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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Answer key: Algebra worksheet — GCSE Higher
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- (d) 7 — Method: set up the equation 2.50 + 1.80m = 15.10, then subtract the booking fee and divide by the cost per mile. Working: 1.80m = 15.10 − 2.50 = 12.60; m = 12.60 ÷ 1.80 = 7. Answer: 7 miles. 8.39 comes from dividing the whole £15.10 by £1.80 without first subtracting the booking fee: 15.10 ÷ 1.80 ≈ 8.39. 5.32 comes from swapping the two amounts round, subtracting £1.80 and dividing by £2.50: (15.10 − 1.80) ÷ 2.50 ≈ 5.32. 9.78 comes from adding the booking fee instead of subtracting it: (15.10 + 2.50) ÷ 1.80 ≈ 9.78.
- (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.
- (c) 26.6 — h² = 1.6² = 2.56, so B = 68 ÷ 2.56 = 26.5625, which rounds to 26.6 (1 d.p.). A candidate who forgets to square the height gets 68 ÷ 1.6 = 42.5. A candidate who squares the mass instead of the height gets 68² ÷ 1.6 = 2890.0. A candidate who truncates 26.5625 instead of rounding it gets 26.5.
- (d) If the graph has two real roots, they are equal and opposite in value, so they sum to zero. — A turning point on the y-axis means the graph's axis of symmetry is the line x = 0, so any two roots must be symmetrical about x = 0 — equal in size but opposite in sign, summing to zero. The graph could still have no real roots, but that is not guaranteed just from the turning point's position, so the option claiming it must have none is too strong. The roots do not have to both be positive — if real, one is positive and one negative (or both are zero). The graph does not have to touch the x-axis at exactly one point either; it could cross at two symmetrical points, touch at one point, or miss the x-axis entirely.
- (b) 5 cm³ — When P = 32, V = 480 ÷ 32 = 15 cm³. When P = 24, V = 480 ÷ 24 = 20 cm³. The increase in volume is 20 − 15 = 5 cm³. Giving only the second volume, 20 cm³, forgets to subtract the first volume. Giving only the first volume, 15 cm³, answers the wrong part of the question. Subtracting the two pressures instead of the two volumes, 32 − 24 = 8, mixes up which quantity the question asks for.
- (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) 40 — From 9am to 12 noon is 3 hours, so the population doubles three times: 5 × 2³ = 40. A candidate who counts the elapsed time as 4 hours (an off-by-one counting error) would reach 5 × 2⁴ = 80. A candidate who counts it as only 2 hours would reach 5 × 2² = 20. A candidate who misreads 'doubles' as 'increases by 2' each hour would compute 5 + 3 × 2 = 11.
- (c) 5 — Method: call the number of years ago t, take t off both ages, and form an equation from the comparison at that time. Working: t years ago the father was 40 − t and Ethan was 10 − t, so 40 − t = 7(10 − t); expanding gives 40 − t = 70 − 7t, adding 7t to both sides gives 40 + 6t = 70, subtracting 40 gives 6t = 30, and dividing by 6 gives t = 5. Checking: five years ago the father was 35 and Ethan was 5, and 35 = 7 × 5. Answer: 5. The distractors: 35 comes from finding the right moment but giving the father's age at that time instead of the number of years; 30 comes from using Ethan's present age on the right-hand side, 40 − t = 7 × 10, which gives t = −30 and is then written as 30; 24 comes from reaching 6t = 30 correctly and subtracting 6 instead of dividing by 6.
- (d) £195 — Substituting x = 3 into y = 45x + 60 gives y = 45 × 3 + 60 = 135 + 60 = 195. A candidate who forgets to add the call-out fee would get only 45 × 3 = £135. A candidate who adds the hours to the fee and the rate instead of multiplying would get 45 + 60 + 3 = £108. A candidate who multiplies both the hourly rate and the call-out fee by the number of hours would get 45 × 3 + 60 × 3 = £315.
- (d) The cost increases by about £10 per extra item — Gradient = change in cost ÷ change in items = (800 − 400) ÷ (70 − 30) = 400 ÷ 40 = 10. The units of the gradient are £ per item, so the cost is increasing by about £10 for every extra item produced, near x = 50. Subtracting in the wrong order, (400 − 800) ÷ (70 − 30) = −10, gives the right size but the wrong sign — check which point comes first each time. Leaving out the division by 40 gives £400 per item; dividing the wrong way round, 40 ÷ 400 = 0.1, gives £0.10 per item.
- (d) x = 4 or x = −4 — Method: divide both sides by 2 to get x² = 16, then take the square root of both sides: x = 4 or x = −4. Distractor origins: x = 4 forgets the negative root; x = 8 comes from halving 16 instead of taking its square root; x = 16 stops at x² = 16 without ever taking the square root.
- (a) 3.5 km/h — Average speed is total distance ÷ total time. Total distance = 5 + 9 = 14 km. Total time, including the rest, = 1 + 1 + 2 = 4 hours. So average speed = 14 ÷ 4 = 3.5 km/h. Leaving out the 1 hour rest and dividing by only the 3 hours of walking gives 14 ÷ 3 = 4.67 km/h. Averaging the two separate speeds, 5 km/h and 4.5 km/h, instead of using total distance over total time, gives (5 + 4.5) ÷ 2 = 4.75 km/h. Dividing only the second leg's distance by the total time, 9 ÷ 4 = 2.25 km/h, ignores the first leg's distance.
- (b) 158 minutes — Rearranging C = 15 + 0.2m for m: subtract 15 from both sides to get C − 15 = 0.2m, then divide by 0.2: m = (C − 15)/0.2. Substituting C = 46.60: m = (46.60 − 15)/0.2 = 31.60/0.2 = 158. Answering 233 minutes comes from dividing the whole £46.60 by 0.2 without first taking off the £15 fixed charge. Answering 308 minutes comes from adding the £15 instead of subtracting it: (46.60 + 15)/0.2. Answering 218 minutes divides first and subtracts 15 afterwards, in the wrong order: 46.60/0.2 − 15 = 233 − 15 = 218. Ali made 158 minutes of calls.
- (c) 2.26 — Method: apply the formula repeatedly, keeping the whole display each time, and stop when two values in a row round to the same 2 decimal places; that shared rounded value is the solution to that accuracy. Working: x₁ = ∛(2 × 2 + 7) = ∛11 = 2.22398…; x₂ = ∛(2 × 2.22398… + 7) = ∛11.44796… = 2.25377…; x₃ = ∛11.50754… = 2.25767…; x₄ = ∛11.51534… = 2.25818…. Now x₃ and x₄ both round to 2.26, so the sequence has settled. Answer: 2.26. The distractors: 2.22 is x₁ rounded, quoted by a candidate who stops after one use of the formula; 2.25 is x₂ rounded, quoted by a candidate who stops as soon as two values look close instead of waiting until two consecutive values round to the same figure; 1.91 is ∛7, which comes from ignoring the 2x term and solving x³ = 7 instead.
- (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.
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