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.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.
- 2.The curve y = −x² + 6x − 5 has a maximum point. Use completing the square to find its coordinates.y = −x² + 6x − 5
- 3.Which expression means 'y squared, multiplied by 3'?
- 4.Solve (3x + 2)/5 = (x + 4)/3
- 5.A car's speed increases steadily while it accelerates. Its speed v m/s after t seconds is given by v = u + at, where u is the starting speed in m/s and a is the acceleration in m/s². A car starts at u = 4 m/s and accelerates at a = 2 m/s² until it reaches v = 20 m/s. Work out t, the time taken in seconds.
- 6.Point P has coordinates (5, 2). Point P is rotated 180° about the origin to point Q. Write down the coordinates of Q.
- 7.A student solves the simultaneous equations 2x + y = 11 and x − y = 1 by elimination, adding the two equations together. Which of these is the correct result of that step?
- 8.The curve y = x² − 3x. Use the chord between x = 1 and x = 4 to estimate the gradient of the curve at x = 2.5.y = x² − 3x
- 9.Solve 2x² = 18.
- 10.The perimeter of a rectangle is given by the formula P = 2(l + w), where l is the length and w is the width. A rectangular garden has a perimeter of 46 m and a length of 14 m. Rearrange the formula to make w the subject, then work out the width of the garden.
- 11.The trapezium rule is used to estimate the area under a curve between x = 2 and x = 6. The curve is concave up (it curves upwards, like the inside of a bowl) throughout this interval. Which statement about the estimate is correct?
- 12.A zip-line runs from a platform 25 m high to a landing point 5 m high, over a horizontal distance of 40 m. Work out the gradient of the zip-line.
- 13.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.
- 14.The nth term of a sequence is n² + 4n. Work out the term number, n, for which the term equals 45.
- 15.A cinema sells adult tickets for £10 and child tickets for £6. A family group buys 9 tickets in total, spending £74. Work out how many adult tickets were bought.
Answer key
- (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) x = 3, y = 4 — −x² + 6x − 5 = −(x² − 6x) − 5 = −[(x − 3)² − 9] − 5 = −(x − 3)² + 9 − 5 = −(x − 3)² + 4. Because the coefficient of x² is negative, −(x − 3)² is at most zero, so this turning point is a maximum. Substituting x = 3 gives (x − 3)² = 0, so y = 4, confirming the maximum value 4 at x = 3: turning point (3, 4). Using 6 instead of half of 6 inside the bracket — forgetting to halve before completing the square — lands on turning point (6, 31), wrong, because only half the coefficient belongs inside the bracket. Reading the bracket's sign directly as the turning point's x-coordinate gives (−3, 4) — wrong, because (x − 3)² is zero at x = 3, not x = −3. Computing 5 − 9 = −4 instead of 9 − 5 = 4 flips the sign of the constant, giving (3, −4) — wrong, since the constant must be evaluated as 9 minus 5, not 5 minus 9. Whenever the leading coefficient is negative, the turning point is a maximum, not a minimum — check by substituting back into the original equation.
- (a) 3y² — 'y squared, multiplied by 3' means the square is applied to y only, and the result is then multiplied by 3, written as 3y². Writing y³ mistakes the multiplication by 3 for an extra factor of y, adding to the power instead of using a coefficient. Writing (3y)² squares the whole of 3y, including the 3, which gives 9y² rather than 3y² — the square should apply to y alone. Writing 3 + y² adds the 3 instead of multiplying by it. The expression for 'y squared, multiplied by 3' is 3y².
- (a) x = 3.5 — Method: clear the fractions by multiplying both sides by 15, expand both brackets, then collect the x terms on one side and the numbers on the other. Working: multiplying both sides by 15 gives 3(3x + 2) = 5(x + 4), which expands to 9x + 6 = 5x + 20; subtracting 5x and 6 from both sides gives 4x = 14, and dividing both sides by 4 gives x = 3.5. Answer: x = 3.5. The distractors: x = 1 comes from collecting the x terms by adding the 5x instead of subtracting it, giving 9x + 5x = 20 − 6 and so 14x = 14; x = 4.5 comes from expanding 3(3x + 2) as 9x + 2, multiplying only the x term by the 3, which leads to 4x = 18; x = 10 comes from reaching 4x = 14 correctly and then subtracting 4 instead of dividing by 4.
- (b) 8 s — Rearranging v = u + at for t: subtract u from both sides to get v − u = at, then divide by a: t = (v − u)/a. Substituting u = 4, a = 2, v = 20: t = (20 − 4)/2 = 16/2 = 8 s. Answering 12 s comes from adding u instead of subtracting it: (20 + 4)/2 = 12. Answering 32 s comes from multiplying (v − u) by a instead of dividing by it: 16 × 2 = 32. Answering 6 s divides v by a first and then subtracts u, in the wrong order: 20/2 − 4 = 10 − 4 = 6. The time taken is 8 s.
- (b) (−5, −2) — A rotation of 180° about the origin reverses the sign of both coordinates, so Q = (−5, −2). A candidate who reverses the sign of only the y-coordinate, as if reflecting in the x-axis, gets (5, −2). A candidate who reverses the sign of only the x-coordinate, as if reflecting in the y-axis, gets (−5, 2). A candidate who swaps the coordinates instead of reversing their signs gets (2, 5).
- (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.
- (c) 2 — At x = 1, y = 1² − 3(1) = 1 − 3 = −2. At x = 4, y = 4² − 3(4) = 16 − 12 = 4. Gradient of the chord = change in y ÷ change in x = (4 − (−2)) ÷ (4 − 1) = 6 ÷ 3 = 2. Writing down the change in y, 6, and stopping there without dividing by the change in x gives 6. Losing the negative sign on y = −2 at x = 1 and treating it as +2 gives (4 − 2) ÷ (4 − 1) = 2 ÷ 3 = 2/3. Dividing the wrong way round, change in x ÷ change in y, gives (4 − 1) ÷ (4 − (−2)) = 3 ÷ 6 = 1/2.
- (d) x = 3 or x = −3 — Method: get x² on its own with a coefficient of 1, then take the square root of both sides and keep both the positive and the negative root. Working: dividing 2x² = 18 by 2 gives x² = 9, and the square root of 9 is 3, so x = 3 or x = −3; both check, because 2 × 9 = 18 either way. Answer: x = 3 or x = −3. The distractors: x = 9 or x = −9 comes from dividing by 2 and then forgetting to take the square root; x = 4 or x = −4 comes from subtracting 2 from 18 instead of dividing by it, giving x² = 16; x = 6 or x = −6 comes from multiplying by 2 instead of dividing, giving x² = 36.
- (a) 9 m — Undo the multiplication by the bracket first: dividing both sides by 2 gives P/2 = l + w. Subtracting the length from both sides gives w = P/2 − l. Substituting the measurements, 46 ÷ 2 = 23, and 23 − 14 = 9, so the width is 9 m. Taking the length off before halving gives (46 − 14) ÷ 2 = 16, which halves the length as well; expanding to P = 2l + 2w and then forgetting to divide by 2 gives 46 − 28 = 18; subtracting the length in the wrong direction gives 23 + 14 = 37.
- (a) Overestimates — trapezium edges lie above the curve. — For a concave-up curve, each chord connecting two points on the curve lies ABOVE the curve between those points, because the curve bends away from the chord underneath it. The trapezium rule uses these chords as the top edges of each trapezium, so the trapeziums cover a larger region than the true area under the curve — the estimate is an overestimate. Saying it underestimates because the edges lie below the curve gets the direction backwards: for a concave-up curve the edges lie above, not below. Saying it overestimates 'because more strips always increase it' gives the right verdict for the wrong reason — using MORE, narrower strips makes a trapezium estimate more accurate and brings it closer to the true area, it does not simply increase the total. Saying the estimate is exact ignores that straight edges cannot follow a curved line perfectly; there is always a gap between the chord and the curve on a bend, which is exactly what makes the rule an estimate rather than an exact calculation.
- (b) −0.5 — Method: gradient = change in height ÷ horizontal distance, and the height decreases so the change is negative. Working: change in height = 5 − 25 = −20, horizontal distance = 40, so gradient = −20 ÷ 40 = −0.5. Answer: the gradient is −0.5. 0.5 comes from dropping the negative sign, ignoring that the zip-line descends. 2 comes from inverting the gradient, dividing the horizontal distance by the drop in height instead of the other way round. −0.8 comes from dividing the drop by the platform height, 25, instead of by the horizontal distance, 40.
- (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.
- (b) 5 — Set n² + 4n = 45, so n² + 4n − 45 = 0. This factorises as (n + 9)(n − 5) = 0, giving n = −9 or n = 5. Since a term number must be positive, n = 5. Taking the magnitude of the rejected negative solution, 9, instead of discarding it, gives 9. Dividing 45 by the coefficient of n and ignoring the n² term entirely, 45 ÷ 4 = 11.25, rounded to the nearest whole number, gives 11. Dropping the linear term 4n and solving n² = 45 instead, the nearest whole number to √45 = 6.708 is 7.
- (c) 5 — Method: let a be the number of adult tickets, so the number of child tickets is 9 − a. Form the equation 10a + 6(9 − a) = 74. Working: expand the bracket: 10a + 54 − 6a = 74, so 4a = 20, giving a = 5. Answer: 5 adult tickets. 4 comes from correctly finding the number of child tickets but reporting it instead of the number of adult tickets asked for. 1.25 comes from a sign error expanding the bracket, writing 10a + 54 + 6a = 74 instead of subtracting, which gives 16a = 20. 7.4 comes from dividing the total takings by the adult ticket price only, ignoring the children entirely, 74 ÷ 10.
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