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
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- 1.A student says that (x + 4)² is equivalent to x² + 16. For which value of x do the two expressions give the SAME result, making it look (misleadingly) like the student could be right?
- 2.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?
- 3.f(x) = x + 2 and g(x) = x². Work out the value of x for which fg(x) = gf(x).y = x + 2
- 4.There are 30 students in a Year 10 maths class. There are 2 more boys than girls. Work out the number of boys and the number of girls.
- 5.The point A(−6, 8) lies on the circle x² + y² = 100, whose centre is the origin O. The tangent to the circle at A crosses the y-axis at the point B. Work out the length of OB.
- 6.Simplify fully: (x + 2)/(3x) × 6x²/(x² − 4)
- 7.Work out the value of 5m² + 3m when m = −2.4
- 8.On a map, where each grid unit represents 1 km, Ravi's house is at (−3, 4) and the library is at (6, 4). Work out the distance between Ravi's house and the library.
- 9.The equation x² − x − 6 = 0 has roots x = 3 and x = −2. It can be rearranged as xₙ₊₁ = xₙ² − 6. This formula is used with starting value x₀ = 2.9, close to the root x = 3. Work out what happens to the sequence of values as n increases.
- 10.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.
- 11.A closed cylinder has radius r cm and height (r + 5) cm. Its volume is 300 cm³, giving the equation πr²(r + 5) = 300, which can be solved using the iterative formula rₙ₊₁ = √(300 ÷ (π(rₙ + 5))). Taking r₀ = 3, work out r₃ correct to 2 decimal places.
- 12.The graph of y = f(x) has x-intercepts at x = −1 and x = 4. Which statement correctly describes the x-intercepts of y = f(2 − x)?
- 13.A runner's journey is shown on a distance-time graph: she runs 8 km in the first hour, rests for 30 minutes with no distance gained, then runs a further 4 km in the next 30 minutes. Work out her average speed for the whole journey, in kilometres per hour.
- 14.Simplify (5x² + 3x − 2) − (2x² − x + 5)
- 15.A rectangular garden has width w metres and length (w + 3) metres. A gardener writes its perimeter as 2w + 3. Which statement corrects the gardener's mistake?
Answer key
- (c) x = 0 — Expand (x + 4)² correctly: (x + 4)² = x² + 8x + 16. This equals x² + 16 only when 8x is zero, i.e. when x = 0 — at every other value of x the two expressions differ by 8x. Choosing x = 4 confuses the constant inside the bracket with the value of x that makes the expressions match. Choosing x = −4 makes the same confusion but with the sign flipped. Choosing x = 8 mistakes the coefficient of the middle term, 8x, for the value of x itself.
- (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) −0.5 — fg(x) = f(g(x)) = f(x²) = x² + 2. gf(x) = g(f(x)) = g(x + 2) = (x + 2)² = x² + 4x + 4. Setting fg(x) = gf(x): x² + 2 = x² + 4x + 4. Subtract x² from both sides: 2 = 4x + 4. Subtract 4 from both sides: −2 = 4x, so x = −0.5. Writing 1.5 comes from adding the 4 instead of subtracting it: 4x = 2 + 4 = 6, giving x = 1.5. Writing 'no solution' comes from expanding (x + 2)² as x² + 4 using (a + b)² = a² + b², losing the middle term — the equation then reads x² + 2 = x² + 4, which has no solution, but the expansion itself is wrong. Writing 0 comes from treating gf(x) as g(x) + f(x) instead of g(f(x)): x² + (x + 2) = x² + 2 gives x = 0, but that adds the two functions rather than composing them.
- (d) 16 boys and 14 girls — Method: write the number of boys in terms of the number of girls, then use the total for the class. Working: if there are g girls then there are g + 2 boys, so g + (g + 2) = 30, that is 2g + 2 = 30, so 2g = 28 and g = 14; the number of boys is 14 + 2 = 16. Answer: 16 boys and 14 girls, which total 30 and differ by 2. The distractors: 14 boys and 16 girls comes from substituting for the wrong group, writing b + (b + 2) = 30 and then calling b the number of boys; 15 boys and 15 girls comes from halving 30 and never using the difference; 17 boys and 13 girls comes from adding the whole 2 to one half of the class and taking the whole 2 off the other half, which leaves a difference of 4.
- (a) 12.5 — Method: the tangent at A is perpendicular to the radius OA, so find the gradient of OA, take its negative reciprocal, write the equation of the tangent and find where it meets the y-axis; the length of OB is then the distance of that crossing from the origin. Working: OA runs from (0, 0) to (−6, 8), so its gradient is 8 ÷ (−6), which cancels to −4/3; the negative reciprocal of −4/3 is 3/4. Substituting into y − 8 = 3/4(x + 6) gives y = 0.75x + 4.5 + 8, so y = 0.75x + 12.5 and B is (0, 12.5). The length OB is therefore 12.5. Answer: 12.5. The distractors: 10 is the radius of the circle, quoted on the assumption that the tangent always meets an axis one radius from the centre, which is only true when the radius itself lies along that axis; 8 is the y-coordinate of A, quoted by treating the tangent as horizontal so that it keeps the height of A; 3.5 comes from turning the gradient of OA upside down without changing its sign, which gives y = −0.75x + 3.5.
- (d) 2x/(x − 2) — Factorise x² − 4 as (x − 2)(x + 2) first — it's a difference of two squares. The (x + 2) in the numerator then cancels with the (x + 2) in the factorised denominator, and 6x² ÷ 3x simplifies to 2x, leaving 2x/(x − 2). Writing 2/(x − 2) comes from over-cancelling 6x² ÷ 3x as 2 instead of 2x, dropping the x that should remain. Writing 2x/(x + 2) comes from factorising x² − 4 as (x + 2)² instead of (x − 2)(x + 2), a difference of two squares always has one plus and one minus bracket. Writing 2x²/(x − 2) comes from simplifying 6x² ÷ 3x as 2x² instead of 2x, dividing the coefficients but not reducing the power of x.
- (c) 21.6 — m² = (−2.4)² = 5.76, since a negative number squared is positive, so 5m² = 28.8. Then 3m = 3 × (−2.4) = −7.2. Total: 28.8 + (−7.2) = 21.6. −36 comes from taking m² as negative, 5 × (−5.76) − 7.2 = −36, forgetting that squaring removes the negative sign. 36 comes from ignoring the negative sign on m altogether and using m = 2.4 throughout. 29.4 comes from treating 3m as 3 + m instead of 3 × m, giving 28.8 + (3 + (−2.4)).
- (a) 9 km — Ravi's house and the library have the same y-coordinate, so the distance between them is the difference between their x-coordinates: 6 − (−3) = 9 km. 3 km comes from dropping the negative sign on Ravi's x-coordinate and working out 6 − 3 instead. 0 km comes from using the y-coordinates, which are equal, instead of the x-coordinates. −9 km comes from working out −3 − 6 = −9 and not converting it to a positive distance.
- (c) The sequence diverges, moving away from x = 3 — Starting from x₀ = 2.9: x₁ = 2.9² − 6 = 2.41, x₂ = 2.41² − 6 = −0.19, x₃ = (−0.19)² − 6 = −5.96, x₄ = (−5.96)² − 6 = 29.56 — the values swing away from 3 and grow rapidly, so the sequence diverges rather than settling anywhere. Choosing 'settles towards x = 3' assumes that starting close to a root is enough for a rearrangement to converge to it, which is not always true — this rearrangement changes values too steeply near x = 3 to stay there. Choosing 'settles towards x = −2' assumes a diverging sequence must eventually land on the other root; instead it runs away to increasingly large values. Choosing 'stays constant at 2.9' ignores that applying the formula changes the value at every step.
- (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.
- (b) 3.38 — r₁ = √(300 ÷ (π × 8)) = √11.9366 = 3.4550. r₂ = √(300 ÷ (π × 8.4550)) = √11.2947 = 3.3608. r₃ = √(300 ÷ (π × 8.3608)) = √11.4232 = 3.3798, which rounds to 3.38. Choosing 3.36 stops at r₂, one iteration too early. Choosing 4.82 leaves out the '+ 5' inside the bracket, dividing by π × rₙ instead of π × (rₙ + 5). Choosing 3.45 comes from using π ≈ 3 instead of the calculator's π key throughout.
- (d) Reflect in the y-axis, +2 in x; roots 3, −2 — f(2 − x) is zero exactly when 2 − x equals one of f's roots: 2 − x = −1 or 2 − x = 4. Solving each correctly (x = 2 − (−1) = 3, and x = 2 − 4 = −2) gives the new roots x = 3 and x = −2. This is the same as reflecting y = f(x) in the y-axis to get f(−x), then translating 2 units in the positive x-direction to get f(−(x − 2)) = f(2 − x). Solving 2 − x = k as x = k − 2 instead of x = 2 − k is a sign slip when rearranging, and gives x = −3 and x = 2. Translating +2 in x with no reflection at all uses f(x − 2), whose roots are the original roots plus 2: x = 1 and x = 6 — this misses the reflection completely. Assuming 'no overall change' wrongly treats a reflection-and-translation pair as always cancelling out, when here the roots genuinely move, from x = −1 and x = 4 to x = 3 and x = −2.
- (c) 6 km/h — Total distance = 8 + 4 = 12 km. Total time = 1 hour + 0.5 hours resting + 0.5 hours = 2 hours. Average speed = total distance ÷ total time = 12 ÷ 2 = 6 km/h. A speed of 8 km/h comes from leaving the resting time out of the total time: 12 ÷ 1.5 = 8. A speed of 4 km/h comes from dividing by too much time, such as double-counting the rest period: 12 ÷ 3 = 4. A speed of 12 km/h simply gives the total distance and forgets to divide by the total time at all.
- (d) 3x² + 4x − 7 — Method: the minus sign in front of the second bracket changes the sign of every term inside it; then collect like terms. Working: removing the brackets gives 5x² + 3x − 2 − 2x² + x − 5; the squared terms give 5x² − 2x² = 3x², the x terms give 3x + x = 4x, and the number terms give −2 − 5 = −7. Answer: 3x² + 4x − 7. The distractors: 7x² + 2x + 3 comes from adding the two brackets instead of subtracting, giving 5x² + 2x², 3x − x and −2 + 5; 3x² + 2x + 3 comes from applying the minus sign to 2x² only, leaving −x and +5 unchanged so that 3x − x = 2x and −2 + 5 = 3; 3x² + 4x + 3 comes from changing the signs of the terms with letters but leaving +5 as it stood, so the number terms give −2 + 5 = 3.
- (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.
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