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.Here are the first five terms of an arithmetic sequence: 2, 5, 8, 11, 14. Work out the 8th term.
- 2.The volume of a cuboid is given by the formula V = lwh. Work out V, in cm³, when l = 2, w = 3 and h = 6.
- 3.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.
- 4.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.
- 5.A company's profit is £500 in its first year. Each year after that, the profit is £300 more than the year before. Work out the profit in the company's 6th year.
- 6.The first four terms of a sequence are 9, 14, 19, 24. Work out an expression, in terms of n, for the nth term.
- 7.Solve x² − 2x − 24 = 0.
- 8.Six years ago Harry was three times as old as his brother Leo. Harry is now 24 years old. Work out Leo's age now.
- 9.The first four terms of a sequence are 5, 8, 11, 14. Work out an expression, in terms of n, for the nth term.
- 10.Line C passes through (0, 0) and (4, 2). Line D passes through (0, 0) and (2, 2). Write down which of the two lines is the steeper.
- 11.The first term of an arithmetic sequence is 5, and each term after that increases by 6. Work out the 8th term of the sequence.
- 12.A water tank empties according to y = −5x + 200, where y is the number of litres left after x minutes. Work out the coordinates of the point where this graph crosses the x-axis.y = -5x + 200
- 13.The solution to an inequality is n ≤ 5. Write down the largest integer value of n that satisfies this inequality.
- 14.A car's speed-time graph shows the following: its speed increases steadily from 0 m/s to 20 m/s over the first 10 seconds, then stays constant at 20 m/s for the next 15 seconds. Work out the total distance travelled in the first 25 seconds.
- 15.Simplify k + k + k + k + k.
Answer key
- (d) 23 — Method: find the number added each time, then add it to the first term once for every step between the first term and the term wanted. Working: 5 − 2 = 3, 8 − 5 = 3 and so on, so 3 is added each time; the 8th term is seven steps on from the 1st term, so it is 2 + 7 × 3 = 2 + 21. Answer: 23, which agrees with counting on from the 5th term: 14, 17, 20 and then one step more. The distractors: 26 comes from adding the common difference eight times instead of seven, 2 + 8 × 3; 20 comes from stopping one term early, at the 7th term; 24 comes from multiplying the position by the common difference, 8 × 3, and ignoring the fact that the sequence starts at 2 rather than at 3.
- (d) 36 cm³ — V = lwh = 2 × 3 × 6 = 36 cm³. A candidate who adds all three numbers instead of multiplying gets 2 + 3 + 6 = 11 cm³. A candidate who multiplies only two of the three numbers, forgetting the length, gets w × h = 3 × 6 = 18 cm³. A candidate who multiplies l × w and w × h separately and adds the two products gets (2 × 3) + (3 × 6) = 6 + 18 = 24 cm³.
- (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) 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.
- (d) £2000 — This is an arithmetic sequence with first term £500 and common difference £300. The 6th term is 500 + 5 × 300 = 2000. A candidate who uses 6 lots of the increase instead of 5 gets 500 + 6 × 300 = 2300. A candidate who forgets to add the first year's profit at all gets 5 × 300 = 1500. A candidate who miscounts the number of increases as 4 instead of 5 gets 500 + 4 × 300 = 1700.
- (d) 5n + 4 — Method: find the common difference, then find the constant that fits the first term. Working: 14 − 9 = 5, 19 − 14 = 5, 24 − 19 = 5, so the terms increase by 5 each time and the nth term has the form 5n + c. Substituting n = 1: 5(1) + c = 9, so c = 4. Answer: the nth term is 5n + 4. The value 5n comes from leaving out the constant. The value 5n + 9 comes from using the first term as the constant directly, without subtracting the common difference first. The value 9n + 5 comes from swapping the roles of the first term and the common difference — using the first term, 9, as the coefficient of n and the difference, 5, as the constant.
- (a) x = 6 or x = −4 — We need two numbers that multiply to −24 and add to −2: these are −6 and 4, since −6 × 4 = −24 and −6 + 4 = −2. So x² − 2x − 24 = (x − 6)(x + 4) = 0, giving x = 6 or x = −4. A candidate who swaps the signs, using 6 and −4 the wrong way round in the brackets, gets x = −6 or x = 4. A candidate who picks the wrong factor pair, 8 and −3 (which multiply to −24 but add to +5, not −2), gets (x + 8)(x − 3) = 0 and answers x = −8 or x = 3. A candidate who makes both factors negative gets x = −6 or x = −4, which would require the constant term to be +24, not −24.
- (a) 12 — Method: write both ages as they were six years ago, call Leo's present age x, and form an equation from the comparison at that time. Working: six years ago Harry was 24 − 6 = 18 and Leo was x − 6, so 18 = 3(x − 6); expanding gives 18 = 3x − 18, adding 18 to both sides gives 36 = 3x, and dividing by 3 gives x = 12. Checking: six years ago Harry was 18 and Leo was 6, and 18 = 3 × 6. Answer: 12. The distractors: 6 comes from solving 18 = 3(x − 6) as far as Leo's age six years ago, 18 ÷ 3 = 6, and giving that as his age now; 8 comes from dividing Harry's present age by 3, 24 ÷ 3 = 8, using the multiple at the wrong moment in time; 36 comes from stopping at 3x = 36 and giving 36 as Leo's age.
- (d) 3n + 2 — The common difference is 3 (8−5=3), so the expression starts 3n. To match the first term when n=1, 3×1+c=5, so c=2: the nth term is 3n+2. A candidate who uses the first term itself as the constant, instead of first term minus the common difference, would write 3n+5 (giving 8, 11, 14, 17 — one term too high throughout). A candidate who omits the constant term altogether would write just 3n (giving 3, 6, 9, 12, not matching the sequence at all). A candidate who adds the common difference to n instead of multiplying would write n+3 (giving 4, 5, 6, 7, far too small).
- (c) Line D — Method: the steeper of two lines is the one whose gradient is greater, and the gradient of a line through two points is the change in y divided by the change in x. Working: line C rises 2 for a run of 4, so its gradient is 2 ÷ 4 = 1/2; the other line rises 2 for a run of 2, so its gradient is 2 ÷ 2 = 1; since 1 is greater than 1/2, it is line D that is the steeper. Answer: Line D. The distractors: Line C is chosen by comparing the x-coordinates and calling the line that reaches further along the x-axis the steeper one; They are equally steep comes from comparing only the y-coordinates, which are both 2, without dividing by the two different runs; There is not enough information comes from believing that a gradient can only be measured off a drawn graph, when two points on a line are enough to work it out.
- (d) 47 — To reach the 8th term from the 1st term, the difference of 6 is added 7 times (8 − 1 = 7): 5 + 7 × 6 = 47. A candidate who multiplies by the term number itself, rather than one less, would compute 5 + 8 × 6 = 53. A candidate who uses one step too few (6 instead of 7) would reach 5 + 6 × 6 = 41. A candidate who forgets to include the first term at all would compute just 8 × 6 = 48.
- (a) (40, 0) — Method: a graph crosses the x-axis where y = 0, which in this context is the moment the tank holds no water, so substituting y = 0 and solving gives the time. Working: 0 = −5x + 200 gives 5x = 200, so x = 200 ÷ 5 = 40 and the crossing point is (40, 0); the tank is empty after 40 minutes. Answer: (40, 0). The distractors: (0, 200) is the y-axis crossing, the 200 litres in the tank at the start, found by substituting x = 0 instead of y = 0; (200, 0) comes from reading the constant 200 as the x-coordinate without dividing by the 5 litres lost each minute; (0, 40) has the right number in the wrong place, with the time written as a y-coordinate.
- (a) 5 — The symbol ≤ means n can equal 5 or any number less than 5, so 5 is included and is the largest integer value. A candidate who treats the inequality as strict, as if it were n < 5, answers 4. A candidate who confuses ≤ with ≥ and looks for a value just above the boundary answers 6. A candidate who makes a sign error and reads the inequality as n ≤ −5 answers −5.
- (d) 400 m — The distance travelled equals the area under the speed-time graph. The first 10 seconds form a triangle with base 10 and height 20, giving an area of 0.5 × 10 × 20 = 100 m. The next 15 seconds form a rectangle with base 15 and height 20, giving an area of 15 × 20 = 300 m. The total distance is 100 + 300 = 400 m. The option 500 m treats the whole 25 seconds as travelled at the constant 20 m/s, ignoring that the speed was building up during the first 10 seconds: 25 × 20 = 500. The option 300 m only counts the constant-speed section and forgets the triangle section entirely. The option 200 m comes from working out the triangle's area without halving it (10 × 20 = 200) and forgetting the rectangle altogether.
- (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.
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