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.A rectangular photo has length (x + 3) cm and width x cm. Its area is 40 cm². Work out the value of x.
- 2.A ball's height, h metres, t seconds after being thrown follows h = (t − 1)(9 − t). Given that the ball is at ground level at t = 1 and t = 9, work out at what time t the ball reaches its maximum height, using symmetry.
- 3.Which of these is an expression?
- 4.A number machine multiplies its input by 2 and then subtracts 5. Work out the output when the input is 6.
- 5.A charity's fundraising total, T pounds, over d days follows T = (d − 3)(30 − d) for 3 ≤ d ≤ 30, where T = 0 marks the start and end of the campaign. Work out how many days the campaign runs for, from start to end.
- 6.The formula for converting a distance in miles, m, to kilometres, k, is k = 8m ÷ 5. Work out k when m = 10.
- 7.The nth term of a sequence is 4n + 2. A term in the sequence has value 30. Work out its position, n, in the sequence.
- 8.Make b the subject of the formula a = 5b.
- 9.A gym charges a reduced price for the first month of membership and a standard price for every month after that. The total cost is £5 for 1 month, £12 for 2 months, £19 for 3 months and £26 for 4 months. Work out an expression, in terms of n, for the total cost, in pounds, of n months of membership.
- 10.Solve x² − x − 12 = 0.
- 11.Solve 2x + 7 = x + 13
- 12.Work out the value of a² − 2b when a = −3 and b = 4.
- 13.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.
- 14.The formula for converting a temperature from degrees Celsius, C, to degrees Fahrenheit, F, is F = 1.8C + 32. Work out F when C = 20, and identify what kind of mathematical statement F = 1.8C + 32 is.
- 15.Make p the subject of the formula n = p/3.
Answer key
- (b) x = 5 — Area = length × width, so x(x + 3) = 40, which rearranges to x² + 3x − 40 = 0. This factorises as (x + 8)(x − 5) = 0: the two numbers in the brackets must multiply to −40 and add to +3, and the pair 8 and −5 does both. This gives x = −8 or x = 5. Since x is a length, it cannot be negative, so x = 5. A candidate who gives both solutions without rejecting the negative one, which cannot be a length, answers x = 5 or x = −8. A candidate who picks the wrong factor pair of 40, such as 10 and −4 instead of 8 and −5, gets (x + 10)(x − 4) = 0 and answers x = 4. A candidate who rejects the wrong root, keeping the negative solution instead of the positive one, answers x = −8.
- (d) t = 5 — The maximum height occurs halfway between the two times when the ball is at ground level: the midpoint of t = 1 and t = 9 is (1 + 9) ÷ 2 = 5, so the ball reaches its maximum height at t = 5 seconds. Getting t = 4 comes from halving the DIFFERENCE between the times, 9 − 1 = 8, then 8 ÷ 2 = 4, instead of finding their midpoint. Getting t = 8 uses that difference, 9 − 1 = 8, as if the gap between the two times were itself the time of the maximum. Getting t = 10 adds the two times, 1 + 9 = 10, but forgets to divide by 2.
- (c) 5y − 2 — An expression is a collection of terms with no equals sign and no inequality sign in it: nothing is being claimed about two sides. Here 5y − 2 on its own fits that description. The line with an equals sign and one unknown letter to find is an equation; the line joined by a comparison sign is an inequality; the line that links two different letters so that one can be worked out from the other is a formula.
- (a) 7 — Multiply the input by 2: 6 × 2 = 12. Then subtract 5: 12 − 5 = 7. A candidate who does the operations in the wrong order, subtracting 5 first and then multiplying by 2, gets (6 − 5) × 2 = 2. A candidate who only carries out the multiplication and forgets to subtract gets 12. A candidate who adds 5 instead of subtracting gets 6 × 2 + 5 = 17.
- (c) 27 days — The campaign starts at d = 3 and ends at d = 30, so it runs for 30 − 3 = 27 days. Getting 33 days comes from adding the two values, 3 + 30 = 33, instead of subtracting them. Getting 30 days uses only the end day and ignores that the campaign did not start at day 0. Getting 24 days comes from subtracting the start day twice, 30 − 3 − 3 = 24, instead of once.
- (a) 16 — k = 8m ÷ 5 = (8 × 10) ÷ 5 = 80 ÷ 5 = 16. A candidate who multiplies 8 by 10 but forgets to divide by 5 gets 80. A candidate who divides 10 by 8 and multiplies by 5, using the fraction upside down, gets 10 × 5 ÷ 8 = 6.25. A candidate who adds instead of substituting into the formula gets 8 + 10 = 18.
- (c) 7 — To reverse the rule, subtract the constant then divide by the coefficient: 30−2=28, then 28÷4=7, so n=7. A candidate who adds the constant instead of subtracting it, a sign error when rearranging, would compute (30+2)÷4=32÷4=8. A candidate who subtracts the constant correctly but then forgets to divide by the coefficient would stop at 30−2=28. A candidate who treats 4n+2 as a single term 6n, adding the coefficient and constant together, would compute 30÷6=5.
- (b) b = a/5 — Writing 5b means 5 × b, so b has been multiplied by 5. The inverse of multiplying by 5 is dividing by 5, and dividing both sides by 5 gives a/5 = b, which is written b = a/5. Multiplying both sides by 5 instead gives 5a, which applies the operation a second time; treating the 5 as though it were added gives a − 5; writing 5/a turns the fraction upside down.
- (d) 7n − 2 — Method: find how much the total cost rises each month, then find the constant that fits the cost for one month. Working: the cost rises by £7 for each extra month (12 − 5 = 7, 19 − 12 = 7, 26 − 19 = 7), so the cost has the form 7n + c. Substituting n = 1: 7(1) + c = 5, so c = −2. Answer: the total cost in pounds is 7n − 2. The value 7n comes from leaving out the constant. The value 7n + 5 comes from using the cost of one month as the constant directly, without subtracting the monthly rise first. The value 5n + 7 comes from swapping the roles of the cost of one month, £5, and the monthly rise, £7 — using the cost of one month as the coefficient of n and the rise as the constant.
- (a) x = 4 or x = −3 — Method: find two numbers that multiply to give −12 and add to give −1 — these are −4 and 3. So x² − x − 12 = (x − 4)(x + 3) = 0, giving x = 4 or x = −3. Distractor origins: x = −4 or x = 3 has the signs the wrong way round; x = 4 or x = 3 makes both roots positive, ignoring the sign of −12; x = 12 or x = −1 comes from reading off the coefficient and the constant directly instead of factorising.
- (d) x = 6 — Method: with an unknown on both sides, first collect the x terms on one side by subtracting the smaller x term from both sides, then deal with the numbers. Working: subtracting x from both sides gives x + 7 = 13, and subtracting 7 from both sides gives x = 6. Answer: x = 6. The distractors: x = 20 comes from adding 7 to 13 instead of subtracting it once the x terms have been collected; x = 2 comes from collecting the x terms by adding them, giving 3x + 7 = 13 and then 3x = 6; x = −6 comes from subtracting 2x from both sides to get 7 = −x + 13, reaching −6 = −x and then copying the sign straight across instead of dividing by −1.
- (a) 1 — a² − 2b = (−3)² − 2(4) = 9 − 8 = 1. A candidate who squares −3 but keeps the negative sign gets −9 − 8 = −17. A candidate who forgets to square a and substitutes it as −3 gets −3 − 8 = −11. A candidate who adds instead of subtracting 2b gets 9 + 8 = 17.
- (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) 68, and it is a formula — Substituting C = 20 gives F = 1.8 × 20 + 32 = 36 + 32 = 68. The statement F = 1.8C + 32 is a formula because it expresses a general relationship between F and C that holds for any value of C, not just C = 20. A candidate who gets the correct value but calls it an equation is confusing "has an equals sign" with "is an equation" — an equation is only true for particular value(s) of the unknown, whereas this relationship is used to calculate rather than to be solved. A candidate who forgets to multiply C by 1.8 and just adds 32 to 20 gets 52. A candidate who multiplies correctly but forgets to add 32 gets 36.
- (b) p = 3n — n = p/3 means p has been divided by 3, so to make p the subject, multiply both sides by 3: p = 3n. Writing p = n/3 leaves the formula exactly as it was, without undoing the division. Writing p = n − 3 mistakes division for subtraction and takes 3 away from n instead of multiplying. Writing p = 3/n incorrectly flips the fraction upside down rather than multiplying n by 3. The correct rearrangement is p = 3n.
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