Printable · GCSE Foundation · ages 14-16
Algebra worksheet — GCSE Foundation
Fifteen questions across the algebra statements at Foundation tier. Choose the non-calculator filter to rehearse Paper 1, which counts for a third of the marks.
Algebra worksheet — GCSE Foundation
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- 1.Work out the value of 3a − 2b when a = 5 and b = 4
- 2.A gym charges a reduced price for the first month of membership and a standard price for every month after that. The total cost is £5 for 1 month, £12 for 2 months, £19 for 3 months and £26 for 4 months. Work out an expression, in terms of n, for the total cost, in pounds, of n months of membership.
- 3.Which of these quadratic graphs does NOT cross the x-axis at all?
- 4.The first five terms of a quadratic sequence are 5, 8, 13, 20, 29. Work out the next term in the sequence.
- 5.The simultaneous equations x = 2 and x + y = 9 are given. Work out the value of y.
- 6.The equation 3(2x − 1) = 4x + 9 is rearranged by expanding the brackets. Which of these is the correctly expanded equation?
- 7.A conversion graph shows that 5 miles is equivalent to 8 kilometres. Use this to work out how many kilometres are equivalent to 20 miles.
- 8.Simplify (5x² + 3x − 2) − (2x² − x + 5)
- 9.A straight line has gradient 3 and passes through the point (1, 4). Work out the equation of the line.
- 10.Solve x² − 6x = 0.
- 11.A market stall's cost of hiring n tables is modelled by two formulas: Formula A: C = 3(2n + 5); Formula B: C = 6n + 15, where C is in pounds. A stallholder says the two formulas always give the same cost. Work out the cost given by each formula when n = 4, and use your results to decide whether the stallholder is correct.
- 12.Point R has coordinates (3, −5). Point R is reflected in the line y = 2 to point S. Write down the coordinates of S.
- 13.A mobile phone tariff charges a fixed £15 plus 20p for every minute of calls made, so the total monthly charge, £C, for m minutes of calls is given by C = 15 + 0.2m. In a month where Ali's charge was £46.60, work out how many minutes of calls he made.
- 14.Solve 5x² − 15x = 0.
- 15.A ball is thrown in the air. Its height, h metres, above the ground after t seconds is given in this table: when t = 0, h = 0; when t = 1, h = 15; when t = 2, h = 20; when t = 3, h = 15; when t = 4, h = 0. Use the table to find the two times, in seconds, at which the ball is at ground level.
Answer key
- (d) 7 — Method: substitute both values, work out the two multiplications first, and only then subtract. Working: 3a = 3 × 5 = 15 and 2b = 2 × 4 = 8, so the expression becomes 15 − 8 = 7. Answer: 7. The distractors: 23 comes from adding the two products instead of subtracting, giving 15 + 8; −7 comes from subtracting the wrong way round and working out 8 − 15; 52 comes from working from left to right instead of following the order of operations, giving 3 × 5 = 15, then 15 − 2 = 13, then 13 × 4.
- (d) 7n − 2 — Method: find how much the total cost rises each month, then find the constant that fits the cost for one month. Working: the cost rises by £7 for each extra month (12 − 5 = 7, 19 − 12 = 7, 26 − 19 = 7), so the cost has the form 7n + c. Substituting n = 1: 7(1) + c = 5, so c = −2. Answer: the total cost in pounds is 7n − 2. The value 7n comes from leaving out the constant. The value 7n + 5 comes from using the cost of one month as the constant directly, without subtracting the monthly rise first. The value 5n + 7 comes from swapping the roles of the cost of one month, £5, and the monthly rise, £7 — using the cost of one month as the coefficient of n and the rise as the constant.
- (a) y = (x − 2)² + 3 — Since (x − 2)² is never negative, (x − 2)² + 3 is always at least 3, so y can never equal 0 and the graph never crosses the x-axis. The other three graphs are all given in a factorised or difference-of-squares form that shows two real roots: y = (x − 2)(x + 3) crosses at x = 2 and x = −3; y = x² − 9 = (x − 3)(x + 3) crosses at x = 3 and x = −3; y = (x + 4)(x − 1) crosses at x = −4 and x = 1.
- (b) 40 — The differences between consecutive terms are 3, 5, 7, 9, increasing by 2 each time, so the next difference is 11, giving 29 + 11 = 40. A candidate who reuses the last difference of 9 without increasing it gets 29 + 9 = 38. A candidate who increases the difference by only 1 instead of 2 gets 29 + 10 = 39. A candidate who reuses the first difference of 3 instead of the pattern gets 29 + 3 = 32.
- (b) y = 7 — Method: one of the equations already gives the value of x, so that value is substituted into the other equation. Working: replacing x with 2 in x + y = 9 gives 2 + y = 9, so y = 9 − 2 = 7. Answer: y = 7, and the pair x = 2 with y = 7 satisfies both equations. The distractors: y = 11 comes from rearranging x + y = 9 as y = 9 + x and adding instead of subtracting; y = 9 comes from ignoring the x term and reading the total as the value of y; y = 4.5 comes from halving 9, as though x and y had to be equal.
- (b) 6x − 3 = 4x + 9 — Method: multiply every term inside the bracket by the number outside it; the right-hand side stays as it is given. Working: 3 × 2x = 6x and 3 × (−1) = −3, so 3(2x − 1) = 6x − 3. Answer: 6x − 3 = 4x + 9. 6x − 1 = 4x + 9 comes from multiplying only the 2x by 3 and leaving the −1 unchanged. 5x − 3 = 4x + 9 comes from adding the 3 to the 2 instead of multiplying, treating 3 × 2x as (3 + 2)x = 5x. 6x − 4 = 4x + 9 comes from working out 3 × (−1) as −1 − 3 = −4 instead of 3 × (−1) = −3.
- (b) 32 km — Method: work out the scale factor connecting the two mile values: 20 miles is 4 times 5 miles (20 ÷ 5 = 4), so multiply the km value by 4 too: 8 × 4 = 32 km. Distractor origins: 24 km comes from subtracting the mile values (20 − 5 = 15), dividing by 5 to get a wrong scale factor of 3, then multiplying 8 × 3; 23 km comes from adding the difference in miles, 15, directly to the km value, 8; 16 km comes from using a scale factor of 2, doubling the km value once instead of scaling by the correct factor of 4.
- (d) 3x² + 4x − 7 — Method: the minus sign in front of the second bracket changes the sign of every term inside it; then collect like terms. Working: removing the brackets gives 5x² + 3x − 2 − 2x² + x − 5; the squared terms give 5x² − 2x² = 3x², the x terms give 3x + x = 4x, and the number terms give −2 − 5 = −7. Answer: 3x² + 4x − 7. The distractors: 7x² + 2x + 3 comes from adding the two brackets instead of subtracting, giving 5x² + 2x², 3x − x and −2 + 5; 3x² + 2x + 3 comes from applying the minus sign to 2x² only, leaving −x and +5 unchanged so that 3x − x = 2x and −2 + 5 = 3; 3x² + 4x + 3 comes from changing the signs of the terms with letters but leaving +5 as it stood, so the number terms give −2 + 5 = 3.
- (c) y = 3x + 1 — Method: a line of known gradient m has equation y = mx + c, and c is found by substituting the coordinates of a point known to lie on it. Working: the gradient is 3, so the line is y = 3x + c; substituting x = 1 and y = 4 gives 4 = 3 × 1 + c, so c = 4 − 3 = 1. Answer: y = 3x + 1. The distractors: y = 3x − 1 comes from working out the constant as mx − y, 3 − 4 = −1, instead of y − mx; y = x + 3 comes from swapping the two numbers over, putting the gradient 3 in the constant position and the x-coordinate 1 in front of x; y = 3x + 4 comes from using the y-coordinate 4 as the constant without substituting.
- (d) x = 0 or x = 6 — Method: factorise by taking out the common factor x: x(x − 6) = 0, so x = 0 or x − 6 = 0, giving x = 0 or x = 6. Distractor origins: x = 6 comes from dividing both sides by x, which loses the solution x = 0; x = 0 stops after finding only one factor; x = 3 comes from halving the coefficient of x instead of factorising.
- (d) Formula A gives £39 and Formula B gives £39, and since 3(2n + 5) expands to 6n + 15 for every value of n, the stallholder is correct. — Formula A: 3(2 × 4 + 5) = 3 × 13 = £39. Formula B: 6 × 4 + 15 = 24 + 15 = £39. Expanding Formula A algebraically gives 3(2n + 5) = 6n + 15, which is identical to Formula B for every value of n, not just n = 4, so the stallholder is correct — this is an identity, not a coincidence. The option giving £29 for Formula A comes from multiplying only the 2n by 3 and forgetting to also multiply the 5, then adding the unmultiplied 5: 3 × 2 × 4 = 24, + 5 = 29. The two options that reach the correct numbers but reject the stallholder's claim both use faulty reasoning — matching values at one value of n, or counting terms, does not decide whether two expressions are identical for every n; expanding the bracket does.
- (c) (3, 9) — Reflecting in the horizontal line y = 2 keeps the x-coordinate the same and maps y to 2 × 2 − y = 4 − (−5) = 9, so S = (3, 9). A candidate who uses k − y instead of 2k − y gets 2 − (−5) = 7, giving (3, 7). A candidate who reflects in the x-axis instead of the line y = 2, simply changing the sign of y, gets (3, 5). A candidate who also changes the sign of the x-coordinate, as if reflecting in both axes, gets (−3, 9).
- (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) x = 0 or x = 3 — Method: take out the common factor 5x: 5x(x − 3) = 0, so 5x = 0 or x − 3 = 0, giving x = 0 or x = 3. Distractor origins: x = 0 or x = 15 comes from forgetting to divide the second term by the common factor 5x correctly, leaving 15 instead of 3; x = 3 loses the solution x = 0 by dividing both sides by x; x = 5 or x = 3 mistakes the coefficient 5 itself for one of the solutions.
- (a) t = 0 or t = 4 — The ball is at ground level exactly when h = 0. From the table, h = 0 at t = 0 and at t = 4, so those are the two times. Distractor origins: t = 1 or t = 3 picks the times with equal (but non-zero) height instead of ground level; t = 2 picks the time of maximum height instead of ground level; t = 0 finds only the starting time and misses the second one.
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