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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- (a) £14.50 — Method: the first 200 minutes are already covered by the £12, so only 250 − 200 = 50 minutes are charged extra. Extra charge = 50 × £0.05 = £2.50. Total cost = £12 + £2.50 = £14.50. Distractor origins: £24.50 charges 5p for all 250 minutes instead of only the 50 minutes over 200 (250 × £0.05 = £12.50, plus £12 = £24.50); £12.50 makes that same slip of charging all 250 minutes but then forgets to add the £12 monthly fee; £13.50 works out 250 − 200 wrongly as 30 extra minutes instead of 50 (30 × £0.05 = £1.50, plus £12 = £13.50).
- (d) −2 — Method: run the machine backwards, undoing the operations in the opposite order and swapping each one for its inverse. Working: the machine added 7 last, so take 7 off the output: 1 − 7 = −6; before that the machine had multiplied by 3, so divide: −6 ÷ 3, and a negative divided by a positive stays negative. Answer: −2, which checks because 3 × (−2) + 7 = −6 + 7 = 1. The distractors: 2 comes from dividing 6 by 3 and losing the minus sign; −6 comes from taking the 7 off and stopping there, never undoing the multiplication; −18 comes from multiplying −6 by 3 instead of dividing by 3.
- (c) t = 1 second and t = 5 seconds — The ball is at ground level when h = 0, which happens when either bracket is zero. t − 1 = 0 gives t = 1, and 5 − t = 0 gives t = 5. The option with t = −1 makes a sign error solving the first bracket. The option with t = −5 makes a sign error solving the second bracket, treating 5 − t = 0 as if it gave a negative solution. The option with t = 4 seconds is an arithmetic slip in solving 5 − t = 0.
- (a) 7 — Method: multiply both sides by 3 to clear the fraction, then solve the resulting equation. Working: 2x + 1 = 5 × 3 = 15. Subtract 1: 2x = 14. Divide by 2: x = 7. Answer: 7. 2 comes from ignoring the denominator altogether, treating the equation as 2x + 1 = 5 without multiplying by 3 first. 8 comes from a sign error, adding 1 to 15 instead of subtracting it, giving 2x = 16. 14 comes from correctly reaching 2x = 14 but stopping there, without dividing by 2 to find x.
- (a) x > 4 — Subtract 2x from both sides: 3x − 3 > 9. Add 3 to both sides: 3x > 12. Divide both sides by 3: x > 4. A candidate who subtracts 3 from 9 instead of adding gets 3x > 6, so x > 2. A candidate who divides correctly but wrongly flips the inequality (as if dividing by a negative) gets x < 4. A candidate who multiplies by 3 instead of dividing gets x > 36.
- (d) y = −2x + 10 — Method: a tangent is perpendicular to the radius drawn to the point where it touches, so work out the gradient of that radius, take its negative reciprocal for the tangent, then substitute into y − y₁ = m(x − x₁). Working: the radius joins (0, 0) to (4, 2), so its gradient is 2 ÷ 4 = 1/2; turning 1/2 upside down gives 2 and changing the sign gives −2. Substituting into y − 2 = −2(x − 4) gives y − 2 = −2x + 8, so y = −2x + 10. Answer: y = −2x + 10. The distractors: y = −0.5x + 4 changes the sign of the radius gradient but never turns it upside down, using −1/2 where −2 belongs; y = 2x − 6 turns the gradient upside down but leaves it positive, using 2 where −2 belongs; y = −2x − 10 has the correct gradient but substitutes the point with both signs reversed, writing y + 2 = −2(x + 4) instead of y − 2 = −2(x − 4).
- (d) 2 — The gradient between two points on a line is the change in y divided by the change in x. Using (5, 2) and (0, −8): (2 − (−8)) ÷ (5 − 0) = 10 ÷ 5 = 2. Inverting the fraction, dividing the change in x by the change in y instead, gives 5 ÷ 10 = 0.5. Reversing the order of the x-values in the denominator, giving (2 − (−8)) ÷ (0 − 5) = 10 ÷ (−5), gives −2. Misreading the y-intercept as 8 instead of −8, giving (2 − 8) ÷ 5 = −6 ÷ 5, gives −1.2.
- (b) Student B's — narrower strips fit the curve more closely. — The trapezium rule replaces the curve with straight-line segments; the narrower each strip, the more closely its straight edge follows the curve, so Student B's estimate with 8 narrower strips is more likely to be closer to the true area. Reasoning from the number of arithmetic steps rather than from how well the straight lines fit the curve gives the wrong claim that Student A makes fewer rounding errors. Believing the trapezium rule is exact, rather than an estimate that improves with narrower strips, gives the wrong claim that the two are always the same. Believing wider strips smooth out the curve better, rather than following it less closely, gives the wrong claim in favour of Student A's wider strips.
- (d) 680 m — First find how long the acceleration takes: acceleration = change in speed ÷ time, so 2.5 = 20 ÷ t, giving t = 20 ÷ 2.5 = 8 seconds. The distance during this phase is the area of a triangle with base 8 and height 20: 1/2 × 8 × 20 = 80 m. The distance during the constant-speed phase is 20 × 30 = 600 m, since distance = speed × time at a constant speed. The total distance is 80 + 600 = 680 m. Using the given 30 seconds for the acceleration phase as well, 1/2 × 30 × 20 = 300, plus the correct 600, gives 900 m — but 30 seconds is only stated for the constant-speed phase. Leaving out the 1/2 and using the full rectangle for the acceleration phase, 8 × 20 = 160, plus the correct 600, gives 760 m — the speed is not constant during acceleration, so this area is a triangle, not a rectangle. Swapping the two times round, and using 8 seconds for the constant-speed distance instead of 30, 20 × 8 = 160, plus the correct triangle area of 80, gives 240 m.
- (d) x = 0, y = −1 and x = 3, y = 5 — Set the two expressions for y equal: 2x − 1 = x² − x − 1. Rearranging, subtracting 2x and adding 1 to both sides: 0 = x² − x − 1 − 2x + 1 = x² − 3x, so x² − 3x = 0. Factorise: x(x − 3) = 0, giving x = 0 or x = 3. Using y = 2x − 1: x = 0 gives y = −1; x = 3 gives y = 5. Distractor routes: x = 0, y = −1 alone stops after the factor x = 0 and never checks the second factor, x − 3 = 0. x = 0, y = −1 and x = −3, y = −7 comes from mis-factorising x² − 3x as x(x + 3), a sign error that gives a second root of −3 instead of 3. x = −2, y = −5 and x = 1, y = 1 comes from adding 2x to both sides instead of subtracting it when rearranging, giving x² + x − 2 = 0 instead of x² − 3x = 0.
- (b) (x + 2)/3 — Start with y = 3x − 2 and swap x and y: x = 3y − 2. Add 2 to both sides: x + 2 = 3y. Divide both sides by 3: y = (x + 2)/3, so f⁻¹(x) = (x + 2)/3. Writing x/3 + 2 comes from dividing only the 3y term by 3 and leaving the +2 outside the division — the 2 must be added before you divide, not after. Writing (x − 2)/3 comes from keeping the subtraction sign instead of flipping it to addition when the −2 is moved across the equals sign. Writing 3x + 2 comes from swapping x and y but never actually solving for y — just changing the sign of the constant term.
- (d) 4.521 — x₁ = √(3 × 4 + 7) = √19 = 4.358898944. x₂ = √(3 × 4.358898944 + 7) = √20.076696833 = 4.480702716. x₃ = √(3 × 4.480702716 + 7) = √20.442108148 = 4.521294964, which rounds to 4.521. Mislabelling the starting value x₀ as x₁, so that the working stops one iteration too early, reports the true x₂ = 4.480702716, which rounds to 4.481. Working out one iteration too many reports the true x₄ = √(3 × 4.521294964 + 7) = 4.534741987, which rounds to 4.535. Applying the formula in the wrong order, calculating √(3xₙ) + 7 at every step instead of √(3xₙ + 7), gives, from x₀ = 4: √12 + 7 = 10.464101615, then √(3 × 10.464101615) + 7 = 12.602883619, then √(3 × 12.602883619) + 7 = 13.148873950, which rounds to 13.149.
- (a) 40 — From 9am to 12 noon is 3 hours, so the population doubles three times: 5 × 2³ = 40. A candidate who counts the elapsed time as 4 hours (an off-by-one counting error) would reach 5 × 2⁴ = 80. A candidate who counts it as only 2 hours would reach 5 × 2² = 20. A candidate who misreads 'doubles' as 'increases by 2' each hour would compute 5 + 3 × 2 = 11.
- (c) x + 2y = 6 — L has gradient 2, so the perpendicular gradient is −1/2. Substituting (4, 1) into y − 1 = −(1/2)(x − 4): y = −(1/2)x + 2 + 1 = −(1/2)x + 3, so 2y = −x + 6, giving x + 2y = 6. Distractor routes: x − 2y = 2 comes from using gradient +1/2 instead of −1/2, forgetting the negative sign the perpendicular rule needs. x + 2y = 9 comes from swapping the point's coordinates, substituting (1, 4) instead of (4, 1). x + 2y = −2 comes from a sign error distributing the negative gradient over the bracket, computing −(1/2)(x − 4) as −(1/2)x − 2 instead of −(1/2)x + 2.
- (c) (3, 9) — Reflecting in the horizontal line y = 2 keeps the x-coordinate the same and maps y to 2 × 2 − y = 4 − (−5) = 9, so S = (3, 9). A candidate who uses k − y instead of 2k − y gets 2 − (−5) = 7, giving (3, 7). A candidate who reflects in the x-axis instead of the line y = 2, simply changing the sign of y, gets (3, 5). A candidate who also changes the sign of the x-coordinate, as if reflecting in both axes, gets (−3, 9).
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