounds, when it makes x thousand items is modelled by P = x² − 8x + 12, for x ≥ 0. Work out the values of x for which the company makes a loss.
- 14.Which of these equations is equivalent to 5x − 2 = 3x + 8?
- 15.Make x the subject of the formula y = (x − 4)/5.
Answer key
- (b) 12 — Method: substitute the total cost into the formula, then subtract the fixed cost and divide by the cost per person. Working: 92 = 6p + 20, so 6p = 92 − 20 = 72, p = 72 ÷ 6 = 12. Answer: 12 people. 15.33 comes from dividing the whole £92 by £6 without first subtracting the £20 fixed cost: 92 ÷ 6 ≈ 15.33. 4.3 comes from swapping the two amounts round, subtracting £6 and dividing by £20: (92 − 6) ÷ 20 = 4.3. 18.67 comes from adding the fixed cost instead of subtracting it: (92 + 20) ÷ 6 ≈ 18.67.
- (b) 235 litres — The flow rate at t = 10 is the gradient of the tangent: change in volume ÷ change in time = (190 − 130) ÷ (12 − 8) = 60 ÷ 4 = 15 litres per minute. Treating this rate as roughly constant for a short interval, the volume 5 minutes after t = 10 is estimated as 160 + 5 × 15 = 235 litres. Using the tangent's own point spacing — 2 minutes, from t = 10 to t = 12 — instead of the 5 minutes actually asked for gives 160 + 2 × 15 = 190, which is just the volume already given at one of the tangent's own points, not an answer to the question asked. Multiplying the gradient by the time WITHOUT adding the starting volume, 5 × 15 = 75, forgets that a rate estimates a CHANGE, which must be added to the starting volume, not given as the answer on its own. Subtracting instead of adding, 160 − 5 × 15 = 85, extrapolates backward in time rather than forward.
- (a) x = (y + 9) / 2 — Method: undo the subtraction of 9 first, then undo the multiplication by 2. Working: y = 2x − 9, so adding 9 to both sides gives y + 9 = 2x, then dividing both sides by 2 gives x = (y + 9) / 2. The value x = (y − 9) / 2 comes from a sign error, keeping the −9 instead of moving it to +9. The value x = 2(y + 9) comes from multiplying by 2 instead of dividing. The value x = y / 2 + 9 comes from dividing by 2 before adding 9, the wrong order of operations, instead of dividing the whole bracket.
- (d) 6 — Method: collect the x-terms on one side and the constants on the other, then divide by the remaining coefficient of x. Working: 6x − 3x = 13 + 5, so 3x = 18, x = 18 ÷ 3 = 6. Answer: x = 6. 2.67 comes from a sign error when moving the 5, subtracting instead of adding: 3x = 13 − 5 = 8, x = 8 ÷ 3 ≈ 2.67. 2 comes from a sign error when moving the x-term, adding instead of subtracting: 9x = 18, x = 2. 18 comes from correctly finding 3x = 18 but forgetting to divide by 3.
- (c) Correct: 3y = 4 − x gives f⁻¹(x) = (4 − x)/3 — Swap x and y: x = 4 − 3y. Add 3y to both sides and subtract x from both sides: 3y = 4 − x. Divide by 3: y = (4 − x)/3, which is exactly what Ben wrote — his rearrangement is correct. Check with a value: f(1) = 4 − 3 = 1, and Ben's formula gives (4 − 1)/3 = 1, which matches. 'Correct, but only because f is its own inverse' gives the right verdict for a false reason — f(f(x)) = 4 − 3(4 − 3x) = 9x − 8, which is not x, so f is not self-inverse; Ben's rearrangement is correct for the ordinary algebraic reason above, not because of any special property of f. 'Wrong: sign kept, giving (−4 − x)/3' comes from not carrying the swap through consistently — testing x = 1 gives (−4 − 1)/3 = −5/3, which does not equal 1, so it is wrong. 'Wrong: correct inverse is (x − 4)/3' comes from writing 3y = x − 4 instead of 3y = 4 − x, a sign slip when isolating y — testing x = 1 gives (1 − 4)/3 = −1, which again does not equal 1.
- (a) 120 m — The gradient of the second line is (0 − V)/(9 − 6) = −V/3, and this equals −8, so V = 24. The distance from t = 0 to t = 6 is the area of a trapezium with parallel sides 4 and 24 and width 6: 1/2 × (4 + 24) × 6 = 84. The distance from t = 6 to t = 9 is the area of a triangle with base 3 and height 24: 1/2 × 3 × 24 = 36. The total distance is 84 + 36 = 120 m. Using V = 8, treating the gradient's number as the missing velocity itself rather than solving −V/3 = −8 for V, gives a trapezium area of 1/2 × (4 + 8) × 6 = 36 and a triangle area of 1/2 × 3 × 8 = 12, a total of 48 m. Leaving out the 1/2 in the trapezium formula, (4 + 24) × 6 = 168, plus the correct triangle of 36, gives 204 m. Using the full 6 seconds as the triangle's base instead of the 3 seconds the second line actually lasts, 1/2 × 6 × 24 = 72, plus the correct trapezium of 84, gives 156 m.
- (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.
- (b) 8 s — Rearranging v = u + at for t: subtract u from both sides to get v − u = at, then divide by a: t = (v − u)/a. Substituting u = 4, a = 2, v = 20: t = (20 − 4)/2 = 16/2 = 8 s. Answering 12 s comes from adding u instead of subtracting it: (20 + 4)/2 = 12. Answering 32 s comes from multiplying (v − u) by a instead of dividing by it: 16 × 2 = 32. Answering 6 s divides v by a first and then subtracts u, in the wrong order: 20/2 − 4 = 10 − 4 = 6. The time taken is 8 s.
- (b) £47, and T=15+8h is a formula (relates T and h). — The charge is £15 fixed plus £8 for each hour: T=15+8h, so for h=4, T=15+8×4=15+32=47. Because T=15+8h relates two different quantities, the total charge T and the number of hours h, it is a formula, not an equation — an equation is solved for one particular value of an unknown, but this relationship holds for every value of h a job might last. A candidate who forgets to include the fixed £15 fee would compute only 8×4=32. A candidate who adds the three numbers in the formula together instead of multiplying the hourly rate by the number of hours would compute 15+8+4=27. A candidate who correctly finds the charge but mistakes the formula for an equation, treating it as something to be solved for one specific value of h rather than a general relationship between T and h, would pick the correct charge with the wrong classification.
- (c) P — Sequence P has a common ratio of 2 (4 × 2 = 8, 8 × 2 = 16, 16 × 2 = 32), so it is geometric. Sequence Q has a common difference of 4, which is arithmetic, not geometric — a candidate who confuses a constant difference with a constant ratio picks Q. Sequence R has increasing differences of 3, 5, 7, a quadratic sequence, not geometric — a candidate who assumes any fast-growing sequence must be geometric picks R. Sequence S is the square numbers from 2² onwards (2², 3², 4², 5²), which is also quadratic, not geometric — a candidate who thinks squaring always means geometric growth picks S.
- (c) x = 0 — Expand (x + 4)² correctly: (x + 4)² = x² + 8x + 16. This equals x² + 16 only when 8x is zero, i.e. when x = 0 — at every other value of x the two expressions differ by 8x. Choosing x = 4 confuses the constant inside the bracket with the value of x that makes the expressions match. Choosing x = −4 makes the same confusion but with the sign flipped. Choosing x = 8 mistakes the coefficient of the middle term, 8x, for the value of x itself.
- (d) y = −2x + 10 — Method: a tangent is perpendicular to the radius drawn to the point where it touches, so work out the gradient of that radius, take its negative reciprocal for the tangent, then substitute into y − y₁ = m(x − x₁). Working: the radius joins (0, 0) to (4, 2), so its gradient is 2 ÷ 4 = 1/2; turning 1/2 upside down gives 2 and changing the sign gives −2. Substituting into y − 2 = −2(x − 4) gives y − 2 = −2x + 8, so y = −2x + 10. Answer: y = −2x + 10. The distractors: y = −0.5x + 4 changes the sign of the radius gradient but never turns it upside down, using −1/2 where −2 belongs; y = 2x − 6 turns the gradient upside down but leaves it positive, using 2 where −2 belongs; y = −2x − 10 has the correct gradient but substitutes the point with both signs reversed, writing y + 2 = −2(x + 4) instead of y − 2 = −2(x − 4).
- (b) 2 < x < 6 — Factorise: x² − 8x + 12 = (x − 2)(x − 6), giving roots at x = 2 and x = 6. Since the coefficient of x² is positive, the graph is a U-shape that is negative, below the axis, between its roots. So P < 0 for 2 < x < 6. Distractor routes: x < 2 or x > 6 takes the region outside the roots, which is where P is positive, a profit, the opposite of a loss. 2 ≤ x ≤ 6 includes the endpoints, where P = 0 exactly, break-even rather than a loss, since the inequality is strict. −6 < x < −2 comes from factorising as (x + 2)(x + 6), reversing the sign of both roots.
- (d) 2x = 10 — Method: collect the x-terms on one side and the constants on the other, keeping the equals sign balanced. Working: 5x − 3x = 8 + 2, giving 2x = 10. Answer: 2x = 10. 2x = 6 comes from a sign error moving the constant, subtracting instead of adding: 8 − 2 = 6. 8x = 6 comes from adding the x-terms instead of subtracting them (5x + 3x = 8x), alongside the constant sign error. 2x = −10 comes from a sign error moving the constant the other way: −2 − 8 = −10.
- (d) x = 5y + 4 — To make x the subject of y = (x − 4)/5, first multiply both sides by 5 to clear the fraction: 5y = x − 4, then add 4 to both sides: x = 5y + 4. Writing x = 5y − 4 multiplies correctly but keeps the minus sign on the 4 instead of changing it to a plus when moving it across. Writing x = y/5 + 4 divides by 5 instead of multiplying, the wrong inverse of the fraction. Writing x = 5(y + 4) adds 4 before multiplying by 5, reversing the correct order of the two steps. The correct rearrangement is x = 5y + 4.
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