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.Expand 4(2x − 3).
- 2.Here are the first four terms of an arithmetic sequence: 5, 9, 13, 17. Work out the 5th term.
- 3.A gardener charges a fixed fee of £15 plus £8 for each hour worked. The total charge, £T, for a job lasting h hours is given by T = 15 + 8h. Work out the charge for a job lasting 4 hours, and identify what kind of statement T = 15 + 8h is.
- 4.a = 3 and b = 5. Work out the value of a²b.
- 5.The nth term of a sequence is 2n − 1. Work out the 6th term of the sequence.
- 6.A rectangular garden has width w metres and length (w + 3) metres. A gardener writes its perimeter as 2w + 3. Which statement corrects the gardener's mistake?
- 7.A caterer uses the formula C = 6p + 20 to work out the total cost, £C, of a buffet for p people, where £20 covers fixed costs. A customer is charged £92. Work out how many people, p, the buffet was for.
- 8.The simple interest, £I, earned on a savings account is given by the formula I = PRT/100, where P is the amount invested in pounds, R is the interest rate per year and T is the time in years. Rearrange the formula to make P the subject.
- 9.Write down the gradient of the line with equation y = 6 + 4x.
- 10.A number machine adds 6 to its input. Work out the output when the input is −4.
- 11.Solve the simultaneous equations 4x + 3y = 25 and 4x − y = 1. Work out the value of y.
- 12.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.
- 13.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.
- 14.A plumber charges a £40 call-out fee plus £25 for each hour worked. The total charge, £C, for a job lasting h hours is given by C = 40 + 25h. Work out the charge for a job lasting 3 hours, and name what kind of statement C = 40 + 25h is.
- 15.The diagram shows the graph of a straight line, drawn on a grid numbered from −5 to 5 on both axes. Which equation could represent the line?
Answer key
- (c) 8x − 12 — Multiply each term inside the bracket by 4: 4 × 2x = 8x and 4 × (−3) = −12, so 4(2x − 3) = 8x − 12. A candidate who forgets to multiply the second term by 4 gets 8x − 3. A candidate who makes a sign error, treating 4 × (−3) as +12, gets 8x + 12. A candidate who adds 4 to the bracket instead of multiplying gets 2x + 1.
- (b) 21 — Method: find the gap between neighbouring terms, then add one gap to the last term that is known. Working: 9 − 5 = 4, 13 − 9 = 4 and 17 − 13 = 4, so 4 is added each time; the 5th term is one step on from the 4th term, so it is 17 + 4. Answer: 21. The distractors: 25 comes from adding the 4 twice and landing on the 6th term; 20 comes from multiplying the position by the common difference, 5 × 4, and ignoring the fact that the sequence starts at 5 rather than at 4; 22 comes from adding the first term, 5, to 17 instead of adding the common difference.
- (b) £47, and T=15+8h is a formula (relates T and h). — The charge is £15 fixed plus £8 for each hour: T=15+8h, so for h=4, T=15+8×4=15+32=47. Because T=15+8h relates two different quantities, the total charge T and the number of hours h, it is a formula, not an equation — an equation is solved for one particular value of an unknown, but this relationship holds for every value of h a job might last. A candidate who forgets to include the fixed £15 fee would compute only 8×4=32. A candidate who adds the three numbers in the formula together instead of multiplying the hourly rate by the number of hours would compute 15+8+4=27. A candidate who correctly finds the charge but mistakes the formula for an equation, treating it as something to be solved for one specific value of h rather than a general relationship between T and h, would pick the correct charge with the wrong classification.
- (a) 45 — a²b means a × a × b, so the index applies to a only and b is multiplied on afterwards. Substituting the values gives 3 × 3 = 9, then 9 × 5 = 45. Reading a²b as (ab)² gives (3 × 5)², which is 15² = 225 and squares b as well. Substituting the two values the wrong way round works out 5 × 5 × 3 = 75, and reading the letters written side by side as an addition gives 9 + 5 = 14.
- (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.
- (a) It is 2(w + (w + 3)) = 4w + 6, not 2w + 3. — The perimeter of a rectangle is twice the width plus twice the length: 2 × w + 2 × (w + 3) = 2w + 2w + 6 = 4w + 6, so the gardener's 2w + 3 is wrong. Writing w + (w + 3) = 2w + 3 forgets to double the sides at all, only adding one width and one length once. Writing 4(w + 3) = 4w + 12 wrongly treats all four sides as equal to the length, as if the garden were a square. Writing 3w + 6 comes from doubling the length correctly but adding the width only once instead of doubling it too.
- (b) 12 — Method: substitute the total cost into the formula, then subtract the fixed cost and divide by the cost per person. Working: 92 = 6p + 20, so 6p = 92 − 20 = 72, p = 72 ÷ 6 = 12. Answer: 12 people. 15.33 comes from dividing the whole £92 by £6 without first subtracting the £20 fixed cost: 92 ÷ 6 ≈ 15.33. 4.3 comes from swapping the two amounts round, subtracting £6 and dividing by £20: (92 − 6) ÷ 20 = 4.3. 18.67 comes from adding the fixed cost instead of subtracting it: (92 + 20) ÷ 6 ≈ 18.67.
- (a) P = 100I / (RT) — Method: undo the operations done to P in reverse order — multiply by 100, then divide by R and by T. Working: I = PRT/100, so 100I = PRT, so P = 100I / (RT). Answer: P = 100I / (RT). P = IRT/100 comes from leaving R and T in the numerator instead of moving them to the denominator. P = 100RT/I comes from swapping P and I when rearranging. P = 100I/R comes from dividing by R only and forgetting to also divide by T.
- (d) 4 — Method: compare the equation with the general form y = mx + c after writing the x term first, then read off the coefficient of x. Working: y = 6 + 4x can be written as y = 4x + 6, so comparing with y = mx + c gives m = 4. Answer: the gradient is 4. The value 6 comes from reading off the constant, written first in the equation, and treating it as the gradient because it comes before the x term. The value 10 comes from adding the two numbers in the equation, 6 + 4, instead of reading off the coefficient of x. The value 24 comes from multiplying the two numbers in the equation, 6 × 4, instead of reading off the coefficient of x.
- (c) 2 — The output is the input plus 6: −4 + 6 = 2. A candidate who subtracts 6 instead of adding it gets −4 − 6 = −10. A candidate who ignores the negative sign on the input and works out 4 + 6 gets 10. A candidate who multiplies the input by 6 instead of adding gets −4 × 6 = −24.
- (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.
- (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) 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.
- (b) £115, and C = 40 + 25h is a formula — Substitute h = 3 into the rule, multiplying before adding. The hours cost 25 × 3 = 75, and adding the call-out fee gives 40 + 75 = 115, so the charge is £115. The rule itself links two different quantities, C and h, and lets one be worked out from the other, so it is a formula; an expression would have no equals sign in it. Adding the fee before multiplying gives 65 × 3 = 195, which charges the call-out fee three times over, and stopping at 40 + 25 leaves £65, the charge for a single hour.
- (d) $y = 2x - 1$ — Method: pick two points the line passes through, work out the gradient as vertical change ÷ horizontal change, and read the y-intercept from where the line crosses the y-axis. Working: the line passes through (0, −1) and (1, 1), so the gradient is (1 − (−1)) ÷ (1 − 0) = 2, and it crosses the y-axis at −1. Answer: y = 2x − 1. Distractor refutation: y = 2x + 1 comes from reading the y-intercept as +1 instead of −1, misreading which side of the origin the line crosses. y = x − 1 comes from taking the gradient as 1, counting the same number of squares across and up instead of checking the rise is twice the run. y = −2x − 1 comes from a sign error on the gradient, treating the line as sloping downward from left to right when it actually rises.
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