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) 6 — Method: rearrange the formula to make d the subject, then substitute C = 133. Working: C = 25 + 18d, so subtracting 25 from both sides gives C − 25 = 18d, then dividing by 18 gives d = (C − 25) / 18. Substituting C = 133: d = (133 − 25) / 18 = 108 / 18 = 6. The value 7.4 comes from dividing 133 by 18 without subtracting the fixed £25 first (133 / 18 ≈ 7.4). The value 4.6 comes from pairing the numbers the wrong way round, working out (133 − 18) / 25 = 4.6. The value 90 comes from subtracting both 25 and 18 from 133 instead of dividing by 18.
- (d) −4 — The coefficient of a letter is the number multiplying it, taken with the sign written in front of that number. The term containing p is being subtracted, so the term is −4p and the number multiplying p is −4. Quoting 4 drops the sign; 7 is a constant term with no letter attached to it; 9 is the number multiplying q, which is a different letter.
- (a) 2x + 24 — Method: multiply each term inside the bracket by the 2 in front of it. Working: 2 × x = 2x and 2 × 12 = 24, and the products are added because the bracket contains an addition. Answer: 2x + 24. The distractors: 2x + 12 comes from multiplying only the x by 2 and copying the 12 across unchanged; 24x comes from multiplying 2 by 12 and then attaching the letter to that product, as though the two terms inside the bracket could be joined into one; 2x + 14 comes from adding 2 and 12 instead of multiplying them.
- (a) (2, 7) — Midpoint = ((x1+x2)/2, (y1+y2)/2) = ((−3+7)/2, (5+9)/2) = (4/2, 14/2) = (2, 7). (4, 14) comes from adding the coordinates correctly but forgetting to divide by 2. (2, 9) comes from correctly averaging the x-coordinates but simply copying the y-coordinate of the second point instead of averaging the y-coordinates. (5, 2) comes from subtracting the coordinates instead of adding them before halving: ((7−(−3))/2, (9−5)/2) = (5, 2).
- (b) 8 s — Rearranging v = u + at for t: subtract u from both sides to get v − u = at, then divide by a: t = (v − u)/a. Substituting u = 4, a = 2, v = 20: t = (20 − 4)/2 = 16/2 = 8 s. Answering 12 s comes from adding u instead of subtracting it: (20 + 4)/2 = 12. Answering 32 s comes from multiplying (v − u) by a instead of dividing by it: 16 × 2 = 32. Answering 6 s divides v by a first and then subtracts u, in the wrong order: 20/2 − 4 = 10 − 4 = 6. The time taken is 8 s.
- (a) 3y² — 'y squared, multiplied by 3' means the square is applied to y only, and the result is then multiplied by 3, written as 3y². Writing y³ mistakes the multiplication by 3 for an extra factor of y, adding to the power instead of using a coefficient. Writing (3y)² squares the whole of 3y, including the 3, which gives 9y² rather than 3y² — the square should apply to y alone. Writing 3 + y² adds the 3 instead of multiplying by it. The expression for 'y squared, multiplied by 3' is 3y².
- (c) r = √(A/π) — A = πr² means r has been squared and then multiplied by π. To make r the subject, first divide both sides by π to get A/π = r², then take the square root of both sides: r = √(A/π). Writing r = A/π stops after dividing by π and forgets that r is still squared — it never undoes the square. Writing r = √A/π takes the square root before dividing by π, which square-roots only the A and not the whole of A/π. Writing r = (A/π)² squares A/π instead of taking its square root — the opposite of what is needed to undo r². The correct rearrangement is r = √(A/π).
- (d) Isla is right, because 5 + 1 = 6 — Method: put the input through the machine yourself, then compare what comes out with the output claimed, and check that the reason given with the verdict is itself true. Working: the machine adds 1 to whatever is put in, so an input of 5 gives 5 + 1 = 6, which is exactly the output claimed. Answer: Isla is right, and the reason is that 5 + 1 = 6. The distractors: the statement that the machine adds 1 to its output describes a different machine — this one adds 1 to what goes in — so the verdict is right but the reason is false; 5 − 1 = 4 comes from running the machine backwards and subtracting instead of adding; 5 × 1 = 5 comes from reading ‘adds 1’ as ‘multiplies by 1’, which leaves the input unchanged.
- (b) x = (y − 3)/5 — To make x the subject of y = 5x + 3, first subtract 3 from both sides to get y − 3 = 5x, then divide both sides by 5: x = (y − 3)/5. Writing x = (y + 3)/5 keeps the division correct but does not change the sign of the 3 when moving it across. Writing x = y/5 − 3 divides only the y term by 5 and leaves the 3 as a separate subtraction, instead of subtracting first and dividing the whole expression. Writing x = 5(y − 3) applies the correct order of subtracting 3 first, but then multiplies by 5 instead of dividing — the inverse of 5x is division, not multiplication. The correct rearrangement is x = (y − 3)/5.
- (c) 2x + 12 — Multiply every term inside the bracket by 2: 2 × x = 2x and 2 × 6 = 12, so 2(x + 6) = 2x + 12. Writing 2x + 6 comes from multiplying only the x-term and leaving the 6 unchanged. Writing x + 12 comes from multiplying only the 6 and leaving the x unchanged. Writing 2x + 8 comes from adding 2 and 6 instead of multiplying them.
- (c) x = 0 or x = 3 — Method: take out the common factor 5x: 5x(x − 3) = 0, so 5x = 0 or x − 3 = 0, giving x = 0 or x = 3. Distractor origins: x = 0 or x = 15 comes from forgetting to divide the second term by the common factor 5x correctly, leaving 15 instead of 3; x = 3 loses the solution x = 0 by dividing both sides by x; x = 5 or x = 3 mistakes the coefficient 5 itself for one of the solutions.
- (c) (−4, 7) — Left of the origin means negative x, and up means positive y, so the point is (−4, 7). (4, 7) comes from forgetting that 'left' means the x-coordinate is negative. (−4, −7) comes from treating 'up' as a negative direction instead of positive. (7, −4) comes from swapping the x- and y-coordinates.
- (c) (2, 6) — Method: substitute the x-coordinate of each point into the rule for the line (y = 3 × x) and compare it with the point's y-coordinate. Working: the line passes through the origin and rises 3 squares for every 1 square across, so at x = 2 the line's y-value is 3 × 2 = 6, giving the point (2, 6). Answer: (2, 6). Distractor refutation: (2, 3) comes from counting only 3 squares up in total between the origin and x = 2, instead of 3 squares up for every 1 square across, halving the true rise. (3, 2) comes from swapping the x-coordinate and the y-coordinate round. (2, 5) comes from a miscounted gridline, landing one square below the line.
- (c) 8a + 2 — The perimeter of a rectangle is twice the length plus twice the width: P = 2(3a + 2) + 2(a − 1) = (6a + 4) + (2a − 2) = 8a + 2. Answering 5a adds the length and width once each but doubles only one of them, missing that a rectangle has two of each side. Answering 8a + 6 distributes the 2 into (a − 1) correctly as far as 2a, but then adds 2 instead of subtracting it, as though the bracket had been (a + 1). Answering 12a + 8 uses the length for all four sides instead of using the length twice and the width twice, as if the field were a square with side (3a + 2). The perimeter of the field is (8a + 2) metres.
- (d) 15b — Method: each balloon costs 15p, so b balloons cost 15 × b pence. Working: 15 × b = 15b. Answer: 15b. b + 15 comes from treating the 15p as a one-off addition rather than a rate applied to every balloon. b − 15 comes from misreading the situation as a discount rather than a cost per balloon. b/15 comes from dividing instead of multiplying, swapping the two operations.
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