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) 1 hour 32 minutes — T = 40 × 1.8 + 20 = 72 + 20 = 92 minutes. Since 92 = 60 + 32, the cooking time is 1 hour 32 minutes. A candidate who rounds the mass to 2 kg before substituting gets 40 × 2 + 20 = 100 minutes = 1 hour 40 minutes. A candidate who forgets to add the 20 minutes gets 40 × 1.8 = 72 minutes = 1 hour 12 minutes. A candidate who multiplies the mass by (40 + 20) = 60 instead of substituting into the formula gets 1.8 × 60 = 108 minutes = 1 hour 48 minutes.
- (a) 36 km/h — Speed is distance ÷ time. Convert 25 minutes to hours: 25 ÷ 60 = 5/12 hours. Then 15 ÷ (5/12) = 15 × 12/5 = 36 km/h. Dividing 15 by 25 without converting minutes to hours gives 0.6 km/h. Multiplying 15 by 25/60 instead of dividing by it gives 6.25 km/h. Working out the speed in km per minute, 0.6, and then multiplying by 24 instead of 60 gives 14.4 km/h.
- (b) 58.8 km/h — Total distance = 100 + 47 = 147 km. Total time = 1.5 + 0.25 + 0.75 = 2.5 hours. Average speed = 147 ÷ 2.5 = 58.8 km/h. Leaving out the 0.25 hours of waiting from the total time gives 147 ÷ 2.25 = 65.3 km/h (1 d.p.) — wrong, because the train is stationary but time is still passing on the whole journey. Using only the first leg gives 100 ÷ 1.5 = 66.7 km/h (1 d.p.) — wrong, because it ignores the second leg of the journey entirely. Working out each leg's own speed (100 ÷ 1.5 = 66.7 km/h and 47 ÷ 0.75 = 62.7 km/h) and then averaging those two speeds gives 64.7 km/h (1 d.p.) — wrong, because the average of two speeds over DIFFERENT times is not the same as total distance divided by total time.
- (a) −2 m/s² — The gradient of a line through two points is the change in the y-value divided by the change in the x-value: (6 − 22)/(16 − 8) = −16/8 = −2 m/s². The negative sign confirms the deceleration mentioned in the question. Writing 2 m/s² comes from subtracting the velocities in the wrong order, using (22 − 6) instead of (6 − 22), which loses the sign that shows the car is slowing down. Writing −0.5 m/s² comes from inverting the fraction, dividing the change in time by the change in velocity instead of the other way round. Writing −16 m/s² comes from finding the change in velocity, −16, but forgetting to divide by the change in time, 8 seconds.
- (b) f(1) = −3 and f(2) = 3 — f(1) = 1³ − 1 − 3 = 1 − 1 − 3 = −3. f(2) = 2³ − 2 − 3 = 8 − 2 − 3 = 3. Since f(1) is negative and f(2) is positive, there is a change of sign, so a solution lies between x = 1 and x = 2. Dropping the −x term entirely gives f(1) = 1 − 3 = −2 and f(2) = 8 − 3 = 5. Using x² instead of x³ throughout gives f(1) = 1 − 1 − 3 = −3, which happens to coincide with the correct value at x = 1, but f(2) = 4 − 2 − 3 = −1, which does not show a change of sign at all. Working out −f(x) instead of f(x), a sign-flip error, gives f(1) = 3 and f(2) = −3, the correct sizes but with both signs reversed.
- (a) 4n + 1 — Method: find the increase in cost per hour, then find the constant by adjusting the 1-hour cost. Working: the cost goes up by £4 for each extra hour (9 − 5 = 4), so the coefficient of n is 4. The constant is the 1-hour cost minus the common difference: 5 − 4 = 1. Answer: the nth term is 4n + 1. 4n + 5 comes from using the 1-hour cost, 5, as the constant without subtracting the common difference. 4n − 3 comes from a slip in working out the constant, subtracting the common difference twice (5 − 4 − 4 = −3) instead of once. n + 4 comes from swapping the hourly increase and the constant.
- (d) Overestimate — the curve bends upward (convex). — The first differences of the speeds are 3, 5, 7 and 9, so the second differences are 2, 2 and 2 — constant and positive, which means the speed-time graph curves upwards (is convex). On a convex curve, each straight chord used by the trapezium rule lies above the curve, so the trapezium rule overestimates the true distance. 'Underestimate — the curve bends upward' states the same correct geometry but gets the conclusion backwards — a chord above the curve means too much area is counted, not too little. 'Overestimate — the speed values are increasing' uses the wrong evidence: increasing speed alone doesn't tell you whether the curve bends up or down, only the second differences do. 'Underestimate — second differences are constant' confuses a constant second difference with a steady rate of change in speed, which isn't what the second difference of a speed-time table measures.
- (a) 1.861 — x₁ = ∛(10 − 2²) = ∛6 = 1.817120593. x₂ = ∛(10 − 1.817120593²) = ∛6.698072751 = 1.885022855. x₃ = ∛(10 − 1.885022855²) = ∛6.446688837 = 1.861139399, which rounds to 1.861. Reporting x₂ instead of x₃ gives 1.885022855, which rounds to 1.885. Stopping after the first iteration and reporting x₁ instead of x₃ gives 1.817120593, which rounds to 1.817. A sign error inside the cube root, using xₙ₊₁ = ∛(10 + xₙ²) instead of ∛(10 − xₙ²), gives x₁ = ∛14 = 2.410142264, x₂ = ∛(10 + 2.410142264²) = 2.509763724, and x₃ = ∛(10 + 2.509763724²) = 2.535437381, which rounds to 2.535.
- (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.
- (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.
- (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.
- (b) Student B's — narrower strips fit the curve more closely. — The trapezium rule replaces the curve with straight-line segments; the narrower each strip, the more closely its straight edge follows the curve, so Student B's estimate with 8 narrower strips is more likely to be closer to the true area. Reasoning from the number of arithmetic steps rather than from how well the straight lines fit the curve gives the wrong claim that Student A makes fewer rounding errors. Believing the trapezium rule is exact, rather than an estimate that improves with narrower strips, gives the wrong claim that the two are always the same. Believing wider strips smooth out the curve better, rather than following it less closely, gives the wrong claim in favour of Student A's wider strips.
- (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.).
- (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.
- (d) x = −2, y = −5 — 3x² + 12x + 7 rewrites as 3(x² + 4x) + 7, then as 3[(x + 2)² − 4] + 7, which simplifies to 3(x + 2)² − 5, since −3 × 4 + 7 = −5. Substituting x = −2: 3 × (−2)² = 12, 12 × (−2) = −24, so 12 − 24 + 7 = −5, confirming the minimum value −5 at x = −2: turning point x = −2, y = −5. Halving b instead of halving b/a — using −4 as the shift instead of −2 — lands on turning point x = −4, y = −41, which is wrong because when a ≠ 1 the shift inside the bracket is b/(2a), not b/a alone. Reading the bracket's sign directly as the turning point's x-coordinate gives x = 2, y = −5 — wrong, because (x + 2)² is zero at x = −2, not x = 2. Computing 12 − 7 = 5 instead of 7 − 12 = −5 flips the sign of the constant, giving x = −2, y = 5 — wrong, since the completed square's constant must be evaluated as 7 minus 12, not 12 minus 7. Whenever a ≠ 1, factor a out of the x² and x terms first, and always check a turning point by substituting back into the original equation.
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