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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- (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) 4 — Method: subtract 3 from both sides, then divide by 4. Working: 4x = 19 − 3 = 16, so x = 16 ÷ 4 = 4. Answer: 4. 4.75 comes from dividing 19 by 4 directly, without subtracting 3 first. 3.75 comes from subtracting 4 instead of 3 before dividing. 5.5 comes from a sign error, adding 3 to 19 instead of subtracting it.
- (c) 4 — Method: the distance of a point from the x-axis is measured vertically, so it is the size of the y-coordinate taken without its sign. Working: the point (−6, 4) has y-coordinate 4, so moving straight down to the x-axis covers 4 units, and a distance is written as a positive number. Answer: 4. The distractors: 6 comes from using the x-coordinate, which measures the distance from the y-axis rather than from the x-axis; −6 comes from that same mistake with the minus sign left in place, although a distance is never negative; 10 comes from adding the two distances, 6 and 4, as though the question asked how far the point is from both axes together.
- (b) x = 4 — Method: every point on a vertical line has the same x-coordinate, so its equation sets x equal to a number, and that number is the x-coordinate of any point known to be on the line. Working: the line is vertical and passes through (4, 7), whose x-coordinate is 4, so every point on the line has x-coordinate 4 and the equation is x = 4. Answer: x = 4. The distractors: y = 7 is the horizontal line through the same point, obtained by fixing the y-coordinate instead of the x-coordinate; x = 7 uses the right form but takes the y-coordinate of the point rather than its x-coordinate; y = 4x treats the 4 as a gradient and gives a sloping line through the origin, which does not pass through (4, 7) at all.
- (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.
- (d) x = 1 — A quadratic graph is symmetrical about its turning point, so the turning point's x-coordinate is the midpoint of the two roots: (−2 + 4) ÷ 2 = 2 ÷ 2 = 1. The option x = 2 comes from adding the two roots but forgetting to divide by 2. The option x = −1 makes a sign error when adding the roots, treating −2 + 4 as −2. The option x = 3 comes from subtracting the roots instead of adding them: (4 − (−2)) ÷ 2 = 3.
- (c) h = V / (lw) — Method: undo the multiplication by lw by dividing both sides by lw. Working: V = lwh, so dividing both sides by lw gives h = V / (lw). The value h = V − lw comes from subtracting lw instead of dividing by it. The value h = Vlw comes from multiplying by lw instead of dividing. The value h = lw / V comes from inverting the fraction, dividing lw by V instead of V by lw.
- (c) S = P + C — Method: S has C subtracted from it, so undo that subtraction by adding C to both sides. Working: P + C = S − C + C, so P + C = S, which is written S = P + C. Answer: S = P + C. S = P − C repeats the formula's own operation instead of its inverse. S = C − P subtracts the wrong way round. S = PC multiplies the two given quantities instead of adding them.
- (c) 5(x + y − 1) — Method: take out the highest common factor of all three terms and divide every term by it, the number term included. Working: the highest common factor of 5x, 5y and −5 is 5; dividing gives 5x ÷ 5 = x, 5y ÷ 5 = y and −5 ÷ 5 = −1, so the bracket holds x + y − 1. Answer: 5(x + y − 1), which multiplies back out to 5x + 5y − 5. The distractors: 5(x + y + 1) comes from dividing −5 by 5 and losing the minus sign; 5(x + y − 5) comes from dividing only the terms containing a letter by 5 and carrying the −5 into the bracket unchanged; 5(xy − 1) comes from collecting the unlike terms 5x and 5y as 5xy before factorising.
- (c) (3, 9) — Reflecting in the horizontal line y = 2 keeps the x-coordinate the same and maps y to 2 × 2 − y = 4 − (−5) = 9, so S = (3, 9). A candidate who uses k − y instead of 2k − y gets 2 − (−5) = 7, giving (3, 7). A candidate who reflects in the x-axis instead of the line y = 2, simply changing the sign of y, gets (3, 5). A candidate who also changes the sign of the x-coordinate, as if reflecting in both axes, gets (−3, 9).
- (b) 20 — F = 4s, so when s = 5, F = 4 × 5 = 20. 9 comes from adding 4 and 5 instead of multiplying, the mistake described in the question. 25 comes from squaring s (5²) instead of multiplying by 4. 45 comes from writing the digits 4 and 5 next to each other instead of carrying out the multiplication.
- (c) y = −x + 30 — Method: in a linear model the value at the start is the constant term and the steady rate of change is the gradient, which is negative when the quantity is falling. Working: at x = 0 the candle is 30 cm tall, so the constant term is 30; it loses 1 cm every hour, so the gradient is −1 and the height after x hours is y = −x + 30. Answer: y = −x + 30. The distractors: y = x + 30 comes from taking the rate as +1 and making the candle grow rather than shrink; y = 30x − 1 comes from building the model correctly as 30 − 1x and then copying it into the form y = mx + c with the two numbers left where they stood, so the starting height 30 ends up multiplying x and the hourly 1 is left behind as the constant being taken away; y = −30x + 1 comes from reading the two numbers the other way round, taking 30 cm per hour as the rate and 1 cm as the starting height, which burns 30 cm an hour from a candle only 1 cm tall.
- (b) 8 kg — The cost above the flat £5 charge is 17 − 5 = £12. At £2 per kg, this covers 12 ÷ 2 = 6 kg above the first 2 kg, so the total weight is 2 + 6 = 8 kg. Dividing the full £17 by £2 per kg without first taking off the £5 flat charge gives 17 ÷ 2 = 8.5 kg. Taking off the £5 flat charge and dividing by £2 per kg, but forgetting to add back the 2 kg that the flat charge covers, gives 12 ÷ 2 = 6 kg. Taking off £2 instead of £5 as the flat charge, (17 − 2) ÷ 2 = 7.5 kg, swaps which number is the fixed fee.
- (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.
- (a) 17 — Substitute a = 3 into 5a + 2: 5 × 3 + 2 = 15 + 2 = 17. 10 comes from treating 5a as 5 + a instead of 5 × a, giving 5 + 3 + 2. 25 comes from adding the 2 to a before multiplying by 5, 5 × (3 + 2). 15 comes from working out 5 × 3 correctly but forgetting to add the 2.
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