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.A plumber charges a call-out fee of £30 plus £25 per hour worked. Work out the total charge for a job that takes 3 hours.
- 2.A mobile phone plan charges a fixed monthly fee plus an amount for each gigabyte of data used. The total cost £C for using g gigabytes in a month is given by C = 15 + 2g. What does the number 2 represent in this formula?
- 3.A table of values is being drawn for the graph of y = x³. Work out the value of y when x = −2.y = x
- 4.The first five terms of a quadratic sequence are 2, 5, 10, 17, 26. Work out the next term in the sequence.
- 5.Solve (2x + 1)/3 = 5
- 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.Here are the first five terms of an arithmetic sequence: 4, 7, 10, 13, 16. Work out the 9th term.
- 8.Two numbers have a sum of 50. The larger number is twice the smaller number. Work out the smaller number.
- 9.Which of these is an inequality?
- 10.A sequence has nth term 3n + 4. Two terms are missing from the list of its first five terms: 7, 10, _, 16, _. Work out the missing 3rd and 5th terms added together.
- 11.Write down the expression that means the same as p + p + p + q.
- 12.The first four terms of a sequence are 4, 6, 8, 10. Work out an expression, in terms of n, for the nth term.
- 13.A student says that (x + 4)² is equivalent to x² + 16. For which value of x do the two expressions give the SAME result, making it look (misleadingly) like the student could be right?
- 14.Expand 2(x − 13)
- 15.The point P has coordinates (0, 12). Write down the axis that P lies on and the y-coordinate of P.
Answer key
- (b) £105 — The hourly charge is 25 × 3 = £75. Adding the call-out fee: £75 + £30 = £105. A candidate who forgets the call-out fee gives just the hourly charge, £75. A candidate who adds the call-out fee to the hourly rate before multiplying by the hours, (30 + 25) × 3, gets £165. A candidate who multiplies the call-out fee by the number of hours instead of the hourly rate, 30 × 3, gets £90.
- (c) The cost, in pounds, for each extra gigabyte of data used — In C = 15 + 2g, the number multiplying g is the gradient, which gives the extra cost for each extra unit of g — here, £2 for each extra gigabyte. 'The fixed monthly fee, in pounds' describes the constant term 15, not the coefficient of g. 'The total number of gigabytes included in the plan' misreads the coefficient as a quantity of data rather than a cost per gigabyte. 'The cost … for each extra 2 gigabytes' doubles the unit the coefficient actually applies to — it is the cost for each single extra gigabyte.
- (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.
- (c) 37 — The first differences are 3, 5, 7, 9 — they increase by 2 each time (the second difference), so the next first difference is 11, giving 26+11=37. A candidate who repeats the last first difference (9) instead of increasing it would reach 26+9=35. A candidate who increases the difference by 4 instead of 2 would reach 26+13=39. A candidate who adds only the second difference (2) to the last term, instead of the next first difference, would reach 26+2=28.
- (a) 7 — Method: multiply both sides by 3 to clear the fraction, then solve the resulting equation. Working: 2x + 1 = 5 × 3 = 15. Subtract 1: 2x = 14. Divide by 2: x = 7. Answer: 7. 2 comes from ignoring the denominator altogether, treating the equation as 2x + 1 = 5 without multiplying by 3 first. 8 comes from a sign error, adding 1 to 15 instead of subtracting it, giving 2x = 16. 14 comes from correctly reaching 2x = 14 but stopping there, without dividing by 2 to find x.
- (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) 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) 50/3 — Method: call the smaller number x, write the larger number in terms of x and use the total. Working: the larger number is 2x, so x + 2x = 50, that is 3x = 50 and x = 50/3. Answer: 50/3, since 50/3 added to 100/3 makes 50 and 100/3 is twice 50/3. The distractors: 100/3 is the larger of the two numbers rather than the smaller one asked for; 25 comes from halving 50 and treating the two numbers as equal; 24 comes from reading 'twice the smaller number' as 'two more than the smaller number' and solving x + (x + 2) = 50.
- (d) 2n + 7 > 15 — An inequality compares two expressions using a symbol other than an equals sign, such as < or >. "2n + 7" is just an expression — it has no equals or inequality sign at all, so a candidate mistaking any bare expression for an inequality would choose this. "2n + 7 = 15" is an equation: it uses an equals sign and is only true for one particular value of n, so a candidate confusing an equation with an inequality would choose this. "P = 2n + 7" is a formula, since it relates two different letters, P and n, with an equals sign, so a candidate confusing a formula with an inequality would choose this. "2n + 7 > 15" is an inequality: it compares an expression with a number using a > sign.
- (c) 32 — Using the rule 3n + 4: the 3rd term is 3 × 3 + 4 = 13, and the 5th term is 3 × 5 + 4 = 19, so their sum is 13 + 19 = 32. Forgetting to add the 4 for the 3rd term, 3 × 3 = 9, and adding the correct 5th term, gives 9 + 19 = 28. Rounding the 19 up to 20 to make the addition easier and then forgetting to take the extra 1 back off, 13 + 20 = 33, gives 33. Using the rule 4n + 3 instead of 3n + 4 gives 4 × 3 + 3 = 15 and 4 × 5 + 3 = 23, summing to 38.
- (a) 3p + q — Repeated addition of the same letter is written as a multiple of that letter, so p + p + p is 3 lots of p, which is 3p. The letter q is added once only, so it stays as a separate term and the result is 3p + q. Writing 3pq multiplies the q by 3 and by p as well; p³ + q records repeated multiplication rather than repeated addition; 3(p + q) multiplies both letters by 3.
- (c) 2n + 2 — Method: find the common difference between consecutive terms, then find the constant that fits the first term. Working: 6 − 4 = 2, 8 − 6 = 2, 10 − 8 = 2, so the terms increase by 2 each time and the nth term has the form 2n + c. Substituting n = 1: 2(1) + c = 4, so c = 2. Answer: the nth term is 2n + 2. The value 2n comes from using only the common difference and leaving out the constant c entirely. The value 2n + 4 comes from using the first term itself as c, without subtracting the common difference to find the true constant. The value 2n − 2 comes from a sign error when working out c, giving −2 instead of +2.
- (c) x = 0 — Expand (x + 4)² correctly: (x + 4)² = x² + 8x + 16. This equals x² + 16 only when 8x is zero, i.e. when x = 0 — at every other value of x the two expressions differ by 8x. Choosing x = 4 confuses the constant inside the bracket with the value of x that makes the expressions match. Choosing x = −4 makes the same confusion but with the sign flipped. Choosing x = 8 mistakes the coefficient of the middle term, 8x, for the value of x itself.
- (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.
- (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.
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