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.Harry will spend at most £150 on a party. The cake costs £60 and each helium balloon costs £3. Solve an inequality to find all the possible numbers of balloons, x, that he can buy.
- 2.Two numbers have a sum of 24. Their difference is 6. Work out the two numbers.
- 3.The point (−4, 7) is moved 4 units to the right and 9 units down. Write down the coordinates of the point it reaches.
- 4.A straight line has equation y = 4 − 3x. Work out the gradient of the line.
- 5.Simplify (5x² + 3x − 2) − (2x² − x + 5)
- 6.Work out the value of (x − 4)/2 + 3 when x = 10
- 7.Solve x² − 100 = 0.
- 8.Write down the coordinates of the point where the line y = 2x + 4 crosses the y-axis.y = 2x + 4
- 9.A box holds n pencils. A shop has 6 full boxes and 4 loose pencils. All of these pencils are shared equally between 2 classes. Write down the expression for the number of pencils each class receives.
- 10.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.
- 11.A rectangle is x cm wide and twice as long as it is wide, so its area A cm² is given by A = 2x². Work out A when x = 3.
- 12.The graph of t = 240 ÷ s shows how the time, t hours, taken for a car journey of 240 miles depends on its average speed, s mph. Use the relationship to work out the time taken when the average speed is 48 mph.
- 13.A sequence has the position-to-term rule: the term is the square of the position number. Its first four terms are 1, 4, 9, 16. Work out the 10th term.
- 14.A rule turns each input x into an output y. An input of 1 gives an output of 1, an input of 2 gives an output of 4 and an input of 3 gives an output of 9. Work out the rule.
- 15.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.
Answer key
- (d) x ≤ 30 — Method: add the fixed cost to the cost of x balloons, set that total against the £150 limit with the sign that 'at most' calls for, then solve. Working: the total spend is 60 + 3x pounds, so 60 + 3x ≤ 150; subtracting 60 from both sides gives 3x ≤ 90; dividing both sides by 3, a positive number, gives x ≤ 30. Answer: x ≤ 30. The distractors: x ≤ 90 comes from taking the cake off the budget and stopping at 3x ≤ 90, reading the 90 as a number of balloons when it is the money left for them; x ≤ 50 comes from dividing the whole £150 by 3 and leaving the cake out of the calculation altogether; x ≥ 30 comes from reading 'at most' as 'at least', which reverses the condition.
- (c) 15 and 9 — Method: write the two facts as two equations in the same pair of letters and add them, because the letter with opposite signs cancels. Working: with x the larger number and y the smaller, x + y = 24 and x − y = 6; adding gives 2x = 30, so x = 15, and substituting into x + y = 24 gives y = 9. Answer: 15 and 9, which add to 24 and differ by 6. The distractors: 18 and 6 come from halving 24 to 12 and then adding and subtracting the whole difference of 6 instead of half of it, which leaves a difference of 12; 15 and 21 come from finding the larger number correctly and then adding 6 to it instead of subtracting; 15 and 6 come from finding the larger number and then taking the 6 in the question to be the smaller number itself.
- (b) (0, −2) — Method: a translation acts on the two coordinates separately: moving right or left changes the x-coordinate only, moving up or down changes the y-coordinate only, and right and up add while left and down subtract. Working: the point starts at (−4, 7); moving 4 units to the right gives an x-coordinate of −4 + 4 = 0; moving 9 units down gives a y-coordinate of 7 − 9 = −2. Answer: (0, −2). The distractors: (5, 3) comes from pairing each number with the wrong coordinate, adding 9 to −4 and taking 4 from 7; (0, 16) comes from treating 'down' as an addition, giving 7 + 9 = 16 for the second coordinate; (−8, −2) comes from treating 'to the right' as a subtraction, giving −4 − 4 = −8 for the first coordinate.
- (d) −3 — Method: rewrite the equation in the form y = mx + c, then read off the gradient. Working: y = 4 − 3x can be written as y = −3x + 4, so comparing with y = mx + c gives m = −3. Answer: the gradient is −3. The value 3 comes from ignoring the negative sign on the x term. The value 4 comes from reading off the y-intercept instead of the gradient. The value −4 comes from a sign error, applying the negative sign to the intercept instead of the gradient.
- (d) 3x² + 4x − 7 — Method: the minus sign in front of the second bracket changes the sign of every term inside it; then collect like terms. Working: removing the brackets gives 5x² + 3x − 2 − 2x² + x − 5; the squared terms give 5x² − 2x² = 3x², the x terms give 3x + x = 4x, and the number terms give −2 − 5 = −7. Answer: 3x² + 4x − 7. The distractors: 7x² + 2x + 3 comes from adding the two brackets instead of subtracting, giving 5x² + 2x², 3x − x and −2 + 5; 3x² + 2x + 3 comes from applying the minus sign to 2x² only, leaving −x and +5 unchanged so that 3x − x = 2x and −2 + 5 = 3; 3x² + 4x + 3 comes from changing the signs of the terms with letters but leaving +5 as it stood, so the number terms give −2 + 5 = 3.
- (c) 6 — Method: substitute the value, work out the top of the fraction first, then the division, and add the 3 last. Working: the top gives 10 − 4 = 6, dividing by 2 gives 6 ÷ 2 = 3, and adding 3 gives 3 + 3 = 6. Answer: 6. The distractors: 11 comes from dividing only the 4 by 2 instead of the whole of the top, giving 10 − 2 + 3; 4.5 comes from dividing the + 3 by 2 as well, giving (10 − 4 + 3) ÷ 2; 0 comes from subtracting the wrong way round on the top, giving (4 − 10) ÷ 2 = −3 and then −3 + 3.
- (a) x = 10 or x = −10 — Rearranging, x² = 100. Taking the square root of both sides gives x = ±10, i.e. x = 10 or x = −10. A candidate who forgets the negative root gives only x = 10. A candidate who halves 100 instead of taking its square root gets x = 50. A candidate who applies the ± sign to 100 itself instead of to its square root gets x = 100 or x = −100.
- (d) (0, 4) — Method: in the form y = mx + c the constant c is the y-coordinate of the point where the line crosses the y-axis, and every point on that axis has x-coordinate 0. Working: comparing y = 2x + 4 with y = mx + c gives c = 4, and substituting x = 0 confirms y = 2 × 0 + 4 = 4, so the crossing point is (0, 4). Answer: (0, 4). The distractors: (4, 0) is the same pair of numbers written in the wrong order and is a point on the x-axis; (0, 2) comes from reading the gradient 2 as the intercept, confusing m with c; (−2, 0) is where the line crosses the x-axis, found by solving 0 = 2x + 4 instead.
- (c) (6n + 4)/2 — Six full boxes hold 6 lots of n pencils, which is 6n, and the 4 loose pencils are added on, so the shop has 6n + 4 pencils altogether. Sharing them equally between 2 classes divides that whole total by 2, and brackets are what show that the division applies to all of it: (6n + 4)/2. Without the brackets, 6n + 4/2 halves only the loose pencils; 6(n + 4)/2 adds the loose pencils to every box before the division; 2(6n + 4) doubles the total instead of halving it.
- (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.
- (a) 18 cm² — Method: substitute the width into the formula, applying the index to the letter before multiplying by the 2 in front of it. Working: x² = 3 × 3 = 9, and then A = 2 × 9 = 18, so the area is 18 cm². Answer: 18 cm². The distractors: 36 cm² comes from multiplying 2 by 3 first and squaring afterwards, giving (2 × 3)²; 12 cm² comes from doubling instead of squaring, so that x² is replaced by 2x and the calculation becomes 2 × 2 × 3; 6 cm² comes from working out 2 × 3 and never applying the index at all.
- (c) 5 hours — Substitute s = 48 into the relationship: t = 240 ÷ 48 = 5 hours. Inverting the relationship, s ÷ 240 = 48 ÷ 240 = 0.2, gives 0.2 hours. Misplacing the decimal point in the division, 240 ÷ 48 read as 4.8, gives 4.8 hours. Halving the speed to 24 mph by mistake before dividing, 240 ÷ 24 = 10, gives 10 hours.
- (c) 100 — Method: apply the position-to-term rule straight to the position asked for, since it does not depend on the term before it. Working: the rule squares the position number and the position is 10, so the term is 10², which means 10 × 10. Answer: 100. The distractors: 81 comes from squaring 9 instead of 10, one position short; 20 comes from multiplying the position by 2 instead of squaring it; 1000 comes from cubing the position, 10 × 10 × 10, instead of squaring it.
- (c) y = x² — Method: test a candidate rule against every pair given, not just one — a rule that fits one pair and fails another is not the rule. Working: the outputs 1, 4, 9 rise by 3 and then by 5, so they are not going up in equal steps and the input is not simply multiplied by a fixed number; comparing each output with its own input gives 1 × 1 = 1, 2 × 2 = 4 and 3 × 3 = 9, and all three pairs fit. Answer: y = x². The distractors: y = 3x comes from fitting only the last pair, where 3 × 3 = 9, and reading that 3 as a multiplier; y = 3x − 2 comes from assuming a multiply-then-add rule and using the first step in the outputs, 4 − 1 = 3, as the multiplier — it fits the first two pairs and fails the third; y = 2x comes from fitting only the pair 2 and 4 and reading every output as double its input.
- (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).
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