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 rectangular photo has length (x + 3) cm and width x cm. Its area is 40 cm². Work out the value of x.
- 2.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.
- 3.An allotment is in the shape of a rectangle. Its length is 5 m more than its width, x metres, and its area is 20 m². This gives x² + 5x − 20 = 0, which can be solved using the iterative formula xₙ₊₁ = 20 ÷ (xₙ + 5). Taking x₀ = 2, so that x₁ is the value found after the formula has been used once, work out x₂ correct to 2 decimal places.
- 4.A plumber charges a call-out fee of £30 plus £25 per hour worked. Work out the total charge for a job that takes 3 hours.
- 5.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.
- 6.The graph of y = 60 × (0.5)ˣ is sketched for x ≥ 0. Which statement correctly describes what happens to the curve as x increases?
- 7.A tap fills a tank at a varying rate. A graph shows the rate of flow, in litres per minute, against time, in minutes. What does the area under this graph represent?
- 8.A tap's flow rate, in litres per minute, is plotted against time, in minutes, on a graph. What are the units of the area under this graph?
- 9.A scout leader is 44 years old and one of the scouts is 16 years old. Work out how many years it will be until the leader is exactly twice as old as the scout.
- 10.Look at the statement 2x + 5 = 17. Considering whether it is an expression, an equation, a formula, or an identity, which classification is correct?
- 11.Solve 5x² − 15x = 0.
- 12.Three consecutive integers add up to 72. Using n for the smallest integer, form an equation and solve it to find the smallest of the three integers.
- 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.A circle has centre (0, 0) and passes through the point (12, 35). Work out the equation of the circle.
- 15.The mass of a chemical sample decays by 15% every hour. It starts at 500 g. Work out the mass remaining after 3 hours, correct to 1 decimal place.
Answer key
- (b) x = 5 — Area = length × width, so x(x + 3) = 40, which rearranges to x² + 3x − 40 = 0. This factorises as (x + 8)(x − 5) = 0: the two numbers in the brackets must multiply to −40 and add to +3, and the pair 8 and −5 does both. This gives x = −8 or x = 5. Since x is a length, it cannot be negative, so x = 5. A candidate who gives both solutions without rejecting the negative one, which cannot be a length, answers x = 5 or x = −8. A candidate who picks the wrong factor pair of 40, such as 10 and −4 instead of 8 and −5, gets (x + 10)(x − 4) = 0 and answers x = 4. A candidate who rejects the wrong root, keeping the negative solution instead of the positive one, answers x = −8.
- (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) 2.55 — x₁ = 20 ÷ (2 + 5) = 20 ÷ 7 = 2.857142857. x₂ = 20 ÷ (2.857142857 + 5) = 20 ÷ 7.857142857 = 2.545454545, which rounds to 2.55. Reporting x₁ instead of x₂ gives 2.857142857, which rounds to 2.86. Dropping the +5 in the denominator, using xₙ₊₁ = 20 ÷ xₙ, gives x₁ = 20 ÷ 2 = 10 and x₂ = 20 ÷ 10 = 2, which is 2.00. A sign error in the denominator, using xₙ₊₁ = 20 ÷ (xₙ − 5), gives x₁ = 20 ÷ (2 − 5) = −6.666666667 and x₂ = 20 ÷ (−6.666666667 − 5) = −1.714285714, which rounds to −1.71.
- (b) £105 — The hourly charge is 25 × 3 = £75. Adding the call-out fee: £75 + £30 = £105. A candidate who forgets the call-out fee gives just the hourly charge, £75. A candidate who adds the call-out fee to the hourly rate before multiplying by the hours, (30 + 25) × 3, gets £165. A candidate who multiplies the call-out fee by the number of hours instead of the hourly rate, 30 × 3, gets £90.
- (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.
- (a) It gets closer to zero but never actually reaches it. — y = 60 × (0.5)ˣ is always positive, however large x is, because a positive number raised to any power stays positive. As x increases, (0.5)ˣ gets smaller and smaller but never equals zero, so y approaches zero without ever reaching it. Believing the curve reaches zero when x = 60, because 60 is the starting value, mixes up the y-intercept with a stopping point — an exponential decay curve like this one never actually reaches the x-axis, so this is wrong. Believing the curve goes negative for large x is wrong because multiplying a positive number by (0.5) any number of times can never produce a negative result. Believing the curve levels off at y = 0.5, confusing the base of the exponential with its eventual level, is wrong: the base only controls how fast the curve falls, not where it settles — this curve settles at y = 0, not y = 0.5. Whenever you sketch an exponential decay curve, draw it approaching the x-axis but never touching it.
- (b) The total volume of water, in litres, that has flowed in. — On a rate-time graph, the y-axis is in litres per minute and the x-axis is in minutes; multiplying a rate by a time gives litres per minute × minutes = litres, a total volume. So the area under the graph represents the total volume of water that has flowed in. Thinking the area itself represents the rate, rather than what the rate accumulates to, gives the wrong claim about the average rate of flow. Confusing the area with the gradient of the graph — which measures how the rate is changing — gives the wrong claim about litres per minute squared. Ignoring the flow-rate axis and focusing only on the time axis gives the wrong claim that the area is simply the total time.
- (a) Litres — The area under a graph is found by multiplying a y-value by an x-value, so its units are the y-axis units multiplied by the x-axis units: litres per minute × minutes = litres, since the 'per minute' cancels with the 'minutes'. Answering 'litres per minute' keeps the y-axis units unchanged, as if multiplying by time did nothing to the units at all. Answering 'litres per minute squared' treats the x-axis as also being measured 'per minute', squaring a unit that should instead cancel. Answering 'minutes' keeps only the x-axis units and drops the rate altogether.
- (b) 12 years — Method: call the number of years t, add t to both ages, and form an equation from the comparison at that future time. Working: in t years the leader will be 44 + t and the scout will be 16 + t, so 44 + t = 2(16 + t); expanding gives 44 + t = 32 + 2t, and subtracting t and 32 from both sides gives t = 12. Checking: in 12 years the leader will be 56 and the scout 28, and 56 = 2 × 28. Answer: 12 years. The distractors: 6 years comes from halving the leader's present age instead, so that the scout has to reach 22, which takes 6 years; 14 years comes from halving the 28-year gap between the two ages; 28 years comes from giving the age gap itself as the number of years.
- (d) An equation, true for one value of x — 2x + 5 = 17 is only true when x = 6, so it is an equation. A candidate who ignores the equals sign and focuses only on the x terms and numbers present picks the expression option. A candidate who sees an equals sign and assumes it must relate two different quantities, as a formula does, picks that option, missing that there is only one quantity, x, involved here. A candidate who wrongly checks whether both sides look similar in structure rather than testing specific values picks the identity option, missing that the two sides are not equal for every value of x.
- (c) x = 0 or x = 3 — Method: take out the common factor 5x: 5x(x − 3) = 0, so 5x = 0 or x − 3 = 0, giving x = 0 or x = 3. Distractor origins: x = 0 or x = 15 comes from forgetting to divide the second term by the common factor 5x correctly, leaving 15 instead of 3; x = 3 loses the solution x = 0 by dividing both sides by x; x = 5 or x = 3 mistakes the coefficient 5 itself for one of the solutions.
- (d) 23 — Method: let the smallest integer be n, so the three consecutive integers are n, n + 1 and n + 2. Form the equation n + (n + 1) + (n + 2) = 72. Working: simplify the left side: 3n + 3 = 72, so 3n = 69, giving n = 23. Answer: 23. 24 comes from dividing 72 by 3 directly, 72 ÷ 3 = 24, which finds the middle integer rather than realising the three numbers differ. 25 comes from correctly finding n = 23 but reading off the largest integer, n + 2, instead of the smallest as asked. 21 comes from dividing first and subtracting after, (72 ÷ 3) − 3, instead of subtracting 3 before dividing by 3.
- (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.
- (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) 307.1 g — The multiplier for one hour is 1 − 0.15 = 0.85. After 3 hours the mass is 500 × 0.85³. Since 0.85 × 0.85 = 0.7225 and 0.85 × 0.7225 = 0.614125, the mass is 500 × 0.614125 = 307.0625 g, which rounds to 307.1 g. Stopping after only 2 hours instead of 3 gives 500 × 0.7225 = 361.25 g, which rounds to 361.3 g — wrong, because the question asks for 3 hours, not 2. Treating the decay as simple (15% × 3 = 45% lost in total, applied once) gives 500 × 0.55 = 275.0 g, which is wrong because the decay compounds hour by hour rather than adding up. Working out the AMOUNT LOST instead of the mass remaining gives 500 − 307.0625 = 192.9375 g, which rounds to 192.9 g — wrong, because the question asks what remains, not what has decayed away. Whenever a percentage decreases repeatedly, multiply by the same factor each period rather than adding the percentages together.
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