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.
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Algebra worksheet — GCSE Foundation
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- 1.Jack is x years old. His brother Kwame is 3 years older than Jack. Write down an expression for Kwame's age.
- 2.The first five square numbers are 1, 4, 9, 16, 25. Write down the next square number in the sequence.
- 3.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.
- 4.A straight line has gradient −2 and passes through the point (3, 1). Work out the equation of the line.
- 5.In a sequence, each term is half the term before it. The sequence begins 100, 50, 25. Write down the next term.
- 6.The point F has x-coordinate −1 and lies 6 units above the x-axis. Write down the coordinates of F.
- 7.Solve the inequality x − 10 < −3.
- 8.The cost, in pounds, of hiring a van for d days is given by the formula C = 25 + 18d. A customer's hire cost £133. Work out how many days, d, they hired the van for, by first rearranging the formula to make d the subject.
- 9.A taxi firm charges a fixed fee of £3.50 plus £2.20 per mile. Work out the total cost of a journey of 6 miles.
- 10.A square has sides of length y cm. Write down an expression for the perimeter of the square.
- 11.Work out the y-intercept of the graph of y = x² + 3x − 10.y = x² + 3x − 10
- 12.The graph of y = x² − 2x − 8 crosses the x-axis at x = −2 and x = 4. Work out the x-coordinate of the turning point of the curve, using the symmetry of the graph.y = x² − 2x − 8
- 13.A lift has a weight limit: it must carry no more than 630 kg. Let w be the total weight, in kg, carried by the lift. Which inequality describes this?
- 14.The nth term of a sequence is 2n − 1. Work out the 6th term of the sequence.
- 15.Solve 3(2x − 1) = 27
Answer key
- (a) x + 3 — Method: "years older" is an addition, and the number of years is added to the age that is known in terms of the letter. Working: Jack's age is x, and Kwame is 3 years older, so Kwame's age is x + 3. Answer: x + 3. The distractors: x − 3 comes from reading "older" as "younger" and subtracting instead of adding; 3x comes from reading "3 years older" as "3 times as old" and multiplying; 3 − x comes from writing the calculation the wrong way round, taking Jack's age from the 3 rather than adding.
- (a) 36 — Square numbers are n² for n = 1, 2, 3, ...; the fifth term, 25, is 5². The sixth square number is 6² = 36. A candidate who mislabels 25 as the sixth square number would compute 7² = 49 instead. A candidate who repeats an earlier difference between terms (5, from 4 to 9) rather than the correct next difference (11, since the differences are the odd numbers 3, 5, 7, 9, 11) would reach 30. A candidate who instead adds 10 would reach 35.
- (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.
- (a) y = −2x + 7 — Using y = −2x + c, and substituting the point (3, 1): 1 = −2(3) + c, so 1 = −6 + c, and c = 7, giving y = −2x + 7. A candidate who uses the y-coordinate of the point directly as the y-intercept, instead of solving for c, would write y = −2x + 1. A candidate who drops the negative sign on the gradient would write y = 2x + 7. A candidate who makes a sign error when isolating c, writing c = 1 − 6 = −5 instead of c = 1 + 6 = 7, would write y = −2x − 5.
- (b) 12.5 — Method: apply the term-to-term rule to the term just written, and keep the exact value even when halving does not give a whole number. Working: the term before the one wanted is 25, and halving it means working out 25 ÷ 2, which is 12 with 1 left over to share, giving a half. Answer: 12.5. The distractors: 12 comes from halving 25 and then cutting the result down to a whole number; 0 comes from treating the sequence as one with a constant difference and taking 25 away from 25; 6.25 comes from halving twice and giving the term after the next one.
- (b) (−1, 6) — '6 units above the x-axis' gives a y-coordinate of 6, and the x-coordinate is given as −1, so F = (−1, 6). (6, −1) comes from swapping the x- and y-coordinates. (−1, −6) comes from treating 'above' as a negative direction instead of positive. (1, 6) comes from dropping the negative sign on the given x-coordinate.
- (d) x < 7 — Method: add 10 to both sides to leave x on its own; adding the same number to both sides never changes the direction of an inequality. Working: adding 10 to both sides of x − 10 < −3 leaves x on the left and −3 + 10 on the right, and −3 + 10 = 7, so x < 7. Answer: x < 7. The distractors: x > 7 comes from turning the sign round while adding, as though every move flipped it; x < −13 comes from subtracting 10 from both sides instead of adding it, giving −3 take away 10; x < 13 comes from ignoring the minus sign on −3 and working out 3 + 10 instead.
- (d) 6 — Method: rearrange the formula to make d the subject, then substitute C = 133. Working: C = 25 + 18d, so subtracting 25 from both sides gives C − 25 = 18d, then dividing by 18 gives d = (C − 25) / 18. Substituting C = 133: d = (133 − 25) / 18 = 108 / 18 = 6. The value 7.4 comes from dividing 133 by 18 without subtracting the fixed £25 first (133 / 18 ≈ 7.4). The value 4.6 comes from pairing the numbers the wrong way round, working out (133 − 18) / 25 = 4.6. The value 90 comes from subtracting both 25 and 18 from 133 instead of dividing by 18.
- (d) £16.70 — The mileage charge is 2.20 × 6 = £13.20. Adding the fixed fee: £13.20 + £3.50 = £16.70. A candidate who forgets the fixed fee gives just the mileage charge, £13.20. A candidate who adds the fixed fee to the per-mile rate before multiplying by the number of miles, (3.50 + 2.20) × 6, gets £34.20. A candidate who rounds £2.20 down to £2 gets 2 × 6 + 3.50 = £15.50.
- (d) 4y — Method: the perimeter is the total distance round the outside of the shape, and a square has four sides all of the same length. Working: the four sides are y, y, y and y, so the perimeter is y + y + y + y, which is written as 4y. Answer: 4y. The distractors: 2y comes from adding only two of the four sides, as if the square were being treated like a rectangle with one length and one width; y² comes from working out the area of the square, y × y, instead of its perimeter; y + 4 comes from adding 4 to the side length instead of multiplying the side length by 4.
- (d) (0, −10) — The y-intercept occurs where x = 0. Substituting x = 0 into y = x² + 3x − 10 gives y = 0 + 0 − 10 = −10, so the graph crosses the y-axis at (0, −10). The option (0, 3) mistakenly uses the coefficient of x instead of the constant term. The option (0, 10) makes a sign error, dropping the negative from the constant term. The option (−10, 0) swaps the x- and y-coordinates, which would instead be a point on the x-axis, not the y-axis.
- (d) x = 1 — A quadratic graph is symmetrical about its turning point, so the turning point's x-coordinate is the midpoint of the two roots: (−2 + 4) ÷ 2 = 2 ÷ 2 = 1. The option x = 2 comes from adding the two roots but forgetting to divide by 2. The option x = −1 makes a sign error when adding the roots, treating −2 + 4 as −2. The option x = 3 comes from subtracting the roots instead of adding them: (4 − (−2)) ÷ 2 = 3.
- (d) w ≤ 630 — 'No more than 630 kg' means the weight can be exactly 630 kg or anything less, so the correct inequality is w ≤ 630, using 'less than or equal to' to include the limit itself. Writing w < 630 excludes 630 kg itself, as though the limit could not be reached exactly. Writing w ≥ 630 reverses the direction, describing a minimum weight rather than a maximum. Writing w > 630 both reverses the direction and excludes the boundary value. The inequality describing the lift's weight limit is w ≤ 630.
- (a) 11 — Substitute n=6 into 2n−1: 2×6−1=11. A candidate who adds 2 and 6 and then subtracts 1, instead of multiplying 2 by 6 first, would compute 2+6−1=7. A candidate who substitutes the wrong term number, n=5, would reach 2×5−1=9. A candidate who forgets to subtract 1 would compute just 2×6=12.
- (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.
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