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 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.
- 2.Solve 5(x + 3) = 40
- 3.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.
- 4.Solve 4x − (2x − 6) = 18
- 5.Solve 3(2x − 1) = 27
- 6.A cafe sells coffees at £c each and pastries at £p each. Three coffees and two pastries cost £9.60. Two coffees and two pastries cost £7.60. Work out the price of one coffee.
- 7.Solve the simultaneous equations 4x + 3y = 25 and 4x − y = 1. Work out the value of y.
- 8.f(x) = x³ − 3x − 5. Given that f(2.2) = −0.952 and f(2.3) = 0.267, work out what this shows about the equation x³ − 3x − 5 = 0.y = x
- 9.The curve y = −x² + 6x − 5 has a maximum point. Use completing the square to find its coordinates.y = −x² + 6x − 5
- 10.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.
- 11.Meera writes the statement 3(x + 4) = 3x + 12. Which of these correctly describes what she has written, with a reason?
- 12.Solve 3x + 14 = 8x − 6
- 13.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.
- 14.The nth term of a sequence is n² + 3. Work out the first term of the sequence that is greater than 50.
- 15.A ball's height, h metres, t seconds after being thrown follows h = (t − 1)(9 − t). Given that the ball is at ground level at t = 1 and t = 9, work out at what time t the ball reaches its maximum height, using symmetry.
Answer key
- (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) x = 5 — Method: expand the bracket by multiplying both terms inside it by 5, then undo the addition and the multiplication in turn. Working: expanding gives 5x + 15 = 40; subtracting 15 from both sides gives 5x = 25; dividing both sides by 5 gives x = 5. Answer: x = 5. The distractors: x = 8 comes from dividing both sides by 5 first, reaching x + 3 = 8 and writing 8 as the value of x without taking the 3 away; x = 11 comes from adding 15 to both sides instead of subtracting it, giving 5x = 55; x = 7.4 comes from expanding 5(x + 3) as 5x + 3, multiplying only the x by the 5, which leads to 5x = 37.
- (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.
- (a) x = 6 — Method: a minus sign in front of a bracket changes the sign of every term inside it, so expand the bracket first and then simplify. Working: expanding gives 4x − 2x + 6 = 18, which simplifies to 2x + 6 = 18; subtracting 6 from both sides gives 2x = 12, and dividing both sides by 2 gives x = 6. Answer: x = 6. The distractors: x = 12 comes from leaving the −6 unchanged when the bracket is removed, giving 4x − 2x − 6 = 18 and so 2x = 24; x = 24 comes from reaching 2x = 12 correctly and then multiplying by 2 instead of dividing by 2; x = 3 comes from dividing by 2 too early, at 2x + 6 = 18, and dividing only the 2x and the 18 while leaving the 6 untouched, which gives x + 6 = 9.
- (b) 5 — Method: expand the bracket first, then solve the resulting linear equation. Working: 3(2x − 1) = 6x − 3, so 6x − 3 = 27. Add 3 to both sides: 6x = 30. Divide by 6: x = 5. Answer: 5. 4 2/3 comes from multiplying only the 2x inside the bracket by 3 and forgetting to multiply the −1 as well, giving 6x − 1 = 27. 4 comes from a sign error when expanding, writing 6x + 3 instead of 6x − 3. 4.5 comes from dividing 27 by 3 to get 2x − 1 = 9, then forgetting to add 1 back before dividing by 2, using 2x = 9 instead of 2x = 10.
- (b) £2.00 — Subtracting the second equation from the first eliminates the pastries: (3c + 2p) − (2c + 2p) = 9.60 − 7.60, so c = 2.00. A candidate who divides the first total by the number of coffees alone, ignoring the pastries, would get 9.60 ÷ 3 = £3.20. A candidate who finds the price of a pastry instead of a coffee — using c = 2.00 in 2c + 2p = 7.60 to get p = 1.80 — would answer £1.80. A candidate who reaches the correct difference of £2.00 but then mistakenly divides again or misplaces the decimal point would get £0.20.
- (a) 6 — Subtracting the second equation from the first: the x-terms, 4x and 4x, cancel; the y-terms combine as 3y − (−y) = 4y; and the right-hand sides give 25 − 1 = 24. This gives 4y = 24, so y = 6. A candidate who subtracts in the wrong order would get 4y = 1 − 25 = −24, so y = −6. A candidate who forgets the sign on the −y term, treating 3y − y as 2y, would get 2y = 24, so y = 12. A candidate who divides 24 by 6 instead of 4 would get y = 4.
- (d) It has a solution between x = 2.2 and x = 2.3 — Since f(2.2) is negative and f(2.3) is positive, the graph of f crosses the x-axis somewhere between x = 2.2 and x = 2.3, so the equation has a solution in that interval. Choosing 'between x = −2.2 and x = −2.3' confuses the negative f-VALUE at 2.2 with a negative x-value. Choosing 'no solution' misreads a change of sign as meaning the opposite of what it shows. Choosing 'exactly two solutions' assumes a single change of sign must give two roots, which is not what the rule guarantees.
- (b) x = 3, y = 4 — −x² + 6x − 5 = −(x² − 6x) − 5 = −[(x − 3)² − 9] − 5 = −(x − 3)² + 9 − 5 = −(x − 3)² + 4. Because the coefficient of x² is negative, −(x − 3)² is at most zero, so this turning point is a maximum. Substituting x = 3 gives (x − 3)² = 0, so y = 4, confirming the maximum value 4 at x = 3: turning point (3, 4). Using 6 instead of half of 6 inside the bracket — forgetting to halve before completing the square — lands on turning point (6, 31), wrong, because only half the coefficient belongs inside the bracket. Reading the bracket's sign directly as the turning point's x-coordinate gives (−3, 4) — wrong, because (x − 3)² is zero at x = 3, not x = −3. Computing 5 − 9 = −4 instead of 9 − 5 = 4 flips the sign of the constant, giving (3, −4) — wrong, since the constant must be evaluated as 9 minus 5, not 5 minus 9. Whenever the leading coefficient is negative, the turning point is a maximum, not a minimum — check by substituting back into the original equation.
- (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.
- (a) An identity, true for every value of x — Expanding the bracket on the left gives 3x + 12, which matches the right-hand side exactly, so the statement is true for every value of x — this makes it an identity. A candidate who reasons that any statement with an equals sign must be an equation picks that option, missing that an equation is only true for particular value(s) of x, not all of them. A candidate who confuses an identity with a formula, because both relate two expressions, picks the formula option — but a formula connects two different quantities, such as area and side length, not two equivalent forms of the same expression. A candidate who assumes it can be solved for a single value of x, as with a normal equation, picks that option, not realising there is no single solution here.
- (c) x = 4 — Method: with an unknown on both sides, subtract the smaller x term from both sides so that all the x is on one side, then collect the numbers on the other. Working: subtracting 3x from both sides gives 14 = 5x − 6; adding 6 to both sides gives 20 = 5x; dividing both sides by 5 gives x = 4. Checking: 3 × 4 + 14 = 26 and 8 × 4 − 6 = 26. Answer: x = 4. The distractors: x = 2.5 comes from taking 3x off the left-hand side only, leaving 14 = 8x − 6 and so 8x = 20; x = −1.6 comes from changing the sign of the 8x when it is moved across but leaving the sign of the 14 unchanged, giving 3x − 8x = −6 + 14 and so −5x = 8; x = 15 comes from reaching 5x = 20 correctly and then subtracting 5 instead of dividing by 5.
- (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.
- (a) 52 — Test successive terms: n=6 gives 6²+3=39, which is not greater than 50. n=7 gives 7²+3=52, which is greater than 50, so the first term greater than 50 is 52. A candidate who stops at n=6, before checking whether 39 actually exceeds 50, would give 39. A candidate who solves n²>50 instead of n²+3>50, ignoring the +3 in the search, would find n=8 is the first value with n²>50 (since 7²=49) and compute 8²+3=67. A candidate who computes n² by doubling n instead of squaring it would compute 2×7+3=17.
- (d) t = 5 — The maximum height occurs halfway between the two times when the ball is at ground level: the midpoint of t = 1 and t = 9 is (1 + 9) ÷ 2 = 5, so the ball reaches its maximum height at t = 5 seconds. Getting t = 4 comes from halving the DIFFERENCE between the times, 9 − 1 = 8, then 8 ÷ 2 = 4, instead of finding their midpoint. Getting t = 8 uses that difference, 9 − 1 = 8, as if the gap between the two times were itself the time of the maximum. Getting t = 10 adds the two times, 1 + 9 = 10, but forgets to divide by 2.
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