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.A point has coordinates (−6, 9). Work out the sum of the x-coordinate and the y-coordinate.
- 2.The graph of y = (x − 4)(x + 1) crosses the x-axis. Which pair gives the correct roots and reasoning?
- 3.A cycle route is 84 km long. Freya sets off along it at a steady 14 km/h. Write down the function for the distance y, in kilometres, that is still to be cycled after x hours.
- 4.A curve has equation y = x³ − 4x. Work out the coordinates of the point where the curve crosses the y-axis.y = x
- 5.A rule turns each input x into an output y. The inputs are x = −1, 0, 1, 2 and the matching outputs are y = 5, 3, 1, −1. Work out the rule.
- 6.The first five terms of a sequence are 2, 5, 8, 11, 14. Work out an expression, in terms of n, for the nth term.
- 7.The formula connecting speed, distance and time is v = s ÷ t. Work out v, in km/h, when s = 180 and t = 4.
- 8.A colony of bacteria doubles in number every hour. At 9am there are 5 bacteria in the colony. Work out how many bacteria there will be at 12 noon.
- 9.The nth term of a sequence is 2n − 1. Work out the 6th term of the sequence.
- 10.Expand 2(x − 13)
- 11.A geometric sequence has first term 5, and each term after that is 3 times the term before it. Work out the first term of the sequence that is greater than 1000.
- 12.Work out the value of 2(x + 2) when x = −5
- 13.A square has vertices at (−2, 3), (−2, −3) and (4, −3). Write down the coordinates of the fourth vertex.
- 14.Yuki is y years old. Her brother is 4 years younger than Yuki. Write down an expression, in terms of y, for the brother's age in 5 years' time.
- 15.Work out the value of 5a + 2 when a = 3
Answer key
- (b) 3 — −6 + 9 = 3. A candidate who ignores the negative sign on the x-coordinate and adds the two positive values gets 6 + 9 = 15. A candidate who treats the y-coordinate as negative too gets −6 + (−9) = −15. A candidate who works out 9 − 6 correctly as 3 but then writes the answer with the wrong sign gets −3.
- (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.
- (b) y = 84 − 14x — Method: the distance still to go is the whole route minus the distance already covered, and at a steady speed the distance covered after x hours is the speed multiplied by x. Working: after x hours Freya has cycled 14x km, so the distance remaining is the 84 km route take away 14x, giving y = 84 − 14x, a straight line with intercept 84 and gradient −14. Answer: y = 84 − 14x. The distractors: y = 84 + 14x comes from adding the distance cycled to the length of the route rather than taking it away, so the ride would grow longer the further she goes; y = 14x − 84 comes from carrying out the subtraction the wrong way round, which is negative for the whole of the ride; y = 14 − 84x comes from swapping the two numbers over, treating 84 km as the hourly speed and 14 km as the length of the route.
- (c) (0, 0) — A curve crosses the y-axis where x = 0. Substituting x = 0 into y = x³ − 4x gives y = 0³ − 4(0) = 0 − 0 = 0, so the curve crosses the y-axis at (0, 0). A candidate who reads off the coefficient of x as the y-intercept, instead of working out the constant term, might write (0, −4). A candidate who swaps the coordinates might write (4, 0). A candidate who takes the coefficient of x but drops its sign might write (0, 4).
- (a) y = −2x + 3 — Method: find the step in the outputs for each step of 1 in the input — falling outputs mean a negative multiplier — then read off the output when the input is 0, because that is the number added on. Working: the outputs 5, 3, 1, −1 fall by 2 each time x rises by 1, so x is multiplied by −2; the output at x = 0 is 3, so 3 is added. Answer: y = −2x + 3, checked at x = 2 by −2 × 2 + 3 = −1. The distractors: y = 2x + 3 comes from taking the size of the step, 2, as the multiplier and ignoring the fact that the outputs are falling; y = −2x − 3 comes from using the correct multiplier but writing the number added on as −3 instead of the output 3 listed at x = 0; y = −x + 4 comes from taking the multiplier as −1, its size read from the step of 1 in the inputs instead of the step of 2 in the outputs and its sign from the fact that the outputs fall, and then fitting the number added on to the pair x = −1, y = 5.
- (a) 3n − 1 — Method: find the common difference between consecutive terms, then find the constant by adjusting the first term. Working: 5 − 2 = 3, 8 − 5 = 3, 11 − 8 = 3, so the common difference is 3 and the coefficient of n is 3. The constant is the first term minus the common difference: 2 − 3 = −1. Answer: the nth term is 3n − 1. 3n + 2 comes from using the first term, 2, as the constant without subtracting the common difference. 3n − 2 comes from a slip when working out the constant, treating 2 − 3 as −2 instead of −1. 2n + 3 comes from swapping the common difference and the first term.
- (d) 45 — v = s ÷ t = 180 ÷ 4 = 45 km/h. 720 comes from multiplying s and t instead of dividing, 180 × 4. 184 comes from adding s and t instead of dividing, 180 + 4. 176 comes from subtracting t from s instead of dividing, 180 − 4.
- (a) 40 — From 9am to 12 noon is 3 hours, so the population doubles three times: 5 × 2³ = 40. A candidate who counts the elapsed time as 4 hours (an off-by-one counting error) would reach 5 × 2⁴ = 80. A candidate who counts it as only 2 hours would reach 5 × 2² = 20. A candidate who misreads 'doubles' as 'increases by 2' each hour would compute 5 + 3 × 2 = 11.
- (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.
- (d) 2x − 26 — Method: multiply each term inside the bracket by the 2 in front of it, and keep the sign that belongs to each term. Working: 2 × x = 2x and 2 × 13 = 26; the bracket contains a subtraction, so the second term is subtracted. Answer: 2x − 26. The distractors: 2x − 13 comes from multiplying only the x by 2 and copying the 13 across unchanged; 2x + 26 comes from multiplying both terms correctly but losing the minus sign that belongs to the second term; 2x − 11 comes from subtracting 2 from 13 instead of multiplying 13 by 2.
- (b) 1215 — Method: generate the terms one at a time with the term-to-term rule and compare each with 1000 as you go, stopping at the first one that passes it. Working: the terms are 5, then 5 × 3 = 15, then 45, then 135, then 405, and 405 × 3 = 1215; 405 is still below 1000 while 1215 is above it. Answer: 1215. The distractors: 405 comes from stopping at the last term that is still below 1000 instead of giving the first one above it; 3645 comes from carrying on one term too far, past the first term that passes 1000; 2187 comes from using the multiplier 3 as the first term as well, generating 3, 9, 27, 81, 243, 729, 2187 instead of the sequence described.
- (c) −6 — Method: substitute the value, work out the bracket first, then multiply what the bracket comes to by 2. Working: the bracket gives (−5) + 2 = −3, and multiplying by 2 gives 2 × (−3) = −6. Answer: −6. The distractors: −3 comes from working out the bracket correctly but stopping there and never multiplying by 2; −1 comes from multiplying the 2 inside the bracket by the 2 outside before adding, giving −5 + 4; −8 comes from multiplying only the x by 2 and leaving the 2 inside the bracket unchanged, giving −10 + 2.
- (d) (4, 3) — The sides are parallel to the axes: the missing vertex must share the y-coordinate 3 with (−2, 3) and the x-coordinate 4 with (4, −3), giving (4, 3). (−4, 3) comes from a sign error on the x-coordinate. (4, −9) comes from continuing the pattern of the given points by subtracting 6 from the y-coordinate again instead of matching it to (−2, 3). (3, 4) comes from swapping the x- and y-coordinates.
- (a) y + 1 — The brother's age now is y − 4. In 5 years' time this becomes y − 4 + 5 = y + 1. A candidate who adds 4 instead of subtracting it, then adds 5, gets y + 4 + 5 = y + 9. A candidate who subtracts 5 instead of adding it gets y − 4 − 5 = y − 9. A candidate who works out the brother's current age but forgets to add on the 5 years gets y − 4.
- (a) 17 — Substitute a = 3 into 5a + 2: 5 × 3 + 2 = 15 + 2 = 17. 10 comes from treating 5a as 5 + a instead of 5 × a, giving 5 + 3 + 2. 25 comes from adding the 2 to a before multiplying by 5, 5 × (3 + 2). 15 comes from working out 5 × 3 correctly but forgetting to add the 2.
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