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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Algebra worksheet — GCSE Foundation
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- 1.The diagram shows a distance–time graph for a cyclist travelling at a constant speed, for the first 6 minutes of a journey. Distance is in kilometres and time is in minutes. Work out how far the cyclist would travel in 20 minutes at the same speed.
- 2.Solve the inequality 9 − 2x ≥ 1.
- 3.Yusuf has £40 saved and adds £6 each week. His sister Aisha has £10 saved and adds £11 each week. After how many weeks will they have saved the same amount?
- 4.Work out the gradient of the straight line that passes through the points (−1, 5) and (3, −7).
- 5.A straight line has gradient −2 and passes through the point (3, 1). Work out the equation of the line.
- 6.A point has coordinates (−6, 9). Work out the sum of the x-coordinate and the y-coordinate.
- 7.A point lies on the y-axis. What can be said about the x-coordinate of that point?
- 8.Solve the inequality 4x + 1 > 2x + 9.
- 9.Which of these is an expression?
- 10.Simplify 5x + 3y − 2x + y.
- 11.Point A has coordinates (4, −5). Point B has coordinates (4, 3). Work out the distance between A and B.
- 12.The simultaneous equations x = 2 and x + y = 9 are given. Work out the value of y.
- 13.Write down the gradient of the line with equation y = −x + 3.y = −x + 3
- 14.Write down the first four terms of the sequence with nth term 6n − 5.
- 15.In the expression 5x² − 3x + 7, which of these is a single term of the expression?
Answer key
- (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.
- (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.
- (d) 6 — Method: write an expression for each person's savings after w weeks, and set them equal. Working: 40 + 6w = 10 + 11w. Subtract 6w from both sides: 40 = 10 + 5w. Subtract 10: 30 = 5w, so w = 6. Answer: 6 weeks. 0 comes from setting only the weekly amounts equal, 6w = 11w, and ignoring the different starting amounts entirely. 10 comes from adding the two starting amounts and dividing by the difference in weekly amounts, (40 + 10) ÷ (11 − 6), instead of forming and solving the correct equation. 1.76 comes from adding the two weekly amounts instead of subtracting them when rearranging, (40 − 10) ÷ (11 + 6).
- (b) −3 — Method: the gradient of a straight line is the change in y divided by the change in x, with the two coordinates taken in the same order in the numerator as in the denominator. Working: going from (−1, 5) to (3, −7), the change in y is −7 − 5 = −12 and the change in x is 3 − (−1) = 4, so the gradient is −12 ÷ 4 = −3. Answer: −3. The distractors: 3 comes from subtracting the y-coordinates in one order and the x-coordinates in the other, giving 12 ÷ 4; −1/3 comes from dividing the change in x by the change in y instead of the other way round, giving 4 ÷ (−12); −6 comes from working out 3 − (−1) as 3 − 1 = 2, so that the change in y is divided by 2 rather than by 4.
- (a) y = −2x + 7 — Using y = −2x + c, and substituting the point (3, 1): 1 = −2(3) + c, so 1 = −6 + c, and c = 7, giving y = −2x + 7. A candidate who uses the y-coordinate of the point directly as the y-intercept, instead of solving for c, would write y = −2x + 1. A candidate who drops the negative sign on the gradient would write y = 2x + 7. A candidate who makes a sign error when isolating c, writing c = 1 − 6 = −5 instead of c = 1 + 6 = 7, would write y = −2x − 5.
- (b) 3 — −6 + 9 = 3. A candidate who ignores the negative sign on the x-coordinate and adds the two positive values gets 6 + 9 = 15. A candidate who treats the y-coordinate as negative too gets −6 + (−9) = −15. A candidate who works out 9 − 6 correctly as 3 but then writes the answer with the wrong sign gets −3.
- (b) x = 0 — Method: the first coordinate of a point measures how far to the left or right of the origin it is, so a point that is neither left nor right of the origin has a first coordinate of zero. Working: the y-axis is the vertical line through the origin; every point on it lies directly above or directly below the origin with no sideways movement at all, so its first coordinate is zero while its second coordinate may take any value. Answer: x = 0. The distractors: y = 0 is the condition for lying on the x-axis, the other axis; x = 1 comes from confusing the y-axis with the vertical line one unit to the right of it; x = y is the condition for the diagonal through the origin, which meets the y-axis at the origin only.
- (d) x > 4 — Method: collect the x terms on one side and the numbers on the other, then divide by the coefficient of x; dividing by a positive number leaves the sign as it is. Working: subtracting 2x from both sides of 4x + 1 > 2x + 9 gives 2x + 1 > 9; subtracting 1 from both sides gives 2x > 8; dividing both sides by 2 gives x > 4. Answer: x > 4. The distractors: x < 4 comes from turning the sign round while dividing by 2; x > 5 comes from adding the 1 to the 9 instead of subtracting it, giving 2x > 10; x < 5 comes from making both of those mistakes together.
- (c) 5y − 2 — An expression is a collection of terms with no equals sign and no inequality sign in it: nothing is being claimed about two sides. Here 5y − 2 on its own fits that description. The line with an equals sign and one unknown letter to find is an equation; the line joined by a comparison sign is an inequality; the line that links two different letters so that one can be worked out from the other is a formula.
- (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.
- (a) 8 — Both points share the x-coordinate, so the distance between them is the difference between the y-coordinates: 3 − (−5) = 8. A candidate who mistakenly uses the equal x-coordinates instead of the y-coordinates gets 4 − 4 = 0. A candidate who adds the y-coordinates instead of subtracting them gets 3 + (−5) = −2. A candidate who reads off only point B's y-coordinate as the distance gets 3.
- (b) y = 7 — Method: one of the equations already gives the value of x, so that value is substituted into the other equation. Working: replacing x with 2 in x + y = 9 gives 2 + y = 9, so y = 9 − 2 = 7. Answer: y = 7, and the pair x = 2 with y = 7 satisfies both equations. The distractors: y = 11 comes from rearranging x + y = 9 as y = 9 + x and adding instead of subtracting; y = 9 comes from ignoring the x term and reading the total as the value of y; y = 4.5 comes from halving 9, as though x and y had to be equal.
- (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.
- (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.
- (c) 5x² — A term is one part of an expression, separated from the others by a + or − sign. The expression 5x² − 3x + 7 has three terms: 5x², −3x and 7. Distractor origins: 5x² − 3x combines two terms together, not a single term; −3x + 7 also combines two terms together; 5x² − 3x + 7 gives the whole expression, not a single term.
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