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
MathsUKwww.geekhero.co.uk
- 1.A theatre has rows of seats arranged so that row 1 has 12 seats, row 2 has 17 seats, row 3 has 22 seats and row 4 has 27 seats, with each row having 5 more seats than the row before. Work out an expression, in terms of n, for the number of seats in row n.
- 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 sequence has the position-to-term rule n² + 1, where n is the position number. Work out how much bigger the 5th term is than the 4th term.
- 4.A hall is set out in rows of chairs for an assembly. Row 1 has 50 chairs, row 2 has 45 chairs, row 3 has 40 chairs, and row 4 has 35 chairs, with each row after the first having 5 fewer chairs than the one before. Work out an expression, in terms of n, for the number of chairs in row n.
- 5.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.
- 6.The first four terms of a sequence are 5, 8, 11, 14. Work out an expression, in terms of n, for the nth term.
- 7.The formula connecting distance (d), speed (s) and time (t) is d = st. Make s the subject of the formula.
- 8.A table of values is being drawn for the graph of y = x³. Work out the value of y when x = −2.y = x
- 9.Solve 4(x − 3) = 2x + 6.
- 10.A printer starts a job with 480 sheets of paper loaded and prints at a steady rate. The number of sheets left, S, after t minutes is given by S = 480 − 24t. Work out how many minutes it takes for the paper to run out completely.
- 11.Which expression means 'add 3 to n, then multiply the result by 4'?
- 12.A car park charges by the hour. Parking for 1 hour costs £5, 2 hours costs £9, 3 hours costs £13 and 4 hours costs £17, with the cost increasing by the same amount for each extra hour. Work out an expression, in terms of n, for the cost, in pounds, of parking for n hours.
- 13.A rule turns each input x into an output y. The inputs are x = 0, 1, 2, 3 and the outputs are y = 4, 7, 10, 13. Work out the output when x = 5.
- 14.Write down the gradient of the line with equation y = 7x − 2.y = 7x − 2
- 15.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.
Answer key
- (c) 5n + 7 — Method: find the common difference between the rows, then find the constant by adjusting the first row's total. Working: each row has 5 more seats than the last, so the coefficient of n is 5. The constant is the first row's total minus the common difference: 12 − 5 = 7. Answer: the nth term is 5n + 7. 5n + 12 comes from using row 1's total, 12, as the constant without subtracting the common difference. 5n + 2 comes from a slip in working out the constant, subtracting the common difference twice (12 − 5 − 5 = 2) instead of once. 7n + 5 comes from swapping the common difference and the constant.
- (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.
- (a) 9 — Method: the gaps in this sequence are not constant, so work out each of the two terms named from the rule and then subtract the earlier from the later. Working: the 5th term is 5² + 1 = 25 + 1 = 26 and the 4th term is 4² + 1 = 16 + 1 = 17, so the difference is 26 − 17. Answer: 9. The distractors: 7 comes from using the 3rd and 4th terms, one position too early, 17 − 10; 11 comes from using the 5th and 6th terms, one position too late, 37 − 26; 1 comes from subtracting the position numbers, 5 − 4, instead of the terms themselves.
- (a) 55 − 5n — The number of chairs decreases by 5 in each row after the first, so the common difference is d=−5, and the first term is a=50. The nth term is a+(n−1)d = 50+(n−1)(−5) = 50−5n+5 = 55−5n. A candidate who uses the common difference as the constant term instead of correctly finding 55, giving the constant as −5 instead, would write −5n−5. A candidate who uses the first term, 50, as the coefficient of n instead of the common difference, would write 50n−5. A candidate who does not multiply the common difference by n at all, treating the nth term as n+d instead of dn+c, would write n−5.
- (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.
- (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).
- (b) s = d / t — Method: undo the multiplication by t by dividing both sides by t. Working: d = st, so dividing both sides by t gives s = d / t. The value s = dt comes from multiplying by t instead of dividing. The value s = t / d comes from inverting the fraction, dividing t by d instead of d by t. The value s = d − t comes from subtracting t instead of dividing by it.
- (c) −8 — (−2)³ = (−2) × (−2) × (−2) = −8, since multiplying three negative numbers gives a negative result. A candidate who forgets the sign of a negative number when cubing it might treat (−2)³ as if it were 2³ = 8. A candidate who multiplies −2 by 3 instead of cubing it might get −2 × 3 = −6. A candidate who combines both mistakes — multiplying by 3 and dropping the sign — might get 2 × 3 = 6.
- (d) 9 — Method: expand the brackets fully first, then collect the x-terms and constants before dividing. Working: 4(x − 3) = 4x − 12, so 4x − 12 = 2x + 6; 4x − 2x = 6 + 12; 2x = 18; x = 9. Answer: x = 9. 4.5 comes from expanding only the x-term in the bracket and forgetting to multiply the 3, giving 4x − 3 = 2x + 6, then 2x = 9, x = 4.5. −3 comes from a sign error when expanding, giving 4x + 12 = 2x + 6, then 2x = −6, x = −3. 3 comes from a sign error collecting the x-terms, adding instead of subtracting: 4x + 2x = 6 + 12, so 6x = 18, x = 3.
- (a) 20 — Method: the paper has run out when S = 0, so substitute S = 0 into the equation and solve for t. Working: 0 = 480 − 24t, so 24t = 480, and t = 480 / 24 = 20. Answer: it takes 20 minutes. The value 480 comes from giving the starting number of sheets, the intercept of the equation, instead of solving for t. The value 456 comes from working out 480 − 24 and stopping after one step instead of solving the equation fully. The value 0.05 comes from inverting the division, working out 24 / 480 instead of 480 / 24.
- (c) 4(n + 3) — 'Add 3 to n' must happen before 'multiply the result by 4', so the addition needs brackets to show it happens first: 4(n + 3). Writing 4n + 3 multiplies n by 4 immediately and only adds the 3 afterwards, which reverses the order the words describe. Writing n + 3 × 4 only multiplies the 3 by 4, and adds n as a separate, unmultiplied term — it treats 'the result' as just the 3, not the whole of n + 3. Writing 3(n + 4) keeps the correct structure but swaps which number is added and which is multiplied. The expression for 'add 3 to n, then multiply the result by 4' is 4(n + 3).
- (a) 4n + 1 — Method: find the increase in cost per hour, then find the constant by adjusting the 1-hour cost. Working: the cost goes up by £4 for each extra hour (9 − 5 = 4), so the coefficient of n is 4. The constant is the 1-hour cost minus the common difference: 5 − 4 = 1. Answer: the nth term is 4n + 1. 4n + 5 comes from using the 1-hour cost, 5, as the constant without subtracting the common difference. 4n − 3 comes from a slip in working out the constant, subtracting the common difference twice (5 − 4 − 4 = −3) instead of once. n + 4 comes from swapping the hourly increase and the constant.
- (a) 19 — Each time x increases by 1, y increases by 3 (4, 7, 10, 13 — a constant difference of 3). So at x = 4, y = 13 + 3 = 16, and at x = 5, y = 16 + 3 = 19. A candidate who stops one step early, giving the value for x = 4 instead of x = 5, answers 16. A candidate who overcounts and adds three steps of 3 instead of two from x = 3 gets 13 + 9 = 22. A candidate who mistakes the y-intercept (4) for the common difference and adds 4 twice from x = 3 gets 13 + 8 = 21.
- (b) 7 — Method: compare the equation with y = mx + c, where m is the gradient. Working: in y = 7x − 2, the coefficient of x is 7. Answer: the gradient is 7. −2 comes from confusing the gradient with the y-intercept. −7 comes from a sign error, treating the coefficient of x as negative. 2 comes from using the y-intercept but dropping its sign.
- (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.
Build your own mix at the worksheet builder.