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
MathsUKwww.geekhero.co.uk
- (b) 1.6 m/s² — Acceleration is the change in velocity divided by the time taken: (11 − 3) ÷ 5 = 8 ÷ 5 = 1.6 m/s². Forgetting to subtract the initial velocity and dividing the final velocity by the time instead gives 11 ÷ 5 = 2.2 m/s². Inverting the fraction, dividing the time by the change in velocity, gives 5 ÷ 8 = 0.625 m/s². Finding the change in velocity, 8 m/s, but stopping without dividing by the time gives 8 m/s².
- (c) 8n + 7 — Method: find the weekly increase, then find the constant by adjusting the week 1 total. Working: the total goes up by £8 each week (23 − 15 = 8), so the coefficient of n is 8. The constant is the week 1 total minus the common difference: 15 − 8 = 7. Answer: the nth term is 8n + 7. 8n + 15 comes from using the week 1 total, 15, as the constant without subtracting the common difference. 8n − 1 comes from a slip in working out the constant, subtracting the common difference twice (15 − 8 − 8 = −1) instead of once. 7n + 8 comes from swapping the weekly increase and the constant.
- (b) (n + 1)² − n² = 2n + 1 — (n + 1)² = n² + 2n + 1, so (n + 1)² − n² = n² + 2n + 1 − n² = 2n + 1, which is odd because it is one more than the even number 2n. Expanding (n + 1)² as n² + 1 uses the false rule (a + b)² = a² + b², and subtracting n² from that leaves just 1 — always expand (a + b)² as a² + 2ab + b². Writing n² + 2n + 1 expands correctly but never carries out the subtraction of n². Writing 2n forgets the constant term left after subtracting.
- (a) 6x + 12 — A regular hexagon has 6 equal sides, so the perimeter is 6(x + 2) = 6x + 12. A candidate who multiplies only the x-term by 6 and forgets to multiply the 2 gets 6x + 2. A candidate who multiplies only the number term by 6 and forgets to multiply the x gets x + 12. A candidate who adds 6 and 2 to make a single coefficient of x instead of expanding the brackets gets 8x.
- (d) £16.70 — The mileage charge is 2.20 × 6 = £13.20. Adding the fixed fee: £13.20 + £3.50 = £16.70. A candidate who forgets the fixed fee gives just the mileage charge, £13.20. A candidate who adds the fixed fee to the per-mile rate before multiplying by the number of miles, (3.50 + 2.20) × 6, gets £34.20. A candidate who rounds £2.20 down to £2 gets 2 × 6 + 3.50 = £15.50.
- (a) 2 — The candle's height falls from 30 cm to 20 cm, a drop of 10 cm, over 5 minutes, so m = 10 ÷ 5 = 2. 10 comes from using the drop in height but forgetting to divide by the time. 0.5 comes from dividing the time by the drop instead of the drop by the time (5 ÷ 10). 4 comes from dividing the final height by the time, 20 ÷ 5, instead of using the drop in height.
- (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) 3 and 4 — f(3) = 3³ − 3 × 3² − 4 = 27 − 27 − 4 = −4, and f(4) = 4³ − 3 × 4² − 4 = 64 − 48 − 4 = 12. Since f(3) is negative and f(4) is positive, there is a change of sign, so the solution lies between 3 and 4. A sign slip when expanding −3x², treating it as +3x², gives f(3) = 27 + 27 − 4 = 50, a positive value; taken with the correctly negative f(2) = 8 − 12 − 4 = −8, this reads as a change of sign between 2 and 3. The same sign slip applied at x = 1 gives f(1) = 1 + 3 − 4 = 0, read as the point the sign changes, against f(0) = 0 + 0 − 4 = −4, giving 0 and 1. Shifting the correctly found interval up by one integer, an indexing slip, reports the change as lying between 4 and 5 instead of 3 and 4.
- (b) 4 — Adding the two equations: the y-terms, −2y and +2y, cancel, and the x-terms combine to 5x + 3x = 8x; the right-hand sides add to 16 + 16 = 32. This gives 8x = 32, so x = 4. A candidate who adds only one of the right-hand sides, instead of both, would get 8x = 16, so x = 2. A candidate who divides 32 by 4 instead of 8 would get x = 8. A candidate who subtracts the equations instead of adding them, getting 2x − 4y = 0, and then wrongly assumes y = 0, would get x = 0.
- (b) 8 — Method: a circle centred on the origin has equation x² + y² = r², where r is the radius, so the number on the right-hand side is the square of the radius and not the radius itself. Working: comparing x² + y² = 64 with x² + y² = r² gives r² = 64, so r = √64 = 8. Answer: the radius is 8. The distractors: 64 is r² read straight off the equation as though the right-hand side were the radius, which is the commonest error on this form; 32 comes from halving 64, treating the right-hand side as a diameter that has to be halved; 16 is the diameter, 2 × 8, quoted in place of the radius.
- (a) y = −2x − 1 — Method: the gradient is the change in y divided by the change in x with both differences taken in the same order, and the constant then comes from substituting either point into y = mx + c. Working: m = (−9 − 3) ÷ (4 − (−2)) = (−12) ÷ 6 = −2, so the line is y = −2x + c; substituting (−2, 3) gives 3 = −2 × (−2) + c = 4 + c, so c = 3 − 4 = −1. Answer: y = −2x − 1. The distractors: y = −2x + 1 comes from rearranging 3 = 4 + c the wrong way round and taking the constant as 4 − 3; y = 2x + 7 comes from losing the minus sign when −12 is divided by 6 and then substituting correctly, 3 = 2 × (−2) + c; y = −(1/2)x + 2 comes from writing the gradient upside down as the change in x over the change in y, 6 ÷ (−12).
- (d) t = 5 — The maximum height occurs halfway between the two times when the ball is at ground level: the midpoint of t = 1 and t = 9 is (1 + 9) ÷ 2 = 5, so the ball reaches its maximum height at t = 5 seconds. Getting t = 4 comes from halving the DIFFERENCE between the times, 9 − 1 = 8, then 8 ÷ 2 = 4, instead of finding their midpoint. Getting t = 8 uses that difference, 9 − 1 = 8, as if the gap between the two times were itself the time of the maximum. Getting t = 10 adds the two times, 1 + 9 = 10, but forgets to divide by 2.
- (a) 2/5 — Method: compare the equation with y = mx + c, where m is the gradient. Working: in y = (2x/5) − 3, the coefficient of x is 2/5. Answer: the gradient is 2/5. −3 comes from confusing the gradient with the y-intercept. 5/2 comes from inverting the fraction that multiplies x. −2/5 comes from wrongly carrying the negative sign from the −3 term onto the coefficient of x.
- (a) 4 — Rectangle 1: P = 2(9 + 4) = 2 × 13 = 26 cm. Rectangle 2: P = 2(6 + 5) = 2 × 11 = 22 cm. Difference = 26 − 22 = 4 cm. 2 comes from finding the difference between (l + w) for each rectangle, 13 − 11 = 2, but forgetting to double it. 26 comes from giving rectangle 1's whole perimeter instead of the difference between the two. 6 comes from doubling only the change in length, 2 × (9 − 6), and ignoring that the width also changed.
- (b) y = 1/x — A graph with two separate curved branches must have a value of x that is excluded from it, creating a break in the curve. For y = 1/x, x = 0 is excluded, so the graph splits into one branch where x and y are both positive, and one where x and y are both negative. y = x² is a single U-shaped curve with no break, even though it has two 'arms'. y = x³ is one continuous curve that passes through the origin without any break. y = 2x + 1 is a single straight line; even though it passes through regions where x and y share the same sign, it is one continuous line, not two separate branches.
Build your own mix at the worksheet builder.