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) (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) The 5 must multiply both terms inside the bracket, so 5(x + 2) expands to 5x + 10, which is never equal to 5x + 2 for any value of x. — Expanding the bracket correctly, 5(x + 2) = 5x + 10, since the 5 multiplies both the x and the 2. This is never equal to 5x + 2, since that would require 10 = 2. The option claiming 5(x + 2) means 5 × x + 2 ignores that the 5 must multiply the whole bracket, not just the x-term. The option about working out the bracket first with a value of x misunderstands algebraic expansion, which holds for every x, not just specific ones. The option about addition before multiplication misapplies the order of operations to bracket expansion, which always distributes the outer factor over every term inside, whatever x is.
- (c) 1, 7, 13, 19 — Method: substitute n = 1, 2, 3, 4 into the rule 6n − 5 in turn. Working: n = 1: 6 − 5 = 1. n = 2: 12 − 5 = 7. n = 3: 18 − 5 = 13. n = 4: 24 − 5 = 19. Answer: 1, 7, 13, 19. 6, 12, 18, 24 comes from using 6n on its own, forgetting to subtract 5. 5, 11, 17, 23 comes from using the rule 6n − 1 instead of 6n − 5, a slip in the constant. 0, 6, 12, 18 comes from using 6(n − 1) instead of 6n − 5, effectively shifting every term one position along.
- (d) x = 4 or x = −9 — Since (x − 4)(x + 9) = 0, either x − 4 = 0, giving x = 4, or x + 9 = 0, giving x = −9. A candidate who takes the sign shown in each bracket rather than its opposite gets x = −4 or x = 9. A candidate who makes both solutions negative gets x = −4 or x = −9. A candidate who drops both signs and makes both solutions positive gets x = 4 or x = 9.
- (b) 60 km/h — Method: the gradient of a distance-time graph is the change in distance divided by the change in time, and for a journey at a steady rate that gradient is the speed. Working: the change in distance is 195 − 15 = 180 km and the change in time is 3 hours, so the gradient is 180 ÷ 3 = 60 km/h. Answer: 60 km/h. The distractors: 180 km/h comes from stopping at the change in distance and never dividing by the 3 hours; 195 km/h comes from reading the final marker as the rate instead of working with the change between the two markers; 3 km/h comes from quoting the time taken, which belongs on the bottom of the fraction, as though it were the value of the fraction itself.
- (d) It is n-shaped, since the x² coefficient is negative. — The coefficient of x² is −2, which is negative, so the quadratic curve opens downward — shaped like an n, with a maximum turning point. Saying it is U-shaped focuses only on x² being non-negative and ignores that the −2 in front of it flips the whole curve to open downward. Saying it is a straight line confuses having a constant term with being linear — any equation with an x² term is a curve, not a line. Saying it repeatedly rises and falls like a wave describes a trigonometric graph such as y = sin x, not a quadratic.
- (c) y = 9x + 3 — Method: in a linear cost model the charge that is paid once is the constant term and the charge made for each item is the coefficient of x. Working: each pizza costs £9, so x pizzas cost 9x pounds; the £3 delivery fee is paid once whatever the value of x, so it is added on as a constant and the total cost is y = 9x + 3. Answer: y = 9x + 3. The distractors: y = 3x + 9 comes from swapping the two charges over, treating the £3 delivery as a per-pizza cost and the £9 as the one-off charge; y = 9x leaves out the delivery fee altogether, which undercharges every order by £3; y = 12x comes from adding the delivery fee to the price of each pizza, £9 + £3 = £12 each, so the fee is charged once per pizza instead of once per order.
- (c) 14 — Substitute b = 6 into 3b − 4: 3 × 6 − 4 = 18 − 4 = 14. 5 comes from treating 3b as 3 + b instead of 3 × b, giving 3 + 6 − 4. 6 comes from subtracting 4 from b before multiplying by 3, 3 × (6 − 4). 18 comes from working out 3 × 6 correctly but forgetting to subtract the 4.
- (d) 7a + 56 — Method: multiplying a single term over a bracket means every term inside the bracket is multiplied by the term outside. Working: 7 × a = 7a and 7 × 8 = 56, and the two products are added because the bracket contains an addition. Answer: 7a + 56. The distractors: 7a + 15 comes from adding 7 and 8 instead of multiplying them; 7a + 8 comes from multiplying only the first term inside the bracket and copying the 8 across unchanged; a + 56 comes from multiplying only the 8 and copying the a across unchanged.
- (a) y = −3x − 5 — Gradient = (−8 − 4) ÷ (1 − (−3)) = −12 ÷ 4 = −3. Using the point (1, −8): −8 = −3(1) + c, so c = −5, giving y = −3x − 5. A candidate who drops the negative sign on the gradient, using m = 3 instead, would then solve −8 = 3(1) + c to get c = −11, writing y = 3x − 11. A candidate who makes a sign error isolating c, writing c = 5 instead of −5, would write y = −3x + 5. A candidate who mixes up both mistakes — keeping the correct gradient but the wrong, positive value of c from the flipped-gradient calculation — would write y = −3x + 11.
- (d) 23 — Method: find the number added each time, then add it to the first term once for every step between the first term and the term wanted. Working: 5 − 2 = 3, 8 − 5 = 3 and so on, so 3 is added each time; the 8th term is seven steps on from the 1st term, so it is 2 + 7 × 3 = 2 + 21. Answer: 23, which agrees with counting on from the 5th term: 14, 17, 20 and then one step more. The distractors: 26 comes from adding the common difference eight times instead of seven, 2 + 8 × 3; 20 comes from stopping one term early, at the 7th term; 24 comes from multiplying the position by the common difference, 8 × 3, and ignoring the fact that the sequence starts at 2 rather than at 3.
- (a) 5 — Method: apply the machine's operation to the input, and treat zero as an input like any other. Working: the machine gives 0 + 5, and counting on 5 from zero leaves the 5 unchanged. Answer: 5. The distractors: 0 comes from assuming that an input of 0 must give an output of 0, which is true for a machine that multiplies but not for one that adds; −5 comes from subtracting 5 instead of adding it; 6 comes from treating the input as 1 rather than 0 and working out 1 + 5.
- (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.
- (c) 8x − 12 — Multiply each term inside the bracket by 4: 4 × 2x = 8x and 4 × (−3) = −12, so 4(2x − 3) = 8x − 12. A candidate who forgets to multiply the second term by 4 gets 8x − 3. A candidate who makes a sign error, treating 4 × (−3) as +12, gets 8x + 12. A candidate who adds 4 to the bracket instead of multiplying gets 2x + 1.
- (d) 3 — Method: add the constant term to both sides first, then divide by the coefficient of x. Working: 3x = 4 + 5 = 9; x = 9 ÷ 3 = 3. Answer: x = 3. −1/3 comes from a sign error when moving the 5, subtracting instead of adding: 3x = 4 − 5 = −1, then x = −1/3. 6 comes from subtracting the coefficient 3 instead of dividing by it: 9 − 3 = 6. 9 comes from correctly finding 3x = 9 but forgetting to divide by 3.
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