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.
Answer key: Algebra worksheet — GCSE Higher
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- (c) 81 m — Split the velocity-time graph into its three phases. The accelerating phase (0 to 4 s) is a triangle: 1/2 × 4 × 6 = 12. The constant phase (4 s at 6 m/s, for 10 s) is a rectangle: 10 × 6 = 60. The decelerating phase (3 s) is a triangle: 1/2 × 3 × 6 = 9. Total distance: 12 + 60 + 9 = 81 m. Forgetting the final decelerating phase entirely and adding only the first two areas gives 12 + 60 = 72 m. Treating all three phases as rectangles, forgetting to halve either triangular phase, gives 4 × 6 + 10 × 6 + 3 × 6 = 102 m. Halving only the deceleration triangle correctly but treating the acceleration phase as a rectangle too gives 4 × 6 + 60 + 9 = 93 m.
- (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.
- (d) x = (y + 8) / 3 — Method: add 8 to both sides, then divide the whole side by 3. Working: y = 3x − 8, so adding 8 to both sides gives y + 8 = 3x, then dividing both sides by 3 gives x = (y + 8) / 3. The value x = y + 8 / 3 is Priya's version, which comes from dividing only the 8 by 3 instead of dividing the whole expression y + 8 by 3. The value x = (y − 8) / 3 comes from a sign error, keeping −8 instead of moving it to +8. The value x = 3(y + 8) comes from multiplying by 3 instead of dividing.
- (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.
- (c) L/5 − 3 — Each of the 5 equal pieces is L/5 metres long, and removing 3 metres from one piece gives L/5 − 3. Subtracting the 3 metres before dividing by 5, (L − 3)/5, divides the removed length between all 5 pieces instead of taking it from just one. Dividing only the 3 by 5 instead of dividing L by 5, L − 3/5, divides the wrong number. Writing 5/L − 3 inverts the fraction, swapping which number is the numerator.
- (d) 6.0 — Subtract 1.8 from both sides: 6.3x ≤ 38.2. Divide both sides by 6.3: x ≤ 6.0634… . The question asks for the greatest value to 1 decimal place that still satisfies the inequality. Testing 6.1: 6.3 × 6.1 + 1.8 = 40.23, which is more than 40, so 6.1 fails. Testing 6.0: 6.3 × 6.0 + 1.8 = 39.6, which is no more than 40, so 6.0 works and is the greatest such value. A candidate who simply rounds 6.0634… to 1 decimal place answers 6.1, without checking that it satisfies the inequality. A candidate who adds 1.8 instead of subtracting works out 41.8 ÷ 6.3 and answers 6.6. A candidate who forgets the 1.8 altogether divides 40 by 6.3 and answers 6.3.
- (d) 30 — Method: give each group a letter, turn each sentence of the stem into an equation in those letters, then eliminate one letter by adding the two equations. Working: let m be the number of men and w the number of women. The 50 adults give m + w = 50, and 10 more men than women gives m − w = 10. Adding the two equations term by term cancels w: (m + w) + (m − w) = 50 + 10, so 2m = 60 and m = 30. Substituting back into m + w = 50 gives w = 20, and 30 − 20 = 10 as required. Answer: 30. The distractors: 20 comes from subtracting the equations instead of adding them, which cancels m and gives 2w = 40, so w = 20 — the other unknown, not the one asked for; 40 comes from forming the second equation as w = 10, reading '10 more men than women' as '10 women', and then getting m = 50 − 10 from the first equation; 35 comes from using m + w = 50 alone with m = w, giving 25 each, and then adding the whole difference of 10 to the men instead of splitting it between the two groups, which leaves a difference of 20 rather than 10.
- (c) (2, 5) — Reflecting in the x-axis keeps the x-coordinate the same and changes the sign of the y-coordinate: Q = (2, 5). (−2, −5) comes from reflecting in the y-axis instead, which changes the sign of the x-coordinate. (−2, 5) comes from reflecting in both axes. (2, −5) comes from not applying the reflection at all.
- (a) x = 3 — Method: factorise x² + 2x − 15 as (x + 5)(x − 3), since 5 × (−3) = −15 and 5 + (−3) = 2. Setting each bracket equal to zero gives x + 5 = 0 or x − 3 = 0, so x = −5 or x = 3. Only x = 3 is offered here. Distractor origins: x = −3 reverses the sign of the factor pair, treating the bracket (x − 3) as giving x = −3 instead of x = 3; x = 5 takes the number from the other factor, (x + 5), but with the wrong sign, giving x = 5 instead of x = −5; x = 15 takes the constant term of the original expression as if it were a root, without factorising at all.
- (d) An inequality, because ≤ compares the two sides — The symbol ≤ means 'is less than or equal to', so the statement compares the sizes of the two sides instead of saying they are equal: that makes it an inequality. Solving it gives n ≤ 5, a whole range of values rather than the single value an equation would give. An identity has to be true for every value of the letter, and this fails at n = 6, so it is not one. A formula works one quantity out from another, and there is only one letter here.
- (d) £16.70 — The mileage charge is 2.20 × 6 = £13.20. Adding the fixed fee: £13.20 + £3.50 = £16.70. A candidate who forgets the fixed fee gives just the mileage charge, £13.20. A candidate who adds the fixed fee to the per-mile rate before multiplying by the number of miles, (3.50 + 2.20) × 6, gets £34.20. A candidate who rounds £2.20 down to £2 gets 2 × 6 + 3.50 = £15.50.
- (d) y = (x + 3)(x − 5) — A root at x = −3 means the matching bracket must be zero when x = −3, so the bracket is (x − (−3)) = (x + 3). A root at x = 5 means the other bracket is (x − 5). So the equation is y = (x + 3)(x − 5); checking the y-intercept, (0 + 3)(0 − 5) = 3 × (−5) = −15, which matches the given value. The option (x − 3)(x + 5) swaps the signs of both roots. The option (x + 3)(x + 5) keeps the correct sign for the first root but gets the second wrong. The option (x − 3)(x − 5) gets the first root's sign wrong.
- (b) −2, −1, 0, 1, 2, 3, 4 — Factorise x² − 2x − 8 = (x − 4)(x + 2), giving roots x = 4 and x = −2. Since the coefficient of x² is positive and the inequality is ≤ 0, the solution is the closed interval between the roots, −2 ≤ x ≤ 4, with the roots included because the inequality is not strict. The integers in this interval are −2, −1, 0, 1, 2, 3, 4. Distractor routes: −1, 0, 1, 2, 3 drops both endpoints, treating ≤ as if it were the strict inequality <. −2, −1, 0, 1, 2, 3, 4, 5 comes from mis-factorising as (x − 5)(x + 2), giving an upper root of 5 instead of 4. −3, −2, −1, 0, 1, 2, 3 comes from mis-factorising as (x − 4)(x + 3), giving a lower root of −3 instead of −2.
- (b) y decreases towards zero but never reaches it — Method: as x gets larger, dividing 20 by a bigger number gives a smaller result, so y decreases; because 20/x can never be exactly zero for any positive x, the curve gets closer to zero without ever reaching it, so y decreases towards zero but never reaches it. Distractor origins: 'y increases towards a limit but never reaches it' has the relationship backwards, treating y as increasing when it is actually decreasing; 'y decreases at a steady rate and reaches zero' wrongly assumes the graph behaves like a straight line that eventually hits zero; 'y stays the same however large x becomes' wrongly assumes there is no change in y at all.
- (d) £2000 — This is an arithmetic sequence with first term £500 and common difference £300. The 6th term is 500 + 5 × 300 = 2000. A candidate who uses 6 lots of the increase instead of 5 gets 500 + 6 × 300 = 2300. A candidate who forgets to add the first year's profit at all gets 5 × 300 = 1500. A candidate who miscounts the number of increases as 4 instead of 5 gets 500 + 4 × 300 = 1700.
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