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.
Answer key: Algebra worksheet — GCSE Higher
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- (d) 40 m — Each trapezium has width 2. Its area is width × the average of its two heights. Strip 1: average of 0 and 5 is 2.5, so area = 2 × 2.5 = 5. Strip 2: average of 5 and 9 is 7, so area = 2 × 7 = 14. Strip 3: average of 9 and 12 is 10.5, so area = 2 × 10.5 = 21. Total distance = 5 + 14 + 21 = 40 m. Leaving out the division by 2 in the averaging step doubles every strip, giving 80 m instead of 40 m. Using only the LEFT height of each strip as a rectangle, 0 × 2 + 5 × 2 + 9 × 2 = 28, or only the RIGHT height, 5 × 2 + 9 × 2 + 12 × 2 = 52, both ignore that the graph curves between the two ends of each strip — always average the two heights of a trapezium, never use just one of them.
- (d) 1 — Method: find the step in the outputs for each step of 1 in the input, write the rule from that step and from one pair of values, then apply the rule to the last input. Working: the outputs −5, −3, −1 rise by 2 while x rises in ones, so x is multiplied by 2; at x = 1, 2 × 1 = 2 while y = −5, so 7 is subtracted, giving y = 2x − 7; at x = 4 the rule gives 2 × 4 = 8 and 8 − 7 = 1. Answer: y = 1. The distractors: 3 comes from carrying the outputs on one step too far, to x = 5; 0 comes from assuming the outputs −5, −3, −1 carry on by adding 1 rather than by adding 2; 8 comes from doubling the input and forgetting to subtract the 7.
- (a) 4 — Method: set up the equation 35 + 20h = 115, then subtract the fixed fee and divide by the hourly rate. Working: 20h = 115 − 35 = 80; h = 80 ÷ 20 = 4. Answer: 4 hours. 5.75 comes from dividing the whole £115 by £20 without first subtracting the fixed fee: 115 ÷ 20 = 5.75. 2.71 comes from swapping the fee and the rate round, subtracting £20 and dividing by £35: (115 − 20) ÷ 35 ≈ 2.71. 7.5 comes from adding the fixed fee instead of subtracting it: (115 + 35) ÷ 20 = 7.5.
- (d) 50/3 — Method: call the smaller number x, write the larger number in terms of x and use the total. Working: the larger number is 2x, so x + 2x = 50, that is 3x = 50 and x = 50/3. Answer: 50/3, since 50/3 added to 100/3 makes 50 and 100/3 is twice 50/3. The distractors: 100/3 is the larger of the two numbers rather than the smaller one asked for; 25 comes from halving 50 and treating the two numbers as equal; 24 comes from reading 'twice the smaller number' as 'two more than the smaller number' and solving x + (x + 2) = 50.
- (a) k = 3 — Method: a solution of an equation makes both sides balance, so substitute it in and solve the equation in k that is left. Working: putting x = 3 gives 3² − (k + 1) × 3 + k = 0, that is 9 − 3k − 3 + k = 0, so 6 − 2k = 0 and k = 3; the equation is then x² − 4x + 3 = 0, whose solutions are 3 and 1. Answer: k = 3. The distractors: k = −3 comes from solving 6 − 2k = 0 as though it gave 2k = −6; k = 4 comes from expanding −3(k + 1) as −3k − 1, multiplying only the k by 3; k = 1.5 comes from working 3² as 3 × 2 = 6, which leaves 3 − 2k = 0.
- (a) 20 — Method: the paper has run out when S = 0, so substitute S = 0 into the equation and solve for t. Working: 0 = 480 − 24t, so 24t = 480, and t = 480 / 24 = 20. Answer: it takes 20 minutes. The value 480 comes from giving the starting number of sheets, the intercept of the equation, instead of solving for t. The value 456 comes from working out 480 − 24 and stopping after one step instead of solving the equation fully. The value 0.05 comes from inverting the division, working out 24 / 480 instead of 480 / 24.
- (d) n/2 + 5 — Half of n is n ÷ 2, which is written as the fraction n/2. 'More than' means add, and the addition happens after the halving, so the expression is n/2 + 5. Writing (n + 5)/2 halves the 5 as well, because everything inside a bracket is divided; writing 2n + 5 doubles n instead of halving it; writing 5n/2 multiplies half of n by 5 instead of adding 5 to it.
- (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) 3 — y = f(x + 2) is f(x) translated 2 units to the LEFT (inside the bracket, adding moves the graph in the negative x-direction). The root moves with the whole graph: 5 − 2 = 3. Moving right instead of left gives 7; assuming a bracket shift leaves the root unchanged gives 5; writing down the shift amount 2 itself skips the translation altogether.
- (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) x = −1.8 and x = 2.8 — Method: the solutions of x² − x − 2 = 3 are the x-coordinates of the points where the curve y = x² − x − 2 meets the line y = 3, read off the grid to 1 decimal place. Working: the curve meets the line y = 3 at approximately x = −1.8 and at x = 2.8. Answer: x ≈ −1.8 and x ≈ 2.8. Distractor refutation: x = −1.0 and x = 2.0 comes from reading where the curve crosses the x-axis (y = 0), solving x² − x − 2 = 0, instead of where it meets the line y = 3. x = −3.0 and x = 4.0 comes from taking the two ends of the drawn curve as the intersection points, instead of finding where it actually crosses the line y = 3. x = 1.8 and x = 2.8 comes from a sign slip on the left-hand intersection, reading it as positive instead of negative.
- (a) 2 — The candle's height falls from 30 cm to 20 cm, a drop of 10 cm, over 5 minutes, so m = 10 ÷ 5 = 2. 10 comes from using the drop in height but forgetting to divide by the time. 0.5 comes from dividing the time by the drop instead of the drop by the time (5 ÷ 10). 4 comes from dividing the final height by the time, 20 ÷ 5, instead of using the drop in height.
- (d) g is one-to-one: f(3) = f(−3), removed by x ≥ 0 — A function has an inverse only if it is one-to-one: every output must come from exactly one input. f(3) = 9 and f(−3) = 9, so two different inputs give the same output, and there is no way to send 9 back to a single input — f is not one-to-one over all real x. Restricting the domain to x ≥ 0 removes one of the two inputs behind every such pair, so g is one-to-one and does have an inverse. 'g's outputs are positive; f's could be negative' is wrong because f(x) = x² also only gives outputs of 0 or more — the outputs of f and g are identical sets; it is the INPUTS that differ, not the outputs. 'Restricting any domain always creates an inverse' is wrong because a restriction only helps if it actually removes the repeated outputs: restricting f(x) = x² to x ≥ −3 still leaves f(1) = f(−1) = 1, so that restricted function is still not one-to-one and still has no inverse. 'Squares can never be reversed, under any conditions' is wrong because √9 = 3 does reverse 3² = 9 once you know the input was non-negative — a square root just cannot tell you WHICH of two inputs you started from unless the domain has already ruled one of them out.
- (d) 400 m — The distance travelled equals the area under the speed-time graph. The first 10 seconds form a triangle with base 10 and height 20, giving an area of 0.5 × 10 × 20 = 100 m. The next 15 seconds form a rectangle with base 15 and height 20, giving an area of 15 × 20 = 300 m. The total distance is 100 + 300 = 400 m. The option 500 m treats the whole 25 seconds as travelled at the constant 20 m/s, ignoring that the speed was building up during the first 10 seconds: 25 × 20 = 500. The option 300 m only counts the constant-speed section and forgets the triangle section entirely. The option 200 m comes from working out the triangle's area without halving it (10 × 20 = 200) and forgetting the rectangle altogether.
- (c) 59 — Substitute n = 7: 7² + 2 × 7 − 4 = 49 + 14 − 4 = 59. A sign error on the +2n term, treating it as −2n, gives 49 − 14 − 4 = 31. Working out 7² + 2 × 7 but forgetting to subtract the final 4 gives 49 + 14 = 63. Using n = 6 instead of n = 7 gives 36 + 12 − 4 = 44.
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