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.Simplify k + k + k + k + k.
- 2.The point P has coordinates (0, 12). Write down the axis that P lies on and the y-coordinate of P.
- 3.A cyclist's distance-time graph shows the following: she travels 12 km in the first 30 minutes at a steady speed, then rests for 30 minutes, then travels a further 8 km in the next 15 minutes. Work out her average speed, in km/h, for the whole journey.
- 4.The graph of y = (x − 4)(x + 1) crosses the x-axis. Which pair gives the correct roots and reasoning?
- 5.Which of these is an expression?
- 6.Solve 2x² = 18.
- 7.Make m the subject of the formula w = m − 8.
- 8.Which of these quadratic graphs does NOT cross the x-axis at all?
- 9.A sequence has the position-to-term rule 5n − 2, where n is the position number. Which of these is a term in the sequence?
- 10.A school trip costs a £15 deposit plus £9 per student for the coach. The total cost for a class is £186. Work out how many students went on the trip.
- 11.Solve the simultaneous equations x + y = 1 and 2x + y = 5.
- 12.Oliver has m stickers and gives 5 of them away. The number he has left is L = m − 5. Work out L when m = 8.
- 13.A company's profit was £2000 in its first year. Each following year, the profit increases by £800. Work out the first year in which the profit is more than £7000.
- 14.Which expression is equivalent to 4(x − 2) + 10?
- 15.Simplify 5a/6 − a/3
Answer key
- (a) 5k — Method: adding a letter to itself repeatedly means counting how many of that letter you have, written as a coefficient in front of the letter. Working: k + k + k + k + k is five lots of k, written 5k. Answer: 5k. k⁵ comes from treating repeated addition as repeated multiplication (raising to a power) instead of counting copies. 5 + k comes from adding the count of terms (5) to a single k instead of multiplying. k/5 comes from dividing instead of counting how many k's there are.
- (d) the y-axis, and 12 — Method: coordinates are written (x, y), so the first number is the distance across and the second the distance up; a point whose first coordinate is 0 has not moved across from the origin and therefore lies on the vertical axis. Working: in (0, 12) the first number is 0, so P is on the y-axis, and the second number, 12, is the y-coordinate of P. Answer: the y-axis, and 12. The distractors: 'the x-axis, and 12' comes from mixing up which axis the condition 'the first coordinate is 0' describes; 'the y-axis, and 0' comes from placing P correctly but reading the pair the wrong way round, so that the first number is quoted as the y-coordinate; 'the x-axis, and 0' comes from making both of those mistakes at once.
- (c) 16 km/h — Method: total distance = 12 + 8 = 20 km. Total time = 30 + 30 + 15 = 75 minutes = 1.25 hours. Average speed = total distance ÷ total time = 20 ÷ 1.25 = 16 km/h. Distractor origins: 20 km/h gives the total distance without ever dividing by the total time; 24 km/h uses only the speed of the first stage (12 km in 30 minutes), ignoring the second stage and the rest; 28 km/h averages the two separate stage speeds, 24 km/h and 32 km/h, instead of using total distance over total time.
- (c) x = 4 and x = −1, because the graph crosses the x-axis where x − 4 = 0 or x + 1 = 0. — The graph crosses the x-axis where y = 0, which happens when either bracket equals zero. Solving x − 4 = 0 gives x = 4, and solving x + 1 = 0 gives x = −1. The option x = −4 and x = 1 incorrectly reverses both signs. The option x = 4 and x = 1 misreads the second bracket, ignoring that x + 1 = 0 requires x to be negative. The option x = −4 and x = −1 wrongly assumes both roots must be the negative of the constants shown, which only matches the second bracket, not the first.
- (c) 5y − 2 — An expression is a collection of terms with no equals sign and no inequality sign in it: nothing is being claimed about two sides. Here 5y − 2 on its own fits that description. The line with an equals sign and one unknown letter to find is an equation; the line joined by a comparison sign is an inequality; the line that links two different letters so that one can be worked out from the other is a formula.
- (d) x = 3 or x = −3 — Method: get x² on its own with a coefficient of 1, then take the square root of both sides and keep both the positive and the negative root. Working: dividing 2x² = 18 by 2 gives x² = 9, and the square root of 9 is 3, so x = 3 or x = −3; both check, because 2 × 9 = 18 either way. Answer: x = 3 or x = −3. The distractors: x = 9 or x = −9 comes from dividing by 2 and then forgetting to take the square root; x = 4 or x = −4 comes from subtracting 2 from 18 instead of dividing by it, giving x² = 16; x = 6 or x = −6 comes from multiplying by 2 instead of dividing, giving x² = 36.
- (b) m = w + 8 — To make m the subject of w = m − 8, add 8 to both sides so the subtraction is undone: m = w + 8. Writing m = w − 8 forgets to change the operation at all — it just relabels the equation without moving the 8 across. Writing m = 8 − w swaps the order and puts the wrong sign on w, as if the equation had been w = 8 − m instead. Writing m = w/8 mistakes subtraction for a scaling relationship and divides by 8 instead of adding it. The correct rearrangement is m = w + 8.
- (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.
- (c) 28 — Method: substitute values of n into 5n − 2 and check which of the options matches. Working: for n = 6, 5 × 6 − 2 = 30 − 2 = 28, so 28 is a term of the sequence, the 6th term. Answer: 28. 27 comes from using the rule 5n − 3 instead of 5n − 2. 30 comes from using 5n on its own, forgetting to subtract 2 at all. 20 comes from applying the subtraction before the multiplication, working out 5 × (n − 2) instead of 5n − 2.
- (c) 19 — Method: set up the equation 15 + 9n = 186, then subtract the deposit and divide by the cost per student. Working: 9n = 186 − 15 = 171; n = 171 ÷ 9 = 19. Answer: 19 students. 20.67 comes from dividing the whole £186 by £9 without first subtracting the deposit: 186 ÷ 9 ≈ 20.67. 11.8 comes from swapping the two amounts round, subtracting £9 and dividing by £15: (186 − 9) ÷ 15 = 11.8. 22.33 comes from adding the deposit instead of subtracting it: (186 + 15) ÷ 9 ≈ 22.33.
- (a) x = 4, y = −3 — Method: both equations contain +y with the same coefficient, so subtracting one equation from the other removes y. Working: (2x + y) − (x + y) = 5 − 1 gives x = 4; substituting x = 4 into x + y = 1 gives 4 + y = 1, so y = −3. Answer: x = 4, y = −3, which also satisfies 8 − 3 = 5. The distractors: x = 4, y = 5 comes from rearranging x + y = 1 as y = 1 + x; x = −2, y = 3 comes from eliminating x by doubling the first equation and then reading −y = 3 as y = 3; x = 6, y = −5 comes from adding the constants instead of subtracting them while eliminating y, taking x as 5 + 1.
- (a) 3 — Method: substitute the given value into the formula and carry out the subtraction in the order the formula is written. Working: replacing m with 8 gives L = 8 − 5, and 8 − 5 = 3. Answer: 3. The distractors: 13 comes from adding 5 to 8 instead of subtracting it; 40 comes from reading m − 5 as a multiplication and working out 8 × 5; −3 comes from subtracting the wrong way round and working out 5 − 8.
- (d) 8 — Method: write the nth term of the sequence, 2000 + 800(n − 1), and find the smallest whole n for which it is greater than 7000. Working: 2000 + 800(n − 1) > 7000, so 800(n − 1) > 5000, giving n − 1 > 6.25. Since n − 1 must be a whole number, the smallest value is 7, so n = 8. Check: year 8's total is 2000 + 800 × 7 = 7600, which is more than £7000, while year 7's total is 2000 + 800 × 6 = 6800, which is not. Answer: year 8. 7 comes from rounding 6.25 to the nearest whole number, 6, and then adding 1, instead of rounding up to the next whole number before adding 1. 6 comes from using 6.25 rounded down to 6 as the year number directly, without adding the 1 needed to convert from the number of increases to the year number. 9 comes from adding one extra year beyond the year that already satisfies the condition.
- (a) 4x + 2 — Expand the bracket: 4(x − 2) = 4x − 8. Then add the 10 that follows the bracket: 4x − 8 + 10 = 4x + 2. The option 4x − 8 comes from expanding the bracket correctly but then forgetting to add the 10. The option 4x + 8 comes from forgetting to multiply the 2 inside the bracket by 4 (treating it as 4x − 2 instead of 4x − 8), then adding 10. The option 4x + 18 comes from a sign error when expanding — treating 4 × (−2) as +8 instead of −8, giving 4x + 8, then adding 10.
- (d) a/2 — Method: write both terms over the same denominator, subtract the numerators, then cancel the fraction down. Working: 6 is a multiple of 3, so a/3 is rewritten as 2a/6; the calculation becomes 5a/6 − 2a/6 = 3a/6, and dividing numerator and denominator by 3 gives a/2. Answer: a/2. The distractors: 4a/3 comes from subtracting the denominators as well as the numerators, giving (5 − 1)a over (6 − 3); 2a/3 comes from subtracting 1 from 5 without first rewriting a/3 as 2a/6, giving 4a/6; 7a/6 comes from adding the two fractions instead of subtracting them, giving 5a/6 + 2a/6.
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