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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Answer key: Algebra worksheet — GCSE Foundation
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- (d) x = 9 or x = −9 — Taking the square root of both sides of x² = 81 gives x = ±9, i.e. x = 9 or x = −9, since both 9² and (−9)² equal 81. A candidate who forgets the negative square root gives only x = 9. A candidate who halves 81 instead of taking its square root gets x = 40.5. A candidate who squares 81 instead of taking its square root gets x = 6561.
- (c) x = 0 or x = −7 — Factorising: x² + 7x = x(x + 7) = 0, so x = 0 or x + 7 = 0, giving x = 0 or x = −7. A candidate who divides both sides of the original equation by x, which loses the solution x = 0, gets only x = −7. A candidate who makes a sign error solving x + 7 = 0 gets x = 0 or x = 7. A candidate who misreads the coefficient and doubles it gets x = 0 or x = −14.
- (a) x = 4, y = −3 — Method: both equations contain +y with the same coefficient, so subtracting one equation from the other removes y. Working: (2x + y) − (x + y) = 5 − 1 gives x = 4; substituting x = 4 into x + y = 1 gives 4 + y = 1, so y = −3. Answer: x = 4, y = −3, which also satisfies 8 − 3 = 5. The distractors: x = 4, y = 5 comes from rearranging x + y = 1 as y = 1 + x; x = −2, y = 3 comes from eliminating x by doubling the first equation and then reading −y = 3 as y = 3; x = 6, y = −5 comes from adding the constants instead of subtracting them while eliminating y, taking x as 5 + 1.
- (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) £195 — Substituting x = 3 into y = 45x + 60 gives y = 45 × 3 + 60 = 135 + 60 = 195. A candidate who forgets to add the call-out fee would get only 45 × 3 = £135. A candidate who adds the hours to the fee and the rate instead of multiplying would get 45 + 60 + 3 = £108. A candidate who multiplies both the hourly rate and the call-out fee by the number of hours would get 45 × 3 + 60 × 3 = £315.
- (b) 3 — Distance from the y-axis is given by the size of the x-coordinate, which is 3. 8 comes from using the size of the y-coordinate instead of the x-coordinate. −8 comes from using the y-coordinate and keeping its negative sign, but a distance is never negative. 11 comes from adding the sizes of both coordinates, 3 + 8, instead of using the x-coordinate alone.
- (a) 3x — Method: "three times as old" is a multiplication by 3, applied to the age that is known in terms of the letter. Working: the brother's age is x, and Leo's age is 3 × x, which is written as 3x. Answer: 3x. The distractors: x + 3 comes from reading "three times as old" as "three years older" and adding instead of multiplying; x − 3 comes from reading it as "three years younger" and subtracting; x/3 comes from dividing by 3 instead of multiplying, which describes the brother in terms of Leo rather than Leo in terms of the brother.
- (c) y = 3x + 2 — Method: divide the change in the outputs by the change in the inputs to find the multiplier, then put one pair of values into the rule to find the number added on. Working: the output rises by 11 − 5 = 6 while the input rises by 3 − 1 = 2, so the multiplier is 6 ÷ 2 = 3; with an input of 1, 3 × 1 = 3 and the output is 5, so 2 is added. Answer: y = 3x + 2, checked against the second pair by 3 × 3 + 2 = 11. The distractors: y = 3x − 2 comes from finding the multiplier 3 and then subtracting the 2 instead of adding it; y = 2x + 3 comes from swapping the multiplier and the number added on; y = x + 4 comes from assuming the input is multiplied by 1 and using 5 − 1 = 4 as the number added on.
- (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.
- (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) The graph never crosses either axis — Since x ≠ 0, there is no point on the graph where x = 0, so it cannot cross the y-axis; likewise 1/x is never equal to 0 for any x, so it cannot cross the x-axis either — the graph never touches either axis. A candidate who forgets the restriction x ≠ 0 might think the graph behaves like other graphs and passes through the origin, (0, 0). A candidate who correctly rules out the x-axis but forgets that x = 0 is also excluded might say the graph crosses the y-axis but never the x-axis. A candidate who only pictures the branch where x and y are both positive might say the graph has only one branch, in quadrant 1, forgetting the second branch where x and y are both negative.
- (c) 2(x + 4) ≡ 2x + 8 — 2(x + 4) ≡ 2x + 8 is an identity because expanding the bracket on the left gives exactly the right-hand side for every value of x — try any number and both sides match. 2(x + 4) = 20 is an equation: expanding gives 2x + 8 = 20, which is only true for the single value x = 6. 2(x + 4) = 2x + 4 is not true for any value of x at all: expanding the left side gives 2x + 8, which can never equal 2x + 4 since 8 ≠ 4. x + 4 = 2x is also an equation, true only for the single value x = 4. The identity is 2(x + 4) ≡ 2x + 8.
- (d) a/2 — Method: write both terms over the same denominator, subtract the numerators, then cancel the fraction down. Working: 6 is a multiple of 3, so a/3 is rewritten as 2a/6; the calculation becomes 5a/6 − 2a/6 = 3a/6, and dividing numerator and denominator by 3 gives a/2. Answer: a/2. The distractors: 4a/3 comes from subtracting the denominators as well as the numerators, giving (5 − 1)a over (6 − 3); 2a/3 comes from subtracting 1 from 5 without first rewriting a/3 as 2a/6, giving 4a/6; 7a/6 comes from adding the two fractions instead of subtracting them, giving 5a/6 + 2a/6.
- (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) 5x² — A term is one part of an expression, separated from the others by a + or − sign. The expression 5x² − 3x + 7 has three terms: 5x², −3x and 7. Distractor origins: 5x² − 3x combines two terms together, not a single term; −3x + 7 also combines two terms together; 5x² − 3x + 7 gives the whole expression, not a single term.
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