OxfordAQA International GCSE Mathematics: Graphs in real contexts, rates of change and variation (9260) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover references A17 and A18 (section 3.2.2, Functions, graphs and calculus) of the OxfordAQA International GCSE Mathematics (9260) specification, Version 5.1 (for exams May/June 2018 onwards). Questions or parts marked “(Extension only)” test areas under graphs, curved-graph work or A18. Learn the methods first in the study guide and revision notes. See the OxfordAQA International GCSE Mathematics course hub and the checklist tracks your progress. Graph paper helps for questions 2, 5 and 6.

Questions

1. A taxi fare is a fixed charge plus a charge per kilometre. The graph of fare (£) against distance (km) is a straight line through (2, 7.40) and (10, 17.00).

(a) Find the gradient and say what it represents. [2] (b) Find the fixed charge, and the fare for a 15 km journey. [2]

2. The depth h metres of water at a harbour entrance t hours after midnight is given in the table.

t	0	1	2	3	4	5	6	7	8	9	10	11	12
h	3.00	4.25	5.17	5.50	5.17	4.25	3.00	1.75	0.83	0.50	0.83	1.75	3.00

(a) Plot the graph for 0 ≤ t ≤ 12. [2] (b) A boat needs a depth of at least 4 m. Use your graph to find the times between which it can enter, for the first period of the day. [2] (c) Find how long this period lasts. [1]

3. Hamza jogs to a park. His distance-time graph is made of straight lines through (0, 0), (20, 2), (35, 2) and (60, 6), with time in minutes and distance in km.

(a) Find his speed in the first 20 minutes, in km/h. [2] (b) For how long does he stop? [1] (c) Find his speed in the last stage, in km/h. [2] (d) Find his average speed over the whole jog. [2]

4. A cyclist’s velocity-time graph is made of straight lines. She speeds up from rest to 8 m/s in 10 s, rides at 8 m/s for 30 s, then slows to rest in 16 s.

(a) Find her acceleration in the first 10 s. [2] (b) Find the size of her deceleration in the last 16 s. [2] (c) (Extension only) Find the total distance she travels. [3] (d) (Extension only) Find her average speed for the whole journey, to 3 significant figures. [2]

5. (Extension only) The speed v m/s of a sprinter at t seconds is shown.

t	0	2	4	6	8	10	12
v	0	5	8	9	8	5	0

(a) Use trapezia with 6 equal strips to estimate the distance travelled. [3] (b) Explain whether your estimate is likely to be too high or too low. [1] (c) Use your estimate to find the average speed. Give your answer to 3 significant figures. [1]

6. (Extension only) On a distance-time graph of a drone, a tangent is drawn at t = 5 s. The tangent passes through (2, 3) and (8, 21), with distance in metres.

(a) Estimate the speed of the drone at t = 5. [2] (b) Convert your answer to km/h. [1] (c) Explain why joining two points on the curve would give a different value. [1]

7. (Extension only) The braking distance d metres of a car is directly proportional to the square of its speed v m/s. When v = 10, d = 20.

(a) Write down a formula for d. [2] (b) Find d when v = 25. [1] (c) Find v when d = 51.2. [2]

8. (Extension only) A stock of feed lasts D days, and D is inversely proportional to the number n of animals fed. The stock lasts 18 days for 40 animals.

(a) Write down a formula for D. [2] (b) Find D when n = 60. [1] (c) Find n when the stock lasts 9 days. [2]

9. (Extension only)

(a) y is directly proportional to x². x goes up by 20%. By what percentage does y change? [2] (b) z is inversely proportional to x. x falls by 20%. By what percentage does z change? [2]

10. (Extension only) Values of x and y are shown.

x	2	4	5	10
y	50	12.5	8	2

(a) Show that y is inversely proportional to x², and write down the formula. [2] (b) Find y when x = 8. [1] (c) Find x when y = 0.5. [2]

11. A tank’s volume V litres against time t minutes is made of straight lines through (0, 40), (10, 160), (30, 160) and (50, 0).

(a) Find the rate at which the tank fills in the first 10 minutes. [2] (b) Describe what happens between t = 10 and t = 30, and how the graph shows it. [1] (c) Find the rate at which the tank empties after t = 30. [2] (d) Find the time when the tank holds 100 litres while emptying. [2] (e) Find the average rate of change of volume over the whole 50 minutes. [2]

Answers

1. (a) Gradient = (17.00 - 7.40)/(10 - 2) [1] = 9.60/8 = 1.2, so £1.20 per km (the cost of each extra kilometre) [1] [2] (b) Fixed charge = 7.40 - 2 × 1.20 = £5.00 [1]. Fare for 15 km = 5 + 1.2 × 15 = £23 [1] [2] Examiner insight: Stating the meaning with units, “pounds per kilometre”, is the kind of interpretation a “what does it represent” question looks for; a bare 1.2 is usually incomplete.

2. (a) Points plotted accurately [1]; smooth curve through them [1] [2] (b) Draw the line h = 4 [1]. It meets the curve at about t = 0.8 and t = 5.2, so between about 0.8 h and 5.2 h (accept 0.7 to 0.9 and 5.1 to 5.3) [1] [2] (c) 5.2 - 0.8 = about 4.4 hours [1] [1] Examiner insight: Draw the horizontal line on your graph so your readings can be seen, and say “approximately”.

3. (a) 2 km in 20 min = 2 km in 1/3 h [1]. Speed = 2 ÷ (1/3) = 6 km/h [1] [2] (b) 35 - 20 = 15 minutes [1] [1] (c) Distance = 6 - 2 = 4 km, time = 25 min [1]. Speed = 4 ÷ (25/60) = 9.6 km/h [1] [2] (d) Total distance 6 km, total time 60 min = 1 h [1]. Average speed = 6/1 = 6 km/h [1] [2] Examiner insight: Show the conversion of minutes to hours before dividing; mixed units give plausible but wrong answers.

4. (a) Acceleration = 8/10 [1] = 0.8 m/s² [1] [2] (b) Deceleration = 8/16 [1] = 0.5 m/s² [1] [2] (c) Trapezium with parallel sides 30 s and 56 s (10 + 30 + 16), height 8 [1]. Area = ½ × (30 + 56) × 8 [1] = 344 m [1] [3] (d) Average speed = 344/56 [1] = 6.14 m/s [1] [2] Examiner insight: Splitting the area into a triangle, rectangle and triangle works too; show each area separately.

5. (a) Strip width h = 2 [1]. Area ≈ (2/2) × [0 + 0 + 2 × (5 + 8 + 9 + 8 + 5)] [1] = 2 × 35 = 70 m [1] [3] (b) The curve bends downwards, so each trapezium lies slightly below it: too low (an underestimate) [1] [1] (c) Average speed = 70/12 = 5.83 m/s [1] [1] Examiner insight: Show the substitution into the trapezium formula, and justify “too high or too low” with the shape of the curve.

6. (a) Gradient = (21 - 3)/(8 - 2) [1] = 18/6 = 3 m/s [1] [2] (b) 3 × 3.6 = 10.8 km/h [1] [1] (c) A line joining two points on the curve gives the average speed between them, not the speed at t = 5 [1] [1] Examiner insight: Use points on the tangent, not points on the curve, and keep the units “m/s” with the answer.

7. (a) d = kv² [1]. 20 = k × 100, so k = 0.2 and d = 0.2v² [1] [2] (b) d = 0.2 × 625 = 125 m [1] [1] (c) 51.2 = 0.2v², so v² = 256 [1], v = 16 m/s (positive speed) [1] [2] Examiner insight: Find k and write the full formula before using the second value; give the positive root for a speed.

8. (a) D = k/n [1]. 18 = k/40, so k = 720 and D = 720/n [1] [2] (b) D = 720/60 = 12 days [1] [1] (c) 9 = 720/n [1], so n = 80 animals [1] [2] Examiner insight: Check the direction: more animals should mean fewer days. If your answer breaks that, the formula has the wrong form.

9. (a) y = kx², new x = 1.2x so new y = k(1.2x)² = 1.44kx² [1]. y increases by 44% [1] [2] (b) z = k/x, new x = 0.8x so new z = k/(0.8x) = 1.25 × (k/x) [1]. z increases by 25% [1] [2] Examiner insight: Use a multiplier such as 1.2 or 0.8 instead of adding percentages; stating whether the change is an increase or decrease is part of the answer.

10. (a) y × x²: 50 × 4 = 200, 12.5 × 16 = 200, 8 × 25 = 200, 2 × 100 = 200 [1]. This is constant, so y = 200/x² [1] [2] (b) y = 200/64 = 3.125 [1] [1] (c) 0.5 = 200/x², so x² = 400 [1], x = 20 [1] [2] Examiner insight: A “show that” needs the check for every pair of values, or a clear statement of the constant product, not just one pair.

11. (a) Gradient = (160 - 40)/10 [1] = 12 litres per minute [1] [2] (b) The volume stays at 160 litres, shown by a horizontal line (gradient zero) [1] [1] (c) Gradient = (0 - 160)/(50 - 30) = -8 [1]. It empties at 8 litres per minute [1] [2] (d) V = 160 - 8(t - 30) = 100, so t - 30 = 7.5 [1], t = 37.5 minutes [1] [2] (e) Change in volume = 0 - 40 = -40 over 50 minutes [1]. Average rate = -0.8 litres per minute [1] [2] Examiner insight: Interpret a negative gradient in words (“losing”, “emptying”); the overall average rate uses the start and end points only, not the individual stages.

Where marks are usually lost

• Giving a gradient without units or without saying what it means in the context.
• Reading a graph carelessly, such as taking a value from the wrong scale.
• Leaving time in minutes while speed is in km/h.
• Dividing total distance by the wrong time, or ignoring stops, when finding average speed.
• Taking a trapezium’s height as a time rather than the speed (or the other way round).
• Using points on the curve rather than on the tangent when estimating a gradient (Extension only).
• Substituting the second value before finding k (Extension only).
• Using y = kx for an inverse relationship, or forgetting the square (Extension only).
• Changing a percentage by adding it to the answer instead of using a multiplier (Extension only).

Next steps

• Revisit the revision notes and study guide.
• Return to the course hub and tick off skills on the checklist.
• Try all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OxfordAQA International GCSE Mathematics (9260), specification Version 5.1 (for exams May/June 2018 onwards), published by OxfordAQA, section 3.2.2 Functions, graphs and calculus, references A17 and A18.