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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- (c) x = −5 — Method: 4 is being subtracted from x, so add 4 to both sides; adding a positive number to a negative one moves to the right along the number line. Working: adding 4 to both sides gives x = −9 + 4, and −9 + 4 = −5. Answer: x = −5. The distractors: x = −13 comes from subtracting 4 from both sides instead of adding it, giving −9 − 4; x = 5 comes from working out −9 + 4 correctly and then writing the result without its minus sign; x = 13 comes from reading the right-hand side as 9 rather than −9 and adding 4 to it.
- (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.
- (a) 5 — The symbol ≤ means n can equal 5 or any number less than 5, so 5 is included and is the largest integer value. A candidate who treats the inequality as strict, as if it were n < 5, answers 4. A candidate who confuses ≤ with ≥ and looks for a value just above the boundary answers 6. A candidate who makes a sign error and reads the inequality as n ≤ −5 answers −5.
- (b) y = 2x − 3 — A straight-line graph has an equation of the form y = mx + c, where x appears only to the power 1. Of the four equations, y = 2x − 3 fits this form, so it gives a straight line. y = x² − 3 has x squared, so it gives a curved (quadratic) graph, not a straight line. y = x³ − 3 has x cubed, so it gives a cubic curve. y = 3/x has x on the bottom of a fraction, so it gives a reciprocal curve with two branches.
- (c) 3 — Substituting y = 3 into 4x − y = 9 gives 4x − 3 = 9, so 4x = 12, and x = 3. A candidate who adds 3 instead of subtracting it, making a sign error when substituting, would get 4x + 3 = 9, so 4x = 6 and x = 1.5. A candidate who forgets to divide by 4 after finding 4x = 12 would write x = 12. A candidate who subtracts 4 instead of dividing by it would get 12 − 4 = 8.
- (a) 9 km — Ravi's house and the library have the same y-coordinate, so the distance between them is the difference between their x-coordinates: 6 − (−3) = 9 km. 3 km comes from dropping the negative sign on Ravi's x-coordinate and working out 6 − 3 instead. 0 km comes from using the y-coordinates, which are equal, instead of the x-coordinates. −9 km comes from working out −3 − 6 = −9 and not converting it to a positive distance.
- (a) 2b — Method: a total cost is the price of one item multiplied by the number of items bought. Working: each pen costs £2 and there are b pens, so the total cost is 2 × b, which is written as 2b. Answer: 2b. The distractors: b + 2 comes from adding the price to the number of pens instead of multiplying, treating £2 as a one-off charge; 2b + 2 comes from multiplying correctly and then adding the £2 a second time as though there were also a fixed charge; b/2 comes from dividing by the price instead of multiplying by it.
- (c) 16 — Let a be the number of adult tickets and c the number of child tickets: a + c = 40 and a = c + 8. Substituting the second equation into the first: (c + 8) + c = 40, so 2c = 32, and c = 16. A candidate who reports the number of adult tickets instead of child tickets would get a = 24. A candidate who splits the 40 tickets evenly, ignoring the difference of 8, would get 20. A candidate who forgets to divide 2c = 32 by 2 would write c = 32.
- (d) x < 7 — Method: add 10 to both sides to leave x on its own; adding the same number to both sides never changes the direction of an inequality. Working: adding 10 to both sides of x − 10 < −3 leaves x on the left and −3 + 10 on the right, and −3 + 10 = 7, so x < 7. Answer: x < 7. The distractors: x > 7 comes from turning the sign round while adding, as though every move flipped it; x < −13 comes from subtracting 10 from both sides instead of adding it, giving −3 take away 10; x < 13 comes from ignoring the minus sign on −3 and working out 3 + 10 instead.
- (b) 7 — Method: compare the equation with y = mx + c, where m is the gradient. Working: in y = 7x − 2, the coefficient of x is 7. Answer: the gradient is 7. −2 comes from confusing the gradient with the y-intercept. −7 comes from a sign error, treating the coefficient of x as negative. 2 comes from using the y-intercept but dropping its sign.
- (c) Second and fourth — If x + y = 0 then y = −x, so x and y always have opposite signs, one positive and one negative. A point with a negative x and a positive y lies in the second quadrant, and a point with a positive x and a negative y lies in the fourth quadrant, so the point lies in the second or the fourth. A candidate who reads x + y = 0 as meaning x and y have the same sign picks First and third, which is where x × y is positive, not where x + y = 0. A candidate who decides that y must be the positive coordinate picks the two quadrants above the x-axis, First and second. A candidate who decides that y must be the negative coordinate picks the two quadrants below the x-axis, Third and fourth.
- (d) t = 5 — The maximum height occurs halfway between the two times when the ball is at ground level: the midpoint of t = 1 and t = 9 is (1 + 9) ÷ 2 = 5, so the ball reaches its maximum height at t = 5 seconds. Getting t = 4 comes from halving the DIFFERENCE between the times, 9 − 1 = 8, then 8 ÷ 2 = 4, instead of finding their midpoint. Getting t = 8 uses that difference, 9 − 1 = 8, as if the gap between the two times were itself the time of the maximum. Getting t = 10 adds the two times, 1 + 9 = 10, but forgets to divide by 2.
- (b) 2(x + 3) = 2x + 6 — An identity is true for every value of x, not just one. Expanding 2(x + 3) gives 2x + 6, which matches the right-hand side exactly — so the equation holds for every value of x, and it is an identity. Each of the other three is only true for one particular value of x: 5x − 3 = 12 gives x = 3, x + 7 = 15 gives x = 8, and 3x = x + 10 gives x = 5 — these are ordinary equations, not identities.
- (d) (4, 3) — The sides are parallel to the axes: the missing vertex must share the y-coordinate 3 with (−2, 3) and the x-coordinate 4 with (4, −3), giving (4, 3). (−4, 3) comes from a sign error on the x-coordinate. (4, −9) comes from continuing the pattern of the given points by subtracting 6 from the y-coordinate again instead of matching it to (−2, 3). (3, 4) comes from swapping the x- and y-coordinates.
- (b) 10 km — Method: find the constant speed from the graph (distance ÷ time for any point on the line), then multiply that speed by 20 minutes. Working: the line passes through (4 minutes, 2 km), so the speed is 2 ÷ 4 = 0.5 km per minute; in 20 minutes the cyclist travels 0.5 × 20 = 10 km. Answer: 10 km. Distractor refutation: 3 km comes from reading off the distance shown at the end of the plotted section (6 minutes) and stopping there, instead of extending the line to 20 minutes. 20 km comes from misreading the speed as 1 km per minute instead of 0.5 km per minute, doubling the true rate. 40 km comes from dividing 20 by the speed instead of multiplying by it, a reciprocal mix-up.
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