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.
Algebra worksheet — GCSE Higher
MathsUKwww.geekhero.co.uk
- 1.Work out the equation of the straight line through the points (−3, 4) and (1, −8).
- 2.The nth term of a sequence is 3n² + 2n − 1. Work out the 10th term.
- 3.A fundraising chain letter starts with 3 people taking part in round 1. Each following round has 4 times as many people taking part as the round before. Work out the number of people taking part in round 4.
- 4.The simple interest, £I, earned on a savings account is given by the formula I = PRT/100, where P is the amount invested in pounds, R is the interest rate per year and T is the time in years. Rearrange the formula to make P the subject.
- 5.Solve 2(x + 3) = 10
- 6.The equation x³ − 5x − 3 = 0 can be rearranged to give an iterative formula of the form xₙ₊₁ = ∛(…). Work out which one of these is a correct rearrangement.
- 7.The curve y = 2x² − 12x + 7 has a minimum point. By completing the square, find the value of x at which the minimum occurs.y = 2x² − 12x + 7
- 8.There are 50 adults on a coach to the Lake District. Every adult is either a man or a woman. There are 10 more men than women. Work out the number of men.
- 9.A rectangle has width x cm. Its length is 3 cm more than its width. The perimeter of the rectangle is 38 cm. Form an equation and solve it to find x.
- 10.Expand and simplify 2(a + 6) − 2(a − 7)
- 11.The formula for the area of a trapezium is A = (a + b)h / 2, where a and b are the parallel side lengths and h is the height. Make h the subject of the formula.
- 12.For the equation 2x + 3 = 11, and the inequality 2x + 3 > 11, which statement correctly compares their solutions?
- 13.A line segment has one endpoint at (−6, 2) and its midpoint at (−1, 5). Work out the coordinates of the other endpoint.
- 14.Make x the subject of the formula y = 4(x − 6).
- 15.A car's speed, in m/s, during a journey is: 0 at t = 0 s, 8 at t = 4 s, and 8 (constant) from t = 4 s to t = 10 s, increasing at a constant rate between t = 0 and t = 4. Estimate the total distance the car travels, using the areas of a triangle and a rectangle.
Answer key
- (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.
- (b) 319 — Substitute n = 10: 3 × 10² + 2 × 10 − 1 = 3 × 100 + 20 − 1 = 300 + 20 − 1 = 319. A sign error on the +2n term, treating it as −2n, gives 300 − 20 − 1 = 279. Working out 3 × 10² + 2 × 10 but forgetting to subtract the final 1 gives 300 + 20 = 320. Using n = 9 instead of n = 10 gives 3 × 81 + 18 − 1 = 243 + 18 − 1 = 260.
- (b) 192 — This is a geometric sequence with first term 3 and common ratio 4: round 2 has 3 × 4 = 12, round 3 has 12 × 4 = 48, round 4 has 48 × 4 = 192. A candidate who applies the ×4 multiplier four times instead of three gets 3 × 4⁴ = 768. A candidate who wrongly treats the growth as arithmetic, taking the round 2 figure of 12 as a fixed amount added each round, gets 3, 15, 27, 39. A candidate who forgets the starting 3 people and just works out 4⁴ gets 256.
- (a) P = 100I / (RT) — Method: undo the operations done to P in reverse order — multiply by 100, then divide by R and by T. Working: I = PRT/100, so 100I = PRT, so P = 100I / (RT). Answer: P = 100I / (RT). P = IRT/100 comes from leaving R and T in the numerator instead of moving them to the denominator. P = 100RT/I comes from swapping P and I when rearranging. P = 100I/R comes from dividing by R only and forgetting to also divide by T.
- (d) x = 2 — Method: expand the bracket by multiplying both terms inside it by 2, then undo the addition and the multiplication in turn. Working: expanding gives 2x + 6 = 10; subtracting 6 from both sides gives 2x = 4; dividing both sides by 2 gives x = 2. Answer: x = 2. The distractors: x = 5 comes from dividing both sides by 2 first, reaching x + 3 = 5 and writing 5 without taking the 3 away; x = 8 comes from adding 6 to both sides instead of subtracting it, giving 2x = 16; x = 3.5 comes from expanding 2(x + 3) as 2x + 3, multiplying only the x by the 2, which leads to 2x = 7.
- (a) xₙ₊₁ = ∛(5xₙ + 3) — Starting from x³ − 5x − 3 = 0, add 5x and 3 to both sides to get x³ = 5x + 3, then take the cube root of both sides: x = ∛(5x + 3), giving the iterative formula xₙ₊₁ = ∛(5xₙ + 3). A sign error when moving the constant term across, treating x³ − 5x − 3 = 0 as x³ = 5x − 3, gives xₙ₊₁ = ∛(5xₙ − 3). Swapping the coefficient of x with the constant term gives xₙ₊₁ = ∛(3xₙ + 5), which does not come from x³ = 5x + 3 at all. Treating cubing as meaning multiply by 3 rather than raise to the power 3, and so undoing it by dividing by 3 instead of taking a cube root, gives xₙ₊₁ = (5xₙ + 3) ÷ 3.
- (d) 3 — 2x² − 12x + 7 rewrites as 2(x² − 6x) + 7, then as 2[(x − 3)² − 9] + 7, which simplifies to 2(x − 3)² − 11, since −2 × 9 + 7 = −11. The bracket (x − 3)² is zero when x = 3, so the minimum occurs at x = 3. Treating the shift as b/a instead of b/(2a) — using 6 instead of 3 — gives x = 6, which is wrong. Reading the bracket's sign directly without negating it gives x = −3, wrong, because (x − 3)² is zero at x = 3, not x = −3. Reading off the coefficient of x itself, −12, and calling that the answer skips the completing-the-square process entirely and gives x = −12, which is wrong because b is not the turning point's x-coordinate under any circumstance. Always check: substituting your value of x should make the bracketed term equal to zero, and nothing else.
- (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.
- (b) 8 — The perimeter is 2(x + (x + 3)) = 4x + 6, so 4x + 6 = 38, which gives 4x = 32 and x = 8. A candidate who forgets the '+3' and treats the rectangle as a square, solving 2(2x) = 38, gets x = 9.5. A candidate who forgets to double the sum of the sides, solving 2x + 3 = 38, gets x = 17.5. A candidate who uses 3x instead of x + 3 for the length, solving 2(x + 3x) = 38, gets x = 4.75.
- (d) 26 — Method: expand both brackets, treating the second as multiplication by −2 so that both of its terms change sign, then collect like terms. Working: 2(a + 6) = 2a + 12 and −2(a − 7) = −2a + 14, so the expression becomes 2a + 12 − 2a + 14; the a terms give 2a − 2a = 0, so no term in a survives, and the numbers give 12 + 14 = 26. Answer: 26. The distractors: −2 comes from expanding the second bracket as −2a − 14, so that the numbers give 12 − 14; 4a − 2 comes from adding 2(a − 7) instead of subtracting it, giving 2a + 12 + 2a − 14; 13 comes from multiplying the 2 over only the first term of each bracket, giving 2a + 6 − 2a + 7.
- (d) h = 2A / (a + b) — Method: undo the multiplication by 1/2 by multiplying both sides by 2, then undo the multiplication by (a + b) by dividing both sides by it. Working: A = (a + b)h / 2, so multiplying both sides by 2 gives 2A = (a + b)h, then dividing both sides by (a + b) gives h = 2A / (a + b). The value h = A / (a + b) comes from forgetting to multiply by 2 to clear the 1/2 first. The value h = 2A / a + b comes from dividing only by a and leaving b outside the fraction, instead of dividing by the whole bracket (a + b). The value h = (a + b) / (2A) comes from inverting the fraction.
- (a) The equation has one solution; the inequality has many. — Method: solve each statement. From 2x + 3 = 11, 2x = 8, so x = 4 — a single value. From 2x + 3 > 11, 2x > 8, so x > 4 — every number greater than 4 makes the inequality true, so there are many solutions. So the equation has one solution and the inequality has many. Distractor origins: swapping the two round gives the range to the equation and the single value to the inequality; saying both have exactly one solution treats the > sign as if it were an = sign; saying both have many solutions treats the equation as if it were an inequality.
- (d) (4, 8) — The other endpoint is found from 2 × midpoint − known endpoint: x = 2 × (−1) − (−6) = −2 + 6 = 4, y = 2 × 5 − 2 = 10 − 2 = 8, giving (4, 8). (−3.5, 3.5) comes from averaging the given endpoint and the midpoint as if they were the two endpoints of a segment, instead of working backwards from the midpoint. (5, 3) comes from working out (−1 − (−6), 5 − 2) instead of doubling the midpoint before subtracting. (4, 5) comes from correctly finding the x-coordinate but copying the midpoint's y-coordinate of 5 instead of doubling it.
- (c) x = (y + 24)/4 — Method: expand the bracket first, then undo the operations done to x in reverse order. Working: y = 4(x − 6) = 4x − 24, so y + 24 = 4x, so x = (y + 24)/4. Answer: x = (y + 24)/4. x = (y + 6)/4 comes from expanding the bracket incorrectly, treating 4(x − 6) as 4x − 6 instead of 4x − 24. x = 4y + 96 comes from multiplying by 4 instead of dividing by 4 to undo the multiplication, giving 4(y + 24) = 4y + 96. x = y/4 − 6 comes from dividing by 4 first without expanding the bracket, then subtracting 6 as if the bracket's operation still applied afterwards.
- (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.
Build your own mix at the worksheet builder.