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.A tree is 2 metres tall. Each year its height increases by 10% of its height at the start of that year. Work out the height of the tree after 3 years, giving your answer to 1 decimal place.
- 2.The tangent to a curve at the point (5, 2) is the line y = mx + c. This tangent crosses the y-axis at (0, −8). Work out the gradient, m, of the tangent.
- 3.A rectangle has length (2x + 5) cm and width (x − 2) cm. Work out an expression, in terms of x, for the perimeter of the rectangle. Give your answer in its simplest form.
- 4.A scout leader is 44 years old and one of the scouts is 16 years old. Work out how many years it will be until the leader is exactly twice as old as the scout.
- 5.Explain why the equation 4x + 3 = 4x + 10 has no solution for x.
- 6.A graph shows the total monthly cost of a mobile phone tariff against the number of minutes of calls. The line starts at £12 and stays level until 200 minutes, then rises by 5p for each further minute of calls. Use the graph to work out the total cost of a month in which 250 minutes of calls are made.
- 7.A circle has centre (0, 0) and passes through the point (12, 35). Work out the equation of the circle.
- 8.The graph of y = 20/x is drawn for x > 0. As the value of x increases, what happens to the value of y?
- 9.A stack of firewood has 3 logs in the top layer. Each layer below has 4 more logs than the layer above it. Work out the number of logs in the 6th layer from the top.
- 10.f(x) = x³ − 3x² − 4. Work out the pair of consecutive integers between which the solution of f(x) = 0 lies.y = x
- 11.A park is drawn on a grid in which 1 unit represents 1 km. The car park is at the point (0, 0) and the lake is at the point (3, 5). Work out the direct distance, in km, from the car park to the lake, giving your answer to 1 decimal place.
- 12.A ferry's distance-time graph is a straight line from (0 minutes, 0 km) to (25 minutes, 15 km). Work out the ferry's speed in kilometres per hour.
- 13.On a distance-time graph, a horizontal line segment shows a period when the graph's gradient is 0. What does this tell you about the journey during that time?
- 14.A phone company works out a monthly bill, £B, using the formula B = 18 + 0.05t, where £18 is the fixed monthly charge and t is the number of extra text messages sent beyond the free allowance, each charged at 5p. Farida's bill for one month is £24.50. Work out the number of extra text messages, t, she sent.
- 15.A taxi company charges a £2.50 booking fee plus £1.80 per mile. A journey costs £15.10 in total. Work out the number of miles travelled.
Answer key
- (d) 2.7 — The height after 3 years of 10% compound growth is 2 × 1.1³ = 2.662 m, which rounds to 2.7 m. A candidate who adds 10% of the original height (0.2 m) in each of the 3 years, instead of compounding on the new height each time, would reach 2 + 3×0.2 = 2.6 m. A candidate who compounds for only 2 years would reach 2 × 1.1² = 2.42 m, rounding to 2.4 m. A candidate who compounds for 4 years instead of 3 would reach 2 × 1.1⁴ = 2.928 m, rounding to 2.9 m.
- (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) 6x + 6 — Perimeter = 2[(2x + 5) + (x − 2)] = 2(3x + 3) = 6x + 6. A candidate who adds the length and width but forgets to double for the perimeter gets 3x + 3. A candidate who makes a sign error and adds 2 instead of subtracting it before doubling gets 2[(2x + 5) + (x + 2)] = 6x + 14. A candidate who multiplies the length and width instead of adding them, confusing the perimeter formula with the area formula, and then doubles that product, gets 2(2x + 5)(x − 2) = 4x² + 2x − 20.
- (b) 12 years — Method: call the number of years t, add t to both ages, and form an equation from the comparison at that future time. Working: in t years the leader will be 44 + t and the scout will be 16 + t, so 44 + t = 2(16 + t); expanding gives 44 + t = 32 + 2t, and subtracting t and 32 from both sides gives t = 12. Checking: in 12 years the leader will be 56 and the scout 28, and 56 = 2 × 28. Answer: 12 years. The distractors: 6 years comes from halving the leader's present age instead, so that the scout has to reach 22, which takes 6 years; 14 years comes from halving the 28-year gap between the two ages; 28 years comes from giving the age gap itself as the number of years.
- (a) Subtracting 4x gives 3 = 10, which is never true. — Method: try to solve the equation as normal and see what happens. Working: subtract 4x from both sides: 4x + 3 − 4x = 4x + 10 − 4x, giving 3 = 10. This statement is false for every value of x, so the equation has no solution. Answer: subtracting 4x gives 3 = 10, which is never true. "x would have to be negative" invents a constraint on x that the equation never states. "It's true for every x" confuses this equation with an identity, where both sides would simplify to the same expression. "x = 7" misreads the false statement 3 = 10 as something to solve for x, rather than recognising it means no solution exists.
- (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).
- (b) x² + y² = 1369 — For a circle centred at the origin, the radius squared equals the sum of the squares of the coordinates of any point on it: r² = 12² + 35² = 144 + 1225 = 1369. The equation is x² + y² = 1369. x² + y² = 2209 comes from adding the coordinates first and then squaring the sum: (12 + 35)² = 47² = 2209, instead of squaring each coordinate separately. x² + y² = 1225 comes from using only 35² and leaving out the 12² term. x² + y² = 144 comes from using only 12² and leaving out the 35² term.
- (b) y decreases towards zero but never reaches it — Method: as x gets larger, dividing 20 by a bigger number gives a smaller result, so y decreases; because 20/x can never be exactly zero for any positive x, the curve gets closer to zero without ever reaching it, so y decreases towards zero but never reaches it. Distractor origins: 'y increases towards a limit but never reaches it' has the relationship backwards, treating y as increasing when it is actually decreasing; 'y decreases at a steady rate and reaches zero' wrongly assumes the graph behaves like a straight line that eventually hits zero; 'y stays the same however large x becomes' wrongly assumes there is no change in y at all.
- (a) 23 — The number of logs increases by 4 for each layer down, starting from 3 in the top layer, so the nth layer has 3+(n−1)×4 logs. For the 6th layer: 3+5×4=3+20=23. A candidate who adds the difference of 4 six times instead of five, treating the top layer as needing an addition too, would compute 3+6×4=27. A candidate who uses 4n instead of 4n−1, omitting the adjustment for the first layer, would compute 4×6=24. A candidate who subtracts the common difference, 4, instead of 1 when adjusting the multiplier would compute 4×6−4=20.
- (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.
- (a) 5.8 — The horizontal distance is 3 and the vertical distance is 5, so using Pythagoras' theorem the distance is √(3² + 5²) = √34 = 5.8 (1 d.p.). A candidate who adds the two differences instead of using Pythagoras gets 3 + 5 = 8.0. A candidate who works out 3² + 5² = 34 but forgets to take the square root gets 34.0. A candidate who subtracts the squares instead of adding them gets √(5² − 3²) = √16 = 4.0.
- (a) 36 km/h — Speed is distance ÷ time. Convert 25 minutes to hours: 25 ÷ 60 = 5/12 hours. Then 15 ÷ (5/12) = 15 × 12/5 = 36 km/h. Dividing 15 by 25 without converting minutes to hours gives 0.6 km/h. Multiplying 15 by 25/60 instead of dividing by it gives 6.25 km/h. Working out the speed in km per minute, 0.6, and then multiplying by 24 instead of 60 gives 14.4 km/h.
- (c) The object was stationary (not moving) — On a distance-time graph, the gradient at any point represents the speed at that point. A gradient of 0 means distance is not changing over time, so the object is stationary. A straight, sloped line (not flat) shows constant nonzero speed; a curve bending one way shows acceleration or deceleration; a flat section is not a maximum speed — it is no speed at all. Read the shape of the graph, not just how steep it looks.
- (a) 130 — Method: substitute the total bill into the formula, then subtract the fixed charge and divide by the cost per text message. Working: 24.50 = 18 + 0.05t, so 0.05t = 24.50 − 18 = 6.50, t = 6.50 ÷ 0.05 = 130. Answer: 130 extra text messages. 490 comes from dividing the whole bill by 0.05 without first subtracting the £18 fixed charge: 24.50 ÷ 0.05 = 490. 13 comes from dividing the £6.50 by 0.5 instead of 0.05, moving the decimal point one place too far: 6.50 ÷ 0.5 = 13. 65 comes from dividing the £6.50 by 0.1 instead of 0.05.
- (d) 7 — Method: set up the equation 2.50 + 1.80m = 15.10, then subtract the booking fee and divide by the cost per mile. Working: 1.80m = 15.10 − 2.50 = 12.60; m = 12.60 ÷ 1.80 = 7. Answer: 7 miles. 8.39 comes from dividing the whole £15.10 by £1.80 without first subtracting the booking fee: 15.10 ÷ 1.80 ≈ 8.39. 5.32 comes from swapping the two amounts round, subtracting £1.80 and dividing by £2.50: (15.10 − 1.80) ÷ 2.50 ≈ 5.32. 9.78 comes from adding the booking fee instead of subtracting it: (15.10 + 2.50) ÷ 1.80 ≈ 9.78.
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