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.
Calculator
Algebra worksheet — GCSE Higher
MathsUKwww.geekhero.co.uk
- 1.The table shows the height, in metres, of a firework rocket at various times, in seconds, during its flight: 40 at t = 2, 54 at t = 3, and 60 at t = 4. Use the chord between t = 2 and t = 4 to estimate the gradient of the height-time graph at t = 3, stating the correct units.
- 2.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.
- 3.A school trip costs a £15 deposit plus £9 per student for the coach. The total cost for a class is £186. Work out how many students went on the trip.
- 4.A tram travels between two stops. Its velocity-time graph consists of straight line segments joining the points (0, 0), (5, 20), (12, 20), (16, 4) and (20, 4), where time is in seconds and velocity is in m/s. Work out the average speed of the tram over the whole 20 seconds. Give your answer to 1 decimal place.
- 5.A photo printing service has two adverts for its price. Advert A: cost in pounds = 3(2n + 4) for n photos. Advert B: cost in pounds = 6n + 12. A customer says the two adverts always charge the same amount. Is the customer correct?
- 6.A graph shows the total monthly cost of a mobile phone tariff against the number of minutes of calls. The line starts at £12 and stays level until 200 minutes, then rises by 5p for each further minute of calls. Use the graph to work out the total cost of a month in which 250 minutes of calls are made.
- 7.A phone company works out a monthly bill, £B, using the formula B = 18 + 0.05t, where £18 is the fixed monthly charge and t is the number of extra text messages sent beyond the free allowance, each charged at 5p. Farida's bill for one month is £24.50. Work out the number of extra text messages, t, she sent.
- 8.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.
- 9.A closed cylinder has radius r cm and height (r + 5) cm. Its volume is 300 cm³, giving the equation πr²(r + 5) = 300, which can be solved using the iterative formula rₙ₊₁ = √(300 ÷ (π(rₙ + 5))). Taking r₀ = 3, work out r₃ correct to 2 decimal places.
- 10.The iterative formula xₙ₊₁ = √(2xₙ + 3) is used repeatedly, starting from x₀ = 1. As n increases, the values of xₙ converge to a limit, L. Work out L.
- 11.A rectangular sheet of metal measures 20 cm by 12 cm. A square of side x cm is cut from each corner and the sides are folded up to make an open box of volume 200 cm³. This gives x³ − 16x² + 60x − 50 = 0, which can be solved using the iterative formula xₙ₊₁ = (16xₙ² − xₙ³ + 50)/60. The starting value is x₀ = 1, so x₁ is the value after the formula has been used once. Work out x₃, and use it to find the longer side of the base of the box correct to 1 decimal place.
- 12.A population of bacteria doubles every 3 hours. It starts at 800. Work out the population after 9 hours.
- 13.A speed-time graph shows a constant speed of 15 m/s for 20 seconds. Work out the distance travelled, using the area under the graph.
- 14.A parcel delivery company's cost graph shows the following: the cost is a flat £5 for parcels weighing up to 2 kg, and then rises by £2 for each additional kg above 2 kg. Use the graph to work out the weight of a parcel that costs £17.
- 15.By completing the square, find the turning point of the curve y = x² + 8x − 3.y = x² + 8x − 3
Answer key
- (c) 10 m/s — A symmetric chord gradient uses the two points either side of t = 3: (2, 40) and (4, 60). The gradient is the change in height divided by the change in time: (60 − 40) ÷ (4 − 2) = 20 ÷ 2 = 10 m/s. Using only the values either side of one gap, (2, 40) and (3, 54), instead of the full symmetric chord, gives (54 − 40) ÷ (3 − 2) = 14 m/s. Getting the correct number but dropping the time unit, leaving only metres, gives 10 m. Dividing time by height instead of height by time inverts the calculation to (4 − 2) ÷ (60 − 40) = 0.1 s/m.
- (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.
- (c) 19 — Method: set up the equation 15 + 9n = 186, then subtract the deposit and divide by the cost per student. Working: 9n = 186 − 15 = 171; n = 171 ÷ 9 = 19. Answer: 19 students. 20.67 comes from dividing the whole £186 by £9 without first subtracting the deposit: 186 ÷ 9 ≈ 20.67. 11.8 comes from swapping the two amounts round, subtracting £9 and dividing by £15: (186 − 9) ÷ 15 = 11.8. 22.33 comes from adding the deposit instead of subtracting it: (186 + 15) ÷ 9 ≈ 22.33.
- (b) 12.7 m/s — Find the total distance from the area under the graph, then divide by the total time. Break the graph into its four straight sections: a triangle from (0, 0) to (5, 20), where 5 × 20 = 100 gives an area of 50 m when halved; a rectangle from (5, 20) to (12, 20), area 7 × 20 = 140 m; a trapezium from (12, 20) to (16, 4), where (20 + 4) × 4 = 96 gives an area of 48 m when halved; and a rectangle from (16, 4) to (20, 4), area 4 × 4 = 16 m — a total distance of 50 + 140 + 48 + 16 = 254 m. Dividing by the 20 seconds gives an average speed of 254 ÷ 20 = 12.7 m/s. Writing 15.1 m/s comes from forgetting to halve the trapezium area for the third section, using 96 instead of 48: a total of 302 m, and 302 ÷ 20 = 15.1 m/s. Writing 11.9 m/s comes from leaving out the final section, from (16, 4) to (20, 4), entirely: a total of 238 m, and 238 ÷ 20 = 11.9 m/s. Writing 12.1 m/s comes from dividing the correct total distance by 21 instead of 20 — a fencepost slip, counting the whole seconds from t = 0 to t = 20 inclusive as 21 seconds of travel rather than reading the journey as a duration of 20 seconds: 254 ÷ 21 = 12.1 m/s (1 d.p.).
- (a) They always charge the same, since 3(2n + 4) = 6n + 12. — Expand Advert A's formula by multiplying both terms inside the bracket by 3: 3 × 2n = 6n, and 3 × 4 = 12, giving 3(2n + 4) = 6n + 12, which is identical to Advert B's formula — so the two adverts always charge the same amount, whatever n is. Getting 6n + 4 comes from multiplying the 2n by 3 but leaving the 4 unmultiplied. Getting 2n + 7 comes from adding 3 to the bracket instead of multiplying by it. Saying it depends on n avoids expanding the bracket at all — once expanded, both formulas are identical for every value of n, so the cost can be compared directly.
- (a) £14.50 — Method: the first 200 minutes are already covered by the £12, so only 250 − 200 = 50 minutes are charged extra. Extra charge = 50 × £0.05 = £2.50. Total cost = £12 + £2.50 = £14.50. Distractor origins: £24.50 charges 5p for all 250 minutes instead of only the 50 minutes over 200 (250 × £0.05 = £12.50, plus £12 = £24.50); £12.50 makes that same slip of charging all 250 minutes but then forgets to add the £12 monthly fee; £13.50 works out 250 − 200 wrongly as 30 extra minutes instead of 50 (30 × £0.05 = £1.50, plus £12 = £13.50).
- (a) 130 — Method: substitute the total bill into the formula, then subtract the fixed charge and divide by the cost per text message. Working: 24.50 = 18 + 0.05t, so 0.05t = 24.50 − 18 = 6.50, t = 6.50 ÷ 0.05 = 130. Answer: 130 extra text messages. 490 comes from dividing the whole bill by 0.05 without first subtracting the £18 fixed charge: 24.50 ÷ 0.05 = 490. 13 comes from dividing the £6.50 by 0.5 instead of 0.05, moving the decimal point one place too far: 6.50 ÷ 0.5 = 13. 65 comes from dividing the £6.50 by 0.1 instead of 0.05.
- (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.
- (b) 3.38 — r₁ = √(300 ÷ (π × 8)) = √11.9366 = 3.4550. r₂ = √(300 ÷ (π × 8.4550)) = √11.2947 = 3.3608. r₃ = √(300 ÷ (π × 8.3608)) = √11.4232 = 3.3798, which rounds to 3.38. Choosing 3.36 stops at r₂, one iteration too early. Choosing 4.82 leaves out the '+ 5' inside the bracket, dividing by π × rₙ instead of π × (rₙ + 5). Choosing 3.45 comes from using π ≈ 3 instead of the calculator's π key throughout.
- (b) 3 — At the limit, L = √(2L + 3). Squaring both sides: L² = 2L + 3, so L² − 2L − 3 = 0, which factorises as (L − 3)(L + 1) = 0, giving L = 3 or L = −1. Since the sequence of iterates stays positive throughout, the limit is L = 3. Taking the other, negative root without rejecting it gives −1. Treating the equation L = 2L + 3 as already linear, forgetting to square both sides first, gives −L = 3, so L = −3. A sign error when factorising, writing (L + 3)(L − 1) = 0 instead of (L − 3)(L + 1) = 0, gives L = 1.
- (c) 17.7 cm — Method: the iteration converges on x, the depth of the box, which is also the side of each square cut away; a square is removed from both ends of the 20 cm side, so the longer side of the base is 20 − 2x. Run the formula three times, then carry out that subtraction. Working: x₁ = (16 × 1² − 1³ + 50) ÷ 60 = 65 ÷ 60 = 1.08333…; x₂ = 67.50636… ÷ 60 = 1.12510…; x₃ = 68.82959… ÷ 60 = 1.14715…. The longer side of the base is 20 − 2 × 1.14715… = 17.70568…, which is 17.7 cm correct to 1 decimal place. Answer: 17.7 cm. The distractors: 1.1 cm is x₃ itself rounded, the depth of the box, given by a candidate who runs the iteration correctly and then stops before the step the question actually asks for; 18.9 cm is 20 − 1.14715…, taking away one square instead of two and forgetting that a corner is cut from each end of that side; 9.7 cm is 12 − 2 × 1.14715…, the shorter side of the base, which measures the wrong edge of the sheet.
- (c) 6,400 — 9 hours contains 9 ÷ 3 = 3 whole periods of doubling, so the population is 800 × 2³. Since 2³ = 8, the population after 9 hours is 800 × 8 = 6,400. Adding 100% growth three times instead of compounding it — treating the growth as simple, not repeated doubling — gives 800 × 4 = 3,200, which is wrong because each period doubles the CURRENT population, not the original one. Using 9 as the power instead of dividing by the 3-hour period first gives 800 × 2⁹ = 409,600, which is wrong because the exponent counts periods, not hours. Giving the growth factor 2³ = 8 on its own, without multiplying by the starting population 800, leaves the answer as 8, which is wrong because the question asks for the population, not the multiplier. Always check that your final number of periods matches the total time divided by the period length.
- (d) 300 m — For a constant speed, the speed-time graph is a horizontal line, and the area underneath is a rectangle: distance = speed × time = 15 × 20 = 300 m. Adding the two numbers instead of multiplying gives 35 m; dividing instead of multiplying gives 1.33 m; halving the product, as you would for a triangle, gives 7.5 m — but this section of the graph is a rectangle, not a triangle, so there is no halving to do.
- (b) 8 kg — The cost above the flat £5 charge is 17 − 5 = £12. At £2 per kg, this covers 12 ÷ 2 = 6 kg above the first 2 kg, so the total weight is 2 + 6 = 8 kg. Dividing the full £17 by £2 per kg without first taking off the £5 flat charge gives 17 ÷ 2 = 8.5 kg. Taking off the £5 flat charge and dividing by £2 per kg, but forgetting to add back the 2 kg that the flat charge covers, gives 12 ÷ 2 = 6 kg. Taking off £2 instead of £5 as the flat charge, (17 − 2) ÷ 2 = 7.5 kg, swaps which number is the fixed fee.
- (d) x = −4, y = −19 — x² + 8x − 3 = (x + 4)² − 4² − 3 = (x + 4)² − 19. Substituting x = −4: (−4)² = 16, 8 × (−4) = −32, so 16 − 32 − 3 = −19, confirming the minimum value −19 at x = −4: turning point (−4, −19). Using 8 instead of half of it inside the bracket gives (x + 8)² − 67, turning point (−8, −67) — wrong, because only half the coefficient of x belongs inside the bracket. Reading the bracket's sign directly as the turning point's x-coordinate gives (4, −19) — wrong, because (x + 4)² is zero at x = −4, not x = 4. Computing 16 + 3 = 19 instead of −3 − 16 = −19 flips the sign of the constant, giving (−4, 19) — wrong, since the completed square's constant must be evaluated as −3 minus 16, not 16 plus 3. Check by substitution whenever the sign of a constant feels uncertain.
Build your own mix at the worksheet builder.