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.The table shows the height, in metres, of a firework rocket at various times, in seconds, during its flight: 40 at t = 2, 54 at t = 3, and 60 at t = 4. Use the chord between t = 2 and t = 4 to estimate the gradient of the height-time graph at t = 3, stating the correct units.
- 2.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?
- 3.A photo printing service has two adverts for its price. Advert A: cost in pounds = 3(2n + 4) for n photos. Advert B: cost in pounds = 6n + 12. A customer says the two adverts always charge the same amount. Is the customer correct?
- 4.On Monday, a runner covers 15 km in 2.5 hours. On Tuesday, she covers 12 km in 1.5 hours. Using the formula speed = distance ÷ time, work out on which day she ran faster, and by how much.
- 5.A car's speed-time graph shows the following: its speed increases steadily from 0 m/s to 20 m/s over the first 10 seconds, then stays constant at 20 m/s for the next 15 seconds. Work out the total distance travelled in the first 25 seconds.
- 6.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.
- 7.The point (9, 12) lies on the circle x² + y² = 225, which has centre (0, 0). The tangent to the circle at (9, 12) crosses the x-axis at the point P. Work out the x-coordinate of P.
- 8.Solve the inequality 6.3x + 1.8 ≤ 40. Hence write down the greatest number with exactly one decimal place that satisfies the inequality.
- 9.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.
- 10.The equation x³ − 2x − 7 = 0 has exactly one solution. It can be found using the iterative formula xₙ₊₁ = ∛(2xₙ + 7), with starting value x₀ = 2, so that x₁ is the value after the formula has been used once. Work out the solution correct to 2 decimal places, iterating until two consecutive values round to the same 2 decimal places.
- 11.A circle has centre (0, 0) and equation x² + y² = 50. Work out the length of the diameter of the circle, correct to 1 decimal place.
- 12.A tap fills a tank at a varying rate. A graph shows the rate of flow, in litres per minute, against time, in minutes. What does the area under this graph represent?
- 13.The iterative formula xₙ₊₁ = √(2xₙ + 3) is used repeatedly, starting from x₀ = 1. As n increases, the values of xₙ converge to a limit, L. Work out L.
- 14.The graph of y = 20/x is drawn for x > 0. As the value of x increases, what happens to the value of y?
- 15.A cyclist's speed rises from rest to a maximum — quickly at first, then more and more slowly — so her speed-time graph is a curve that is concave down (its gradient decreases as time goes on). A trapezium estimate is made for the distance travelled during this phase, using two points on the curve. Is the trapezium estimate an overestimate or an underestimate of the true distance, and why?
Answer key
- (c) 10 m/s — A symmetric chord gradient uses the two points either side of t = 3: (2, 40) and (4, 60). The gradient is the change in height divided by the change in time: (60 − 40) ÷ (4 − 2) = 20 ÷ 2 = 10 m/s. Using only the values either side of one gap, (2, 40) and (3, 54), instead of the full symmetric chord, gives (54 − 40) ÷ (3 − 2) = 14 m/s. Getting the correct number but dropping the time unit, leaving only metres, gives 10 m. Dividing time by height instead of height by time inverts the calculation to (4 − 2) ÷ (60 − 40) = 0.1 s/m.
- (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) They always charge the same, since 3(2n + 4) = 6n + 12. — Expand Advert A's formula by multiplying both terms inside the bracket by 3: 3 × 2n = 6n, and 3 × 4 = 12, giving 3(2n + 4) = 6n + 12, which is identical to Advert B's formula — so the two adverts always charge the same amount, whatever n is. Getting 6n + 4 comes from multiplying the 2n by 3 but leaving the 4 unmultiplied. Getting 2n + 7 comes from adding 3 to the bracket instead of multiplying by it. Saying it depends on n avoids expanding the bracket at all — once expanded, both formulas are identical for every value of n, so the cost can be compared directly.
- (a) Tuesday, by 2 km/h — Monday's speed is 15 ÷ 2.5 = 6 km/h and Tuesday's speed is 12 ÷ 1.5 = 8 km/h, so Tuesday was faster, by 8 − 6 = 2 km/h. A candidate who works out the correct speeds but mislabels which day is faster gets Monday, by 2 km/h. A candidate who divides 15 ÷ 2.5 incorrectly as 5 instead of 6 gets a difference of 8 − 5 = 3 km/h, still crediting Tuesday. A candidate who forgets to find Monday's speed and gives Tuesday's speed itself as the difference states Tuesday, by 8 km/h.
- (d) 400 m — The distance travelled equals the area under the speed-time graph. The first 10 seconds form a triangle with base 10 and height 20, giving an area of 0.5 × 10 × 20 = 100 m. The next 15 seconds form a rectangle with base 15 and height 20, giving an area of 15 × 20 = 300 m. The total distance is 100 + 300 = 400 m. The option 500 m treats the whole 25 seconds as travelled at the constant 20 m/s, ignoring that the speed was building up during the first 10 seconds: 25 × 20 = 500. The option 300 m only counts the constant-speed section and forgets the triangle section entirely. The option 200 m comes from working out the triangle's area without halving it (10 × 20 = 200) and forgetting the rectangle altogether.
- (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.
- (b) 25 — The tangent at (9, 12) is 9x + 12y = 225. Setting y = 0 (the x-axis): 9x = 225, so x = 25. Choosing 18.75 comes from swapping the coefficients in the tangent equation (using 12x + 9y = 225) before setting y = 0. Choosing 15 is where the circle itself meets the x-axis (from x² = 225), not where the tangent does. Choosing 9 is just the x-coordinate of the original point (9, 12), not the point P.
- (d) 6.0 — Subtract 1.8 from both sides: 6.3x ≤ 38.2. Divide both sides by 6.3: x ≤ 6.0634… . The question asks for the greatest value to 1 decimal place that still satisfies the inequality. Testing 6.1: 6.3 × 6.1 + 1.8 = 40.23, which is more than 40, so 6.1 fails. Testing 6.0: 6.3 × 6.0 + 1.8 = 39.6, which is no more than 40, so 6.0 works and is the greatest such value. A candidate who simply rounds 6.0634… to 1 decimal place answers 6.1, without checking that it satisfies the inequality. A candidate who adds 1.8 instead of subtracting works out 41.8 ÷ 6.3 and answers 6.6. A candidate who forgets the 1.8 altogether divides 40 by 6.3 and answers 6.3.
- (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.
- (c) 2.26 — Method: apply the formula repeatedly, keeping the whole display each time, and stop when two values in a row round to the same 2 decimal places; that shared rounded value is the solution to that accuracy. Working: x₁ = ∛(2 × 2 + 7) = ∛11 = 2.22398…; x₂ = ∛(2 × 2.22398… + 7) = ∛11.44796… = 2.25377…; x₃ = ∛11.50754… = 2.25767…; x₄ = ∛11.51534… = 2.25818…. Now x₃ and x₄ both round to 2.26, so the sequence has settled. Answer: 2.26. The distractors: 2.22 is x₁ rounded, quoted by a candidate who stops after one use of the formula; 2.25 is x₂ rounded, quoted by a candidate who stops as soon as two values look close instead of waiting until two consecutive values round to the same figure; 1.91 is ∛7, which comes from ignoring the 2x term and solving x³ = 7 instead.
- (c) 14.1 — Method: in x² + y² = r² the right-hand side is the square of the radius, so take its square root to find the radius, then double the radius because the diameter is twice the radius. Working: r² = 50, so r = √50 = 7.07106…, and the diameter is 2 × 7.07106… = 14.14213…, which is 14.1 correct to 1 decimal place. Answer: 14.1. The distractors: 7.1 is the radius, worked out correctly but never doubled, so it answers a question about the radius rather than the diameter; 100.0 comes from doubling the 50 on the right-hand side, treating r² as though it were already the radius; 25.0 comes from halving the 50, treating r² as though it were already the diameter.
- (b) The total volume of water, in litres, that has flowed in. — On a rate-time graph, the y-axis is in litres per minute and the x-axis is in minutes; multiplying a rate by a time gives litres per minute × minutes = litres, a total volume. So the area under the graph represents the total volume of water that has flowed in. Thinking the area itself represents the rate, rather than what the rate accumulates to, gives the wrong claim about the average rate of flow. Confusing the area with the gradient of the graph — which measures how the rate is changing — gives the wrong claim about litres per minute squared. Ignoring the flow-rate axis and focusing only on the time axis gives the wrong claim that the area is simply the total time.
- (b) 3 — At the limit, L = √(2L + 3). Squaring both sides: L² = 2L + 3, so L² − 2L − 3 = 0, which factorises as (L − 3)(L + 1) = 0, giving L = 3 or L = −1. Since the sequence of iterates stays positive throughout, the limit is L = 3. Taking the other, negative root without rejecting it gives −1. Treating the equation L = 2L + 3 as already linear, forgetting to square both sides first, gives −L = 3, so L = −3. A sign error when factorising, writing (L + 3)(L − 1) = 0 instead of (L − 3)(L + 1) = 0, gives L = 1.
- (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) Underestimate: the chord lies below the curve — For a curve that is concave down (bending downward), the straight line joining any two points on the curve lies BELOW the curve. The trapezium's top edge is this straight line, so the trapezium's area is smaller than the true area under the curve — the trapezium rule UNDERESTIMATES the distance in this case. This is because a curve whose gradient keeps decreasing bends away from any chord joining two of its points, dropping the chord below every point of the curve in between. The reverse, a chord lying above the curve, is the rule for a concave-up curve, not this one — check which way the curve bends before deciding. The trapezium rule does not always overestimate or underestimate, and it is not exact except when the graph really is a straight line.
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