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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- (a) 7 — Method: substitute the given value of x into the rule and carry out the operation written in it. Working: y = x + 4 with x = 3 gives y = 3 + 4, an addition of two whole numbers. Answer: y = 7. The distractors: 12 comes from multiplying by 4 instead of adding 4; 4 comes from writing down the number in the rule and never using the input at all; 1 comes from subtracting, 4 − 3, instead of adding.
- (a) −7 — The coefficient of xy is the number multiplying that exact term, including its sign: in 4x²y − 7xy + 2, the term in xy is −7xy, so the coefficient is −7. Answering 7 correctly picks out the number but drops the minus sign that belongs to it. Answering 4 picks the coefficient of the x²y term instead of the xy term — x²y and xy are different terms because their powers of x differ. Answering −5 combines −7 with the constant term +2, as though the coefficient of xy included the number with no letter attached to it. The coefficient of xy is −7.
- (a) (−6, −3) — Reflecting in the y-axis changes the sign of the x-coordinate and keeps the y-coordinate the same: (−6, −3). (6, 3) comes from reflecting in the x-axis instead, which changes the sign of the y-coordinate. (−6, 3) comes from reflecting in both axes. (6, −3) comes from not applying the reflection at all.
- (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.
- (c) 5n + 7 — Method: find the common difference between the rows, then find the constant by adjusting the first row's total. Working: each row has 5 more seats than the last, so the coefficient of n is 5. The constant is the first row's total minus the common difference: 12 − 5 = 7. Answer: the nth term is 5n + 7. 5n + 12 comes from using row 1's total, 12, as the constant without subtracting the common difference. 5n + 2 comes from a slip in working out the constant, subtracting the common difference twice (12 − 5 − 5 = 2) instead of once. 7n + 5 comes from swapping the common difference and the constant.
- (a) 50 — Method: in ab², only the b is squared, so square b first, then multiply by a. Working: b² = 5² = 25, then a × b² = 2 × 25 = 50. Answer: 50. 100 comes from squaring the product ab instead of just b: (2 × 5)² = 100. 20 comes from squaring a instead of b: a² × b = 4 × 5 = 20. 10 comes from ignoring the square altogether and working out a × b = 2 × 5 = 10.
- (a) x = 2 — Method: undo the subtraction first, then undo the multiplication, doing the same to both sides each time. Working: adding 2 to both sides gives 2x = 4, and dividing both sides by 2 gives x = 2. Answer: x = 2. The distractors: x = 4 comes from stopping at 2x = 4 and writing 4 as the value of x; x = 0 comes from subtracting 2 from both sides instead of adding it, giving 2x = 0; x = 8 comes from multiplying 4 by 2 instead of dividing by 2.
- (c) Line D — Method: the steeper of two lines is the one whose gradient is greater, and the gradient of a line through two points is the change in y divided by the change in x. Working: line C rises 2 for a run of 4, so its gradient is 2 ÷ 4 = 1/2; the other line rises 2 for a run of 2, so its gradient is 2 ÷ 2 = 1; since 1 is greater than 1/2, it is line D that is the steeper. Answer: Line D. The distractors: Line C is chosen by comparing the x-coordinates and calling the line that reaches further along the x-axis the steeper one; They are equally steep comes from comparing only the y-coordinates, which are both 2, without dividing by the two different runs; There is not enough information comes from believing that a gradient can only be measured off a drawn graph, when two points on a line are enough to work it out.
- (c) 3x + 4y — Collect the x terms: 5x − 2x = 3x. Collect the y terms: 3y + y = 4y. So 5x + 3y − 2x + y = 3x + 4y. A candidate who subtracts the y terms instead of adding them (3y − y) gets 3x + 2y. A candidate who adds 2x instead of subtracting it (5x + 2x) gets 7x + 4y. A candidate who wrongly combines the x and y terms into a single term gets 6xy.
- (d) y = 4 — Method: substitute x = 6 into y = 24/x: y = 24 ÷ 6 = 4. Distractor origins: y = 18 subtracts instead of dividing (24 − 6 = 18); y = 144 multiplies instead of dividing (24 × 6 = 144); y = 30 adds instead of dividing (24 + 6 = 30).
- (b) y decreases towards zero but never reaches it — Method: as x gets larger, dividing 20 by a bigger number gives a smaller result, so y decreases; because 20/x can never be exactly zero for any positive x, the curve gets closer to zero without ever reaching it, so y decreases towards zero but never reaches it. Distractor origins: 'y increases towards a limit but never reaches it' has the relationship backwards, treating y as increasing when it is actually decreasing; 'y decreases at a steady rate and reaches zero' wrongly assumes the graph behaves like a straight line that eventually hits zero; 'y stays the same however large x becomes' wrongly assumes there is no change in y at all.
- (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.
- (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.
- (c) x = 2 and x = 3 — x² − 5x + 6 factorises as (x − 2)(x − 3), since −2 and −3 multiply to give 6 and add to give −5. The graph crosses the x-axis where each bracket equals zero: x − 2 = 0 gives x = 2, and x − 3 = 0 gives x = 3. The option x = −2 and x = −3 comes from reading the signs inside the brackets directly instead of solving x − 2 = 0 and x − 3 = 0. The option x = 2 and x = −3 mixes up the sign of only one root. The option x = −1 and x = −6 comes from picking the wrong pair of factors of 6 (1 and 6 instead of 2 and 3) and then reading their signs directly from the brackets.
- (c) x ≤ 4 — Subtract 9 from both sides: −2x ≥ 1 − 9 = −8. Divide both sides by −2, flipping the inequality: x ≤ 4. A candidate who divides by −2 but forgets to flip the inequality gets x ≥ 4. A candidate who divides −8 by −2 but keeps a negative sign gets x ≤ −4. A candidate who miscalculates 1 − 9 as −10 gets x ≤ 5.
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