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.The graph of y = 24/x is drawn for x > 0. Work out the value of y when x = 6.
- 2.A ball is dropped and bounces. The height of each bounce after the first is 8 cm less than the bounce before it. The first bounce reaches 60 cm. Work out the height of the 5th bounce.
- 3.A point lies on the y-axis. What can be said about the x-coordinate of that point?
- 4.The formula for the energy stored in a stretched spring is E = (1/2)kx², where E is in joules, k is the spring constant in N/m and x is the extension in m. A spring with a spring constant of 40 N/m is stretched so that its extension is 0.3 m. Work out the energy stored in the spring.
- 5.The simultaneous equations y = 3 and 4x − y = 9 are given. Work out the value of x.
- 6.Here are the first five terms of an arithmetic sequence: 4, 7, 10, 13, 16. Work out the 9th term.
- 7.A sequence begins at 50, and each term after that is found by subtracting 8 from the term before it. Priya says the 8th term of the sequence is negative. Is Priya correct?
- 8.Priya buys b balloons for a party. Each balloon costs 15p. Write down an expression for the total cost, in pence.
- 9.Factorise fully 5x + 5y − 5
- 10.Solve the simultaneous equations 4x + 3y = 25 and 4x − y = 1. Work out the value of y.
- 11.A parcel delivery company's cost graph shows the following: the cost is a flat £5 for parcels weighing up to 2 kg, and then rises by £2 for each additional kg above 2 kg. Use the graph to work out the weight of a parcel that costs £17.
- 12.A car's speed increases steadily while it accelerates. Its speed v m/s after t seconds is given by v = u + at, where u is the starting speed in m/s and a is the acceleration in m/s². A car starts at u = 4 m/s and accelerates at a = 2 m/s² until it reaches v = 20 m/s. Work out t, the time taken in seconds.
- 13.Which expression is equivalent to 2(x + 6)?
- 14.Work out the value of 3a − 2b when a = 5 and b = 4
- 15.A geometric sequence starts at 3, and each term after that is found by multiplying the term before it by 2. Write down the first four terms.
Answer key
- (d) y = 4 — Method: substitute x = 6 into y = 24/x: y = 24 ÷ 6 = 4. Distractor origins: y = 18 subtracts instead of dividing (24 − 6 = 18); y = 144 multiplies instead of dividing (24 × 6 = 144); y = 30 adds instead of dividing (24 + 6 = 30).
- (d) 28 — The height decreases by 8 cm at each bounce after the first, so the nth bounce reaches 60−(n−1)×8 cm. For the 5th bounce: 60−4×8=60−32=28. A candidate who subtracts 8 one time too many, five times instead of four, would compute 60−5×8=20. A candidate who adds the decrease instead of subtracting it, a sign error, would compute 60+4×8=92. A candidate who works out only the total decrease and forgets to include the starting height of 60 cm would compute just 5×8=40.
- (b) x = 0 — Method: the first coordinate of a point measures how far to the left or right of the origin it is, so a point that is neither left nor right of the origin has a first coordinate of zero. Working: the y-axis is the vertical line through the origin; every point on it lies directly above or directly below the origin with no sideways movement at all, so its first coordinate is zero while its second coordinate may take any value. Answer: x = 0. The distractors: y = 0 is the condition for lying on the x-axis, the other axis; x = 1 comes from confusing the y-axis with the vertical line one unit to the right of it; x = y is the condition for the diagonal through the origin, which meets the y-axis at the origin only.
- (a) 1.8 — E = (1/2)kx² = 0.5 × 40 × 0.3² = 0.5 × 40 × 0.09 = 1.8 joules. 3.6 comes from forgetting the (1/2) at the front, 40 × 0.09. 72 comes from squaring kx together instead of squaring only x, 0.5 × (40 × 0.3)² = 0.5 × 144 = 72. 6 comes from using x instead of x², 0.5 × 40 × 0.3.
- (c) 3 — Substituting y = 3 into 4x − y = 9 gives 4x − 3 = 9, so 4x = 12, and x = 3. A candidate who adds 3 instead of subtracting it, making a sign error when substituting, would get 4x + 3 = 9, so 4x = 6 and x = 1.5. A candidate who forgets to divide by 4 after finding 4x = 12 would write x = 12. A candidate who subtracts 4 instead of dividing by it would get 12 − 4 = 8.
- (b) 28 — Method: find the common difference, then use the position-to-term rule (or extend the sequence) to reach the 9th term. Working: the common difference is 7 − 4 = 3, so the nth term is 4 + 3(n − 1). For n = 9: 4 + 3 × 8 = 4 + 24 = 28. Answer: 28. 27 comes from using 3n instead of 3n + 1, dropping the constant term from the rule, 3 × 9 = 27. 31 comes from extending from the 5th term by one step too many, adding 3 five times instead of four, 16 + 3 × 5. 25 comes from extending by one step too few, adding 3 three times instead of four, 16 + 3 × 3.
- (d) Yes: the 8th term is 50 − 8 × 7 = −6, which is negative. — Method: find the 8th term by subtracting 8 a total of 7 times from the first term, since the 1st term itself needs 0 subtractions. Working: 8th term = 50 − 8 × 7 = 50 − 56 = −6, which is negative, so Priya is correct. Answer: Yes, the 8th term is 50 − 8 × 7 = −6, which is negative. The "50 − 8 × 6 = 2" option subtracts 8 only six times instead of seven, an off-by-one error in counting the steps. The "50 − 8 × 8 = −14" option subtracts 8 eight times instead of seven, the opposite off-by-one error. The claim that repeated subtraction "can never go negative" ignores that subtracting enough times from any starting value eventually gives a negative result.
- (d) 15b — Method: each balloon costs 15p, so b balloons cost 15 × b pence. Working: 15 × b = 15b. Answer: 15b. b + 15 comes from treating the 15p as a one-off addition rather than a rate applied to every balloon. b − 15 comes from misreading the situation as a discount rather than a cost per balloon. b/15 comes from dividing instead of multiplying, swapping the two operations.
- (c) 5(x + y − 1) — Method: take out the highest common factor of all three terms and divide every term by it, the number term included. Working: the highest common factor of 5x, 5y and −5 is 5; dividing gives 5x ÷ 5 = x, 5y ÷ 5 = y and −5 ÷ 5 = −1, so the bracket holds x + y − 1. Answer: 5(x + y − 1), which multiplies back out to 5x + 5y − 5. The distractors: 5(x + y + 1) comes from dividing −5 by 5 and losing the minus sign; 5(x + y − 5) comes from dividing only the terms containing a letter by 5 and carrying the −5 into the bracket unchanged; 5(xy − 1) comes from collecting the unlike terms 5x and 5y as 5xy before factorising.
- (a) 6 — Subtracting the second equation from the first: the x-terms, 4x and 4x, cancel; the y-terms combine as 3y − (−y) = 4y; and the right-hand sides give 25 − 1 = 24. This gives 4y = 24, so y = 6. A candidate who subtracts in the wrong order would get 4y = 1 − 25 = −24, so y = −6. A candidate who forgets the sign on the −y term, treating 3y − y as 2y, would get 2y = 24, so y = 12. A candidate who divides 24 by 6 instead of 4 would get y = 4.
- (b) 8 kg — The cost above the flat £5 charge is 17 − 5 = £12. At £2 per kg, this covers 12 ÷ 2 = 6 kg above the first 2 kg, so the total weight is 2 + 6 = 8 kg. Dividing the full £17 by £2 per kg without first taking off the £5 flat charge gives 17 ÷ 2 = 8.5 kg. Taking off the £5 flat charge and dividing by £2 per kg, but forgetting to add back the 2 kg that the flat charge covers, gives 12 ÷ 2 = 6 kg. Taking off £2 instead of £5 as the flat charge, (17 − 2) ÷ 2 = 7.5 kg, swaps which number is the fixed fee.
- (b) 8 s — Rearranging v = u + at for t: subtract u from both sides to get v − u = at, then divide by a: t = (v − u)/a. Substituting u = 4, a = 2, v = 20: t = (20 − 4)/2 = 16/2 = 8 s. Answering 12 s comes from adding u instead of subtracting it: (20 + 4)/2 = 12. Answering 32 s comes from multiplying (v − u) by a instead of dividing by it: 16 × 2 = 32. Answering 6 s divides v by a first and then subtracts u, in the wrong order: 20/2 − 4 = 10 − 4 = 6. The time taken is 8 s.
- (c) 2x + 12 — Multiply every term inside the bracket by 2: 2 × x = 2x and 2 × 6 = 12, so 2(x + 6) = 2x + 12. Writing 2x + 6 comes from multiplying only the x-term and leaving the 6 unchanged. Writing x + 12 comes from multiplying only the 6 and leaving the x unchanged. Writing 2x + 8 comes from adding 2 and 6 instead of multiplying them.
- (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) 3, 6, 12, 24 — Method: a term-to-term rule is applied to the term just written, so start with the first term and use the rule three more times. Working: the first term is 3; 3 × 2 = 6 is the second; 6 × 2 = 12 is the third; 12 × 2 = 24 is the fourth. Answer: 3, 6, 12, 24. The distractors: 3, 5, 7, 9 comes from adding 2 each time instead of multiplying by 2; 3, 9, 27, 81 comes from multiplying by the first term, 3, instead of by the multiplier 2 the rule gives; 6, 12, 24, 48 comes from doubling before writing anything down, so the list starts one term too late and the given first term is missing.
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