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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Answer key: Algebra worksheet — GCSE Higher
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- (c) 25 — Method: read off a, b and c with their signs and substitute them into b² − 4ac. Working: for 2x² + 3x − 2 = 0, a = 2, b = 3 and c = −2, so b² − 4ac = 3² − 4 × 2 × (−2) = 9 − (−16) = 9 + 16 = 25. Answer: 25. The distractors: −7 comes from taking c as +2, which gives 9 − 16; 22 comes from working b² as 2 × 3 = 6 and then 6 + 16; 13 comes from leaving the 4 out of 4ac and working 9 − 2 × (−2).
- (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.
- (c) 64 m — The distance travelled is the area under the speed-time graph. From t = 0 to t = 4, the shape is a triangle with base 4 and height 8, area 1/2 × 4 × 8 = 16. From t = 4 to t = 10, the shape is a rectangle with base 6 and height 8, area 6 × 8 = 48. Total distance: 16 + 48 = 64 m. Treating the whole 10 seconds as a single trapezium with parallel sides 0 and 8 and width 10, instead of splitting it into the triangle and rectangle, gives 40 m. Ignoring the acceleration phase completely and assuming the car travels at a constant 8 m/s for all 10 seconds gives 8 × 10 = 80 m. Misreading the second interval as running from t = 4 to t = 8 instead of t = 4 to t = 10 gives a rectangle area of 4 × 8 = 32, plus the correct triangle of 16, totalling 48 m.
- (a) It is 2(w + (w + 3)) = 4w + 6, not 2w + 3. — The perimeter of a rectangle is twice the width plus twice the length: 2 × w + 2 × (w + 3) = 2w + 2w + 6 = 4w + 6, so the gardener's 2w + 3 is wrong. Writing w + (w + 3) = 2w + 3 forgets to double the sides at all, only adding one width and one length once. Writing 4(w + 3) = 4w + 12 wrongly treats all four sides as equal to the length, as if the garden were a square. Writing 3w + 6 comes from doubling the length correctly but adding the width only once instead of doubling it too.
- (a) 5p + 7q — Tom's total spend combines like terms: (2p + 6q) + (3p + q) = 5p + 7q, adding the notebook terms (2p + 3p = 5p) and the pencil terms (6q + q = 7q) separately. Answering 12pq adds every coefficient together (2 + 6 + 3 + 1 = 12) and multiplies the letters, combining unlike terms as though notebooks and pencils were the same item. Answering 5p + 6q correctly combines the notebook terms but forgets to add Tuesday's extra pencil to the 6q. Answering 2p + 7q correctly combines the pencil terms but forgets to add Tuesday's 3 extra notebooks to the 2p. The total amount Tom has spent is 5p + 7q pence.
- (a) 3x = 12 — Adding the two equations: the y-terms, +y and −y, have opposite signs, so they cancel; the x-terms combine to 2x + x = 3x; and the right-hand sides add to 11 + 1 = 12. This gives 3x = 12. A candidate who forgets that the y-terms cancel, and instead adds them as if they had the same sign, would write 3x + 2y = 12. A candidate who subtracts the right-hand sides instead of adding them would get 3x = 10. A candidate who correctly reaches 3x = 12 but then treats 12 itself as the value of x, skipping the final division, would write x = 12.
- (a) 120 m — The gradient of the second line is (0 − V)/(9 − 6) = −V/3, and this equals −8, so V = 24. The distance from t = 0 to t = 6 is the area of a trapezium with parallel sides 4 and 24 and width 6: 1/2 × (4 + 24) × 6 = 84. The distance from t = 6 to t = 9 is the area of a triangle with base 3 and height 24: 1/2 × 3 × 24 = 36. The total distance is 84 + 36 = 120 m. Using V = 8, treating the gradient's number as the missing velocity itself rather than solving −V/3 = −8 for V, gives a trapezium area of 1/2 × (4 + 8) × 6 = 36 and a triangle area of 1/2 × 3 × 8 = 12, a total of 48 m. Leaving out the 1/2 in the trapezium formula, (4 + 24) × 6 = 168, plus the correct triangle of 36, gives 204 m. Using the full 6 seconds as the triangle's base instead of the 3 seconds the second line actually lasts, 1/2 × 6 × 24 = 72, plus the correct trapezium of 84, gives 156 m.
- (c) x = 3, y = 1 — Method: the y terms are +y and −y, so adding the two equations removes y and leaves an equation in x alone. Working: adding 2x + y = 7 and x − y = 2 gives 3x = 9, so x = 3; substituting x = 3 into x − y = 2 gives 3 − y = 2, so y = 1. Answer: x = 3, y = 1, which also satisfies 2 × 3 + 1 = 7. The distractors: x = 1, y = 3 comes from finding the two values correctly and then writing them against the wrong letters; x = 3, y = 2 comes from substituting x = 3 into 2x + y = 7 as 2 + 3 + y = 7, adding the coefficient instead of multiplying by it; x = 3, y = −1 comes from substituting into x − y = 2 as though it read x + y = 2.
- (a) 52 — Test successive terms: n=6 gives 6²+3=39, which is not greater than 50. n=7 gives 7²+3=52, which is greater than 50, so the first term greater than 50 is 52. A candidate who stops at n=6, before checking whether 39 actually exceeds 50, would give 39. A candidate who solves n²>50 instead of n²+3>50, ignoring the +3 in the search, would find n=8 is the first value with n²>50 (since 7²=49) and compute 8²+3=67. A candidate who computes n² by doubling n instead of squaring it would compute 2×7+3=17.
- (d) a = 4 — Expand the left-hand side: (2x + 3)(x + a) = 2x² + 2ax + 3x + 3a = 2x² + (2a + 3)x + 3a. For this to match 2x² + 11x + 12 for every value of x, the x-coefficients must be equal and the constants must be equal: 2a + 3 = 11 and 3a = 12. Both give a = 4, so a = 4. Writing a = 12 comes from the constant-term equation 3a = 12: reading it as saying a itself is 12, rather than dividing both sides by 3. Writing a = 8 comes from the x-coefficient equation 2a + 3 = 11: working out 11 − 3 = 8 correctly but then stopping, without dividing by the 2 in front of a. Writing a = −4 comes from rearranging 2a + 3 = 11 the wrong way round, as 2a = 3 − 11 = −8, which gives a = −4 instead of a = 4.
- (d) It crosses the x-axis at x = 3 and x = −3. — y = x² − 9 factorises as (x − 3)(x + 3), since 9 = 3², so the curve crosses the x-axis at x = 3 and x = −3. Saying it crosses once at x = 9 mistakes the constant term for a root directly, without taking its square root. Saying it crosses at x = 9 and x = −9 makes the same mistake but adds a sign either way. Saying it does not cross the x-axis confuses the y-intercept, which is negative at (0, −9), with the number of times the curve meets the x-axis — a negative y-intercept combined with an upward-opening curve guarantees it crosses the x-axis twice.
- (a) −10 — Method: multiply each bracket out, treating the second bracket as being multiplied by −3 because it is subtracted, then collect like terms. Working: 4(2x − 1) = 8x − 4 and −3(x + 2) = −3x − 6, so the expression becomes 8x − 4 − 3x − 6 − 5x; the x terms give 8x − 3x − 5x = 0, so no term in x survives, and the numbers give −4 − 6 = −10. Answer: −10. The distractors: 2 comes from expanding −3(x + 2) as −3x + 6, leaving the numbers −4 + 6; 5x − 10 comes from forgetting the final −5x, so the x terms give 8x − 3x = 5x; −7 comes from multiplying the 4 over only the first term of its bracket, giving 8x − 1 and so the numbers −1 − 6.
- (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) $y = x^2 + 3$ — Method: a point lies on a curve only if substituting its x-coordinate into the equation gives back its y-coordinate, so read one or two points off curve B and test each equation. Working: curve B crosses the y-axis at (0, 3), and its lowest point is also (0, 3); substituting x = 0 into y = x² + 3 gives 0² + 3 = 3, which matches. Checking a second point: at x = 2 curve B is at y = 7, and 2² + 3 = 4 + 3 = 7, which matches as well. Answer: curve B has equation y = x² + 3. Distractor refutation: y = x² − 3 comes from reading the 3 as a move down instead of a move up; substituting x = 0 gives −3, so that curve would cross the y-axis three squares below the origin, while curve B crosses it three squares above. y = (x − 3)² comes from putting the 3 inside the brackets; substituting x = 0 gives (−3)² = 9, and that curve's lowest point is at (3, 0), three squares to the right along the x-axis, whereas curve B has its lowest point on the y-axis. y = x² + 3x comes from attaching the 3 to the x term instead of writing it on its own; substituting x = 0 gives 0² + 3 × 0 = 0, so that curve passes through the origin, and curve B does not pass through the origin.
- (c) 12 cm — For a square, area = side². So side² = 144, giving side = ±12. Since a length must be positive, the side length is 12 cm. A candidate who gives both square roots without rejecting the negative one, which cannot be a length, answers 12 cm or −12 cm. A candidate who halves 144 instead of taking its square root gets 72 cm. A candidate who divides 144 by 4, confusing the area formula with a perimeter calculation, gets 36 cm.
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