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.
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.
- (b) −1 — Method: in the form y = mx + c the gradient m is the number multiplying x, and an x term written with no number in front of it has a coefficient of 1. Working: y = −x + 3 is the same equation as y = (−1)x + 3, so comparing it with y = mx + c gives m = −1 and c = 3. Answer: −1. The distractors: 1 comes from taking the coefficient as 1 and leaving the minus sign out of the answer; 3 comes from reading the constant as the gradient, confusing m with c; −3 comes from moving the minus sign across to the constant term and then reading the gradient off that term instead of the x term.
- (d) (−1, 4) — In parallelogram ABCD the side DC is parallel and equal to the side AB, so D = C − AB. The vector from A to B is (2 − (−3), 1 − 1) = (5, 0), so D = (4 − 5, 4 − 0) = (−1, 4). A candidate who adds this vector to C instead of subtracting it gets (4 + 5, 4 + 0) = (9, 4). A candidate who subtracts A's coordinates from C's rather than the vector AB, and drops the minus sign on −3 while doing so, works out (4 − 3, 4 − 1) and gets (1, 3). A candidate who makes only the y-part of that slip, working out 4 − 1 instead of 4 − 0, gets (−1, 3).
- (d) x² − 49 — Area = length × width = (x + 7)(x − 7). Expanding: x × x = x², x × (−7) = −7x, 7 × x = 7x, 7 × (−7) = −49. The two middle terms −7x and 7x cancel, leaving x² − 49. A candidate who misremembers the difference-of-two-squares result as a sum gets x² + 49. A candidate who makes a sign error and treats both middle terms as −7x instead of cancelling gets x² − 14x − 49. A candidate who confuses area with perimeter and simply adds the length and width gets 2x.
- (b) x ≥ 5 — Expand the bracket: 2(3x − 1) = 6x − 2, so the inequality is 6x − 2 ≥ 4x + 8. Subtract 4x from both sides and add 2 to both sides: 2x ≥ 10. Divide both sides by 2: x ≥ 5. A candidate who only multiplies the 3x by 2 and forgets to multiply the −1 gets 6x − 1 ≥ 4x + 8, leading to x ≥ 4.5. A candidate who adds 4x instead of subtracting it gets 10x ≥ 10, leading to x ≥ 1. A candidate who multiplies by 2 instead of dividing gets x ≥ 20.
- (c) 100 — Method: apply the position-to-term rule straight to the position asked for, since it does not depend on the term before it. Working: the rule squares the position number and the position is 10, so the term is 10², which means 10 × 10. Answer: 100. The distractors: 81 comes from squaring 9 instead of 10, one position short; 20 comes from multiplying the position by 2 instead of squaring it; 1000 comes from cubing the position, 10 × 10 × 10, instead of squaring it.
- (d) 23 — Method: let the smallest integer be n, so the three consecutive integers are n, n + 1 and n + 2. Form the equation n + (n + 1) + (n + 2) = 72. Working: simplify the left side: 3n + 3 = 72, so 3n = 69, giving n = 23. Answer: 23. 24 comes from dividing 72 by 3 directly, 72 ÷ 3 = 24, which finds the middle integer rather than realising the three numbers differ. 25 comes from correctly finding n = 23 but reading off the largest integer, n + 2, instead of the smallest as asked. 21 comes from dividing first and subtracting after, (72 ÷ 3) − 3, instead of subtracting 3 before dividing by 3.
- (a) 5.8 — The horizontal distance is 3 and the vertical distance is 5, so using Pythagoras' theorem the distance is √(3² + 5²) = √34 = 5.8 (1 d.p.). A candidate who adds the two differences instead of using Pythagoras gets 3 + 5 = 8.0. A candidate who works out 3² + 5² = 34 but forgets to take the square root gets 34.0. A candidate who subtracts the squares instead of adding them gets √(5² − 3²) = √16 = 4.0.
- (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.
- (c) the fourth quadrant — Method: read the sign of each coordinate in turn, then count round the regions, which are numbered anticlockwise from the one where both coordinates are positive. Working: the x-coordinate 5 is positive, so A is to the right of the y-axis; the y-coordinate −3 is negative, so A is below the x-axis; right of the y-axis and below the x-axis is the fourth of the four regions. Answer: the fourth quadrant. The distractors: 'the first quadrant' comes from ignoring the minus sign on the y-coordinate and treating the point as (5, 3); 'the second quadrant' comes from writing the pair the wrong way round and locating (−3, 5) instead; 'the third quadrant' comes from assuming that any point with a negative coordinate belongs to the region where the minus signs are, without checking the other coordinate.
- (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.
- (a) −2 and 2 — Method: the graph crosses the x-axis where y = 0, so read the two crossing points straight off the curve. Working: the curve meets the horizontal axis two squares to the left of the origin and two squares to the right, at x = −2 and x = 2. Answer: −2 and 2. Distractor refutation: −4 and 4 comes from reading the value where the curve crosses the y-axis, −4, and using that value and its positive partner as the x-axis crossings instead. −2 and 3 comes from misreading the right-hand crossing point one square out along the grid, taking it where the curve is already above the axis. 0 and 2 comes from confusing the lowest point of the curve, which sits on the y-axis at x = 0, with one of the crossing points.
- (d) x = (y + 3)/4 — Two operations have been applied to x: it has been multiplied by 4 and then 3 has been subtracted. Undo them in the opposite order. Adding 3 to both sides gives y + 3 = 4x. Dividing both sides by 4 then gives (y + 3)/4 = x, so x = (y + 3)/4. The bracket matters: writing y/4 + 3 divides only the y by 4 and leaves the 3 untouched. Writing (y − 3)/4 subtracts the 3 instead of adding it, and writing 4(y + 3) multiplies by 4 rather than dividing.
- (a) 1 — Method: the value of y when x = 0 is where the line meets the y-axis, which is the constant c in y = mx + c, so the gradient is worked out from the two given points first and the constant follows by substituting one of them. Working: m = (16 − 7) ÷ (5 − 2) = 9 ÷ 3 = 3, so the line is y = 3x + c; substituting x = 2 and y = 7 gives 7 = 3 × 2 + c, so c = 7 − 6 = 1, and the value of y when x = 0 is that constant. Answer: 1. The distractors: 3 comes from stopping at the gradient and offering it as the intercept; 4 comes from stepping back from x = 2 to x = 0 by one unit of x instead of two, 7 − 3 = 4; −1 comes from working the constant out as mx − y, 3 × 2 − 7 = −1, instead of y − mx.
- (d) 21 − 3n — Method: find the common difference, then find the constant that fits the first term. Working: 15 − 18 = −3, 12 − 15 = −3, 9 − 12 = −3, so the terms decrease by 3 each time and the nth term has the form c − 3n. Substituting n = 1: c − 3(1) = 18, so c = 21. Answer: the nth term is 21 − 3n. The value 18 − 3n comes from using the first term as c directly, without adding back the difference that was removed. The value 3n − 21 comes from a sign error that flips the whole expression. The value −3n comes from using only the common difference and leaving out the constant.
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