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 ferry's distance-time graph is a straight line from (0 minutes, 0 km) to (25 minutes, 15 km). Work out the ferry's speed in kilometres per hour.
- 2.A photo printing service has two adverts for its price. Advert A: cost in pounds = 3(2n + 4) for n photos. Advert B: cost in pounds = 6n + 12. A customer says the two adverts always charge the same amount. Is the customer correct?
- 3.A gas company charges a standing charge plus a rate per unit used. The total cost, y in pounds, for using x units is given by y = 0.15x + 20. Work out the cost of using 200 units.y = 0.15x + 20
- 4.Explain why the equation 4x + 3 = 4x + 10 has no solution for x.
- 5.Solve the inequality 6.3x + 1.8 ≤ 40. Hence write down the greatest number with exactly one decimal place that satisfies the inequality.
- 6.A taxi charges a fixed fee of £3 plus £2 for every mile travelled. The journey is m miles. Write down an expression for the total cost, in pounds.
- 7.A square tile has an area of 144 cm². Work out the side length of the tile.
- 8.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.
- 9.The graph of P = 480 ÷ V shows how the pressure, P pascals, of a gas depends on its volume, V cm³, at constant temperature. Use the relationship to work out the volume when the pressure is 32 pascals, and then work out by how much the volume increases when the pressure drops to 24 pascals.
- 10.A colony of bacteria doubles in number every hour. At 9am there are 5 bacteria in the colony. Work out how many bacteria there will be at 12 noon.
- 11.Amelia's mother is three times as old as Amelia. Four years ago her mother was five times as old as Amelia was then. Work out Amelia's age now.
- 12.A graph shows the total monthly cost of a mobile phone tariff against the number of minutes of calls. The line starts at £12 and stays level until 200 minutes, then rises by 5p for each further minute of calls. Use the graph to work out the total cost of a month in which 250 minutes of calls are made.
- 13.Write down the gradient of the line with equation y = 6 + 4x.
- 14.A phone company works out a monthly bill, £B, using the formula B = 18 + 0.05t, where £18 is the fixed monthly charge and t is the number of extra text messages sent beyond the free allowance, each charged at 5p. Farida's bill for one month is £24.50. Work out the number of extra text messages, t, she sent.
- 15.A ball is thrown in the air. Its height, h metres, above the ground after t seconds is given in this table: when t = 0, h = 0; when t = 1, h = 15; when t = 2, h = 20; when t = 3, h = 15; when t = 4, h = 0. Use the table to find the two times, in seconds, at which the ball is at ground level.
Answer key
- (a) 36 km/h — Speed is distance ÷ time. Convert 25 minutes to hours: 25 ÷ 60 = 5/12 hours. Then 15 ÷ (5/12) = 15 × 12/5 = 36 km/h. Dividing 15 by 25 without converting minutes to hours gives 0.6 km/h. Multiplying 15 by 25/60 instead of dividing by it gives 6.25 km/h. Working out the speed in km per minute, 0.6, and then multiplying by 24 instead of 60 gives 14.4 km/h.
- (a) They always charge the same, since 3(2n + 4) = 6n + 12. — Expand Advert A's formula by multiplying both terms inside the bracket by 3: 3 × 2n = 6n, and 3 × 4 = 12, giving 3(2n + 4) = 6n + 12, which is identical to Advert B's formula — so the two adverts always charge the same amount, whatever n is. Getting 6n + 4 comes from multiplying the 2n by 3 but leaving the 4 unmultiplied. Getting 2n + 7 comes from adding 3 to the bracket instead of multiplying by it. Saying it depends on n avoids expanding the bracket at all — once expanded, both formulas are identical for every value of n, so the cost can be compared directly.
- (a) £50 — Substituting x = 200 into y = 0.15x + 20 gives y = 0.15 × 200 + 20 = 30 + 20 = £50. A candidate who forgets to add the standing charge would get only 0.15 × 200 = £30. A candidate who misplaces the decimal point in the rate, using 1.5 instead of 0.15, would get 1.5 × 200 + 20 = £320. A candidate who swaps the roles of the rate and the number of units would work out 0.15 × 20 + 200 = £203.
- (a) Subtracting 4x gives 3 = 10, which is never true. — Method: try to solve the equation as normal and see what happens. Working: subtract 4x from both sides: 4x + 3 − 4x = 4x + 10 − 4x, giving 3 = 10. This statement is false for every value of x, so the equation has no solution. Answer: subtracting 4x gives 3 = 10, which is never true. "x would have to be negative" invents a constraint on x that the equation never states. "It's true for every x" confuses this equation with an identity, where both sides would simplify to the same expression. "x = 7" misreads the false statement 3 = 10 as something to solve for x, rather than recognising it means no solution exists.
- (d) 6.0 — Subtract 1.8 from both sides: 6.3x ≤ 38.2. Divide both sides by 6.3: x ≤ 6.0634… . The question asks for the greatest value to 1 decimal place that still satisfies the inequality. Testing 6.1: 6.3 × 6.1 + 1.8 = 40.23, which is more than 40, so 6.1 fails. Testing 6.0: 6.3 × 6.0 + 1.8 = 39.6, which is no more than 40, so 6.0 works and is the greatest such value. A candidate who simply rounds 6.0634… to 1 decimal place answers 6.1, without checking that it satisfies the inequality. A candidate who adds 1.8 instead of subtracting works out 41.8 ÷ 6.3 and answers 6.6. A candidate who forgets the 1.8 altogether divides 40 by 6.3 and answers 6.3.
- (b) 3 + 2m — Method: add the fixed fee to the cost of the miles travelled. Working: the miles cost 2 × m = 2m pounds, and the fixed fee is £3, so the total is 3 + 2m. Answer: 3 + 2m. 2 + 3m comes from swapping which number is the fixed fee and which is the per-mile rate. 5 + m comes from adding the fixed fee and the rate together into one number, 3 + 2 = 5, before adding the miles, instead of multiplying the rate by the number of miles. 5m comes from multiplying that combined total by the miles, treating the whole £5 as if it were a per-mile rate.
- (c) 12 cm — For a square, area = side². So side² = 144, giving side = ±12. Since a length must be positive, the side length is 12 cm. A candidate who gives both square roots without rejecting the negative one, which cannot be a length, answers 12 cm or −12 cm. A candidate who halves 144 instead of taking its square root gets 72 cm. A candidate who divides 144 by 4, confusing the area formula with a perimeter calculation, gets 36 cm.
- (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.
- (b) 5 cm³ — When P = 32, V = 480 ÷ 32 = 15 cm³. When P = 24, V = 480 ÷ 24 = 20 cm³. The increase in volume is 20 − 15 = 5 cm³. Giving only the second volume, 20 cm³, forgets to subtract the first volume. Giving only the first volume, 15 cm³, answers the wrong part of the question. Subtracting the two pressures instead of the two volumes, 32 − 24 = 8, mixes up which quantity the question asks for.
- (a) 40 — From 9am to 12 noon is 3 hours, so the population doubles three times: 5 × 2³ = 40. A candidate who counts the elapsed time as 4 hours (an off-by-one counting error) would reach 5 × 2⁴ = 80. A candidate who counts it as only 2 hours would reach 5 × 2² = 20. A candidate who misreads 'doubles' as 'increases by 2' each hour would compute 5 + 3 × 2 = 11.
- (d) 8 — Method: write both present ages in terms of one letter, take four years off each of them, and form an equation from the comparison four years ago. Working: let Amelia be y, so her mother is 3y; four years ago they were y − 4 and 3y − 4, giving 3y − 4 = 5(y − 4); expanding gives 3y − 4 = 5y − 20; adding 20 and subtracting 3y from both sides gives 16 = 2y, and dividing by 2 gives y = 8. Checking: Amelia is 8 and her mother 24; four years ago they were 4 and 20, and 20 = 5 × 4. Answer: 8. The distractors: 4 comes from solving correctly but giving Amelia's age four years ago instead of her age now; 16 comes from stopping at 2y = 16 and giving 16 as her age; 2 comes from collecting the y terms by adding the 5y instead of subtracting it, giving 3y + 5y = 20 − 4 and so 8y = 16.
- (a) £14.50 — Method: the first 200 minutes are already covered by the £12, so only 250 − 200 = 50 minutes are charged extra. Extra charge = 50 × £0.05 = £2.50. Total cost = £12 + £2.50 = £14.50. Distractor origins: £24.50 charges 5p for all 250 minutes instead of only the 50 minutes over 200 (250 × £0.05 = £12.50, plus £12 = £24.50); £12.50 makes that same slip of charging all 250 minutes but then forgets to add the £12 monthly fee; £13.50 works out 250 − 200 wrongly as 30 extra minutes instead of 50 (30 × £0.05 = £1.50, plus £12 = £13.50).
- (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.
- (a) 130 — Method: substitute the total bill into the formula, then subtract the fixed charge and divide by the cost per text message. Working: 24.50 = 18 + 0.05t, so 0.05t = 24.50 − 18 = 6.50, t = 6.50 ÷ 0.05 = 130. Answer: 130 extra text messages. 490 comes from dividing the whole bill by 0.05 without first subtracting the £18 fixed charge: 24.50 ÷ 0.05 = 490. 13 comes from dividing the £6.50 by 0.5 instead of 0.05, moving the decimal point one place too far: 6.50 ÷ 0.5 = 13. 65 comes from dividing the £6.50 by 0.1 instead of 0.05.
- (a) t = 0 or t = 4 — The ball is at ground level exactly when h = 0. From the table, h = 0 at t = 0 and at t = 4, so those are the two times. Distractor origins: t = 1 or t = 3 picks the times with equal (but non-zero) height instead of ground level; t = 2 picks the time of maximum height instead of ground level; t = 0 finds only the starting time and misses the second one.
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