Printable · GCSE Higher · ages 14-16
Algebra worksheet — GCSE Higher
Fifteen questions across the algebra statements at Higher 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 Higher
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- 1.A straight line has equation 4y = 12x + 20. Work out the gradient of the line.y = 12x + 20
- 2.Work out the equation of the straight line through the points (−3, 4) and (1, −8).
- 3.Line P has equation y = −2x + 5 and line Q has equation 2y = x + 6. Are lines P and Q perpendicular to each other?y = -2x + 5y = x + 6
- 4.A student claims: 'For every positive integer n, n² + n + 1 is a prime number.' Which value of n shows that this claim is false?
- 5.Solve the simultaneous equations y = 2x − 1 and y = x² − x − 1, giving both pairs of solutions.y = 2x − 1y = x²
- 6.A rule turns each input x into an output y. An input of 1 gives an output of 1, an input of 2 gives an output of 4 and an input of 3 gives an output of 9. Work out the rule.
- 7.Two brothers have ages that add up to 30. The elder brother is 6 years older than the younger brother. Work out the age of the younger brother.
- 8.A geometric sequence has first term 3 and common ratio 2. Work out the 5th term of the sequence.
- 9.A ball is dropped and bounces. The height of each bounce after the first is 8 cm less than the bounce before it. The first bounce reaches 60 cm. Work out the height of the 5th bounce.
- 10.The line y = 3x + 6 meets the x-axis at one point. Work out the coordinates of that point.y = 3x + 6
- 11.A company's weekly profit, in £, is modelled by y = f(x), where x is the number of weeks since launch. The graph of y = f(x) has a maximum at (10, 45000). A rival company uses the same marketing strategy but starts trading 6 weeks later and has fixed costs £8000 higher every week, so its profit is modelled by y = f(x − 6) − 8000. In which week does the rival's maximum weekly profit occur, and what is it?
- 12.x = 5. Work out the value of 3x² − 4.
- 13.Which expression is equivalent to 6x − (2x − 5)?
- 14.Solve (2x + 1)/3 = 5
- 15.A circle has centre (0, 0) and equation x² + y² = 50. Work out the radius of the circle, giving your answer as a surd in its simplest form.
Answer key
- (b) 3 — Method: rearrange the equation into the form y = mx + c first, then read off the gradient. Working: dividing 4y = 12x + 20 by 4 gives y = 3x + 5, so the gradient is 3. The value 12 comes from reading the coefficient of x before dividing the equation by 4. The value 20 comes from using the constant term of the unsimplified equation instead of the gradient. The value 5 comes from finding the y-intercept, 20 / 4 = 5, and giving that instead of the gradient.
- (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.
- (a) Yes — the gradients multiply to −2 × 1/2 = −1. — Rearrange Q into the form y = mx + c: 2y = x + 6 gives y = (1/2)x + 3, so Q has gradient 1/2. P has gradient −2. Two lines are perpendicular exactly when the product of their gradients is −1: −2 × 1/2 = −1. Since this holds, P and Q are perpendicular. Distractor routes: "the product is −1, but perpendicular needs 1" works out the product correctly but misremembers the condition — the perpendicular test is a product of exactly −1, and parallel lines are spotted by their gradients being equal, not by a product of 1. "Q's gradient is 2, and −2 × 2 = −4" comes from reading the 2 in front of y in 2y = x + 6 as the gradient, instead of dividing the whole equation by 2 first to reach y = (1/2)x + 3, where the gradient is 1/2. "Both equations have a negative x-term" is not a valid test at all — P's equation does have a negative x-term, but Q's, once rearranged, does not, and matching signs say nothing about the actual gradients.
- (c) n = 4 — Testing n = 4: 4² = 16, and 16 + 4 + 1 = 21, and 21 = 3 × 7 is not a prime number, so this value disproves the claim. Testing n = 1: 1² = 1, and 1 + 1 + 1 = 3, which is prime, so it does not disprove the claim. Testing n = 2: 2² = 4, and 4 + 2 + 1 = 7, which is prime. Testing n = 3: 3² = 9, and 9 + 3 + 1 = 13, which is also prime — a counterexample has to give a result that isn't prime, and only n = 4 does that.
- (d) x = 0, y = −1 and x = 3, y = 5 — Set the two expressions for y equal: 2x − 1 = x² − x − 1. Rearranging, subtracting 2x and adding 1 to both sides: 0 = x² − x − 1 − 2x + 1 = x² − 3x, so x² − 3x = 0. Factorise: x(x − 3) = 0, giving x = 0 or x = 3. Using y = 2x − 1: x = 0 gives y = −1; x = 3 gives y = 5. Distractor routes: x = 0, y = −1 alone stops after the factor x = 0 and never checks the second factor, x − 3 = 0. x = 0, y = −1 and x = −3, y = −7 comes from mis-factorising x² − 3x as x(x + 3), a sign error that gives a second root of −3 instead of 3. x = −2, y = −5 and x = 1, y = 1 comes from adding 2x to both sides instead of subtracting it when rearranging, giving x² + x − 2 = 0 instead of x² − 3x = 0.
- (c) y = x² — Method: test a candidate rule against every pair given, not just one — a rule that fits one pair and fails another is not the rule. Working: the outputs 1, 4, 9 rise by 3 and then by 5, so they are not going up in equal steps and the input is not simply multiplied by a fixed number; comparing each output with its own input gives 1 × 1 = 1, 2 × 2 = 4 and 3 × 3 = 9, and all three pairs fit. Answer: y = x². The distractors: y = 3x comes from fitting only the last pair, where 3 × 3 = 9, and reading that 3 as a multiplier; y = 3x − 2 comes from assuming a multiply-then-add rule and using the first step in the outputs, 4 − 1 = 3, as the multiplier — it fits the first two pairs and fails the third; y = 2x comes from fitting only the pair 2 and 4 and reading every output as double its input.
- (d) 12 — Method: call the younger brother's age x, write the elder brother's age in terms of x, and form an equation from the total. Working: the elder brother is x + 6, so x + (x + 6) = 30; simplifying gives 2x + 6 = 30, subtracting 6 from both sides gives 2x = 24, and dividing by 2 gives x = 12. Checking: 12 and 18 add up to 30 and differ by 6. Answer: 12. The distractors: 18 comes from solving correctly and then giving the elder brother's age, which is not the age asked for; 15 comes from halving 30 and ignoring the 6-year difference altogether; 24 comes from taking 6 off the total, 30 − 6 = 24, and giving that as an age.
- (d) 48 — Each term is found by multiplying the previous term by the common ratio, 2: 3, 6, 12, 24, 48 — the 5th term is 48. A candidate who mistakes the common ratio for a common difference, and adds 2 four times, would reach 3+4×2=11. A candidate who works out the multiplier 2⁴=16 but forgets to multiply it by the first term would give 16. A candidate who multiplies one time too many (finding the 6th term instead of the 5th) would reach 3×2⁵=96.
- (d) 28 — The height decreases by 8 cm at each bounce after the first, so the nth bounce reaches 60−(n−1)×8 cm. For the 5th bounce: 60−4×8=60−32=28. A candidate who subtracts 8 one time too many, five times instead of four, would compute 60−5×8=20. A candidate who adds the decrease instead of subtracting it, a sign error, would compute 60+4×8=92. A candidate who works out only the total decrease and forgets to include the starting height of 60 cm would compute just 5×8=40.
- (b) (−2, 0) — Method: a graph meets the x-axis where the y-value is 0, so setting y = 0 turns the equation into a linear equation in x. Working: 0 = 3x + 6 gives 3x = −6, so x = (−6) ÷ 3 = −2 and the meeting point is (−2, 0). Answer: (−2, 0). The distractors: (2, 0) comes from solving 3x = −6 and then dropping the minus sign from the result; (0, 6) is the y-axis crossing, found by substituting x = 0 instead of y = 0; (6, 0) comes from reading the constant 6 straight off as the x-coordinate, without dividing by 3 and without changing its sign.
- (c) Week 16, £37,000 — y = f(x − 6) − 8000 combines a horizontal translation of 6 units RIGHT (subtracting 6 inside the brackets) with a vertical translation of £8000 DOWN (subtracting 8000 outside). Applying both to the maximum (10, 45000): 10 + 6 = 16, so the new maximum is in week 16. And 45000 − 8000 = 37000, so the maximum weekly profit is £37,000.
- (c) 71 — Method: square x first, then multiply by 3, then subtract 4, following the order of operations. Working: x² = 5² = 25; 3 × 25 = 75; 75 − 4 = 71. Answer: 71. 221 comes from squaring (3x) as a whole first: (3 × 5)² = 225, then − 4 = 221, squaring the coefficient along with x. 75 comes from correctly working out 3x² but forgetting to subtract the 4. 3 comes from subtracting the 4 from x before squaring: (5 − 4)² × 3 = 3, doing the operations in the wrong order.
- (d) 4x + 5 — A minus sign directly before a bracket changes the sign of both terms inside it: 6x − (2x − 5) = 6x − 2x + 5 = 4x + 5. The option 4x − 5 comes from only changing the sign of the 2x term and not the −5, giving 6x − 2x − 5. The option 8x − 5 comes from adding 2x instead of subtracting it, as if the minus sign did not apply to the bracket, giving 6x + 2x − 5. The option 8x + 5 repeats that same addition mistake and also changes the sign of the −5 term.
- (a) 7 — Method: multiply both sides by 3 to clear the fraction, then solve the resulting equation. Working: 2x + 1 = 5 × 3 = 15. Subtract 1: 2x = 14. Divide by 2: x = 7. Answer: 7. 2 comes from ignoring the denominator altogether, treating the equation as 2x + 1 = 5 without multiplying by 3 first. 8 comes from a sign error, adding 1 to 15 instead of subtracting it, giving 2x = 16. 14 comes from correctly reaching 2x = 14 but stopping there, without dividing by 2 to find x.
- (d) 5√2 — For a circle x² + y² = r², the 50 on the right-hand side is r², not r, so the radius is √50. Writing 50 as 25 × 2, the largest square factor times what remains, gives √50 = √25 × √2 = 5√2. Forgetting to square-root 50 at all and giving the value of r² instead gives 50. Halving 50 instead of taking its square root gives 25. Using 25 as the number left outside the square root sign, instead of as the number under it, gives the wrongly simplified 25√2.
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