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Secondary 1 Mathematics Calculus Quiz
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Questions
Secondary 1 Mathematics Quiz - Calculus
Name: _______________________________ Class: _______________________________ Date: _______________________________ Score: _______ / 40
Duration: 40 minutes Total Marks: 40
Instructions:
- Answer ALL questions.
- Show your working clearly in the spaces provided.
- Non-exact answers should be given correct to 2 decimal places unless otherwise stated.
- The use of calculators is not allowed for this quiz.
- Marks are awarded for correct working, not just the final answer.
Section A: Rate of Change and Gradient (Questions 1–5)
Questions 1 to 5 carry 2 marks each.
1. The table below shows the distance travelled by a car at different times.
| Time (s) | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| Distance (m) | 0 | 5 | 20 | 45 | 80 |
(a) Find the average rate of change of distance from t = 1 s to t = 3 s.
(b) Is the rate of change constant over the entire journey? Explain your answer.
2. A container is being filled with water. The volume of water V (in cm³) after t seconds is given by the formula V = 15t + 10.
(a) Find the volume of water in the container when t = 0.
(b) Find the rate at which the volume of water is increasing per second.
3. The line y = 4x − 7 is drawn on a coordinate grid.
(a) State the gradient of the line.
(b) Find the change in y when x increases by 3 units.
4. A straight line passes through the points A(1, 3) and B(5, 11).
(a) Calculate the gradient of the line AB.
(b) Write down the equation of the line in the form y = mx + c.
5. The table shows the cost C (in dollars) of buying n apples.
| n | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| C ($) | 1.50 | 3.00 | 4.50 | 6.00 | 7.50 |
(a) Find the rate of change of cost with respect to the number of apples.
(b) State what this rate of change represents in context.
Section B: Gradient of Curves and Tangents (Questions 6–10)
Questions 6 to 10 carry 2 marks each.
6. The distance d (in metres) of a particle from a fixed point at time t (in seconds) is given by d = t² + 3t.
(a) Copy and complete the table below.
| t | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| d |
(b) Find the average rate of change of d from t = 1 to t = 3.
7. The height h (in metres) of a ball thrown vertically upwards after t seconds is given by h = 20t − 5t².
(a) Find the height of the ball when t = 1.
(b) Find the average rate of change of height from t = 0 to t = 2.
8. A curve has equation y = x² − 2x + 1.
(a) Find the coordinates of the point on the curve where x = 3.
(b) By considering the gradient of the chord joining the points where x = 2 and x = 4, estimate the gradient of the curve at x = 3.
9. The table shows the area A (in cm²) of a growing circular oil spill at time t minutes.
| t (min) | 0 | 2 | 4 | 6 | 8 |
|---|---|---|---|---|---|
| A (cm²) | 0 | 12.6 | 50.3 | 113.1 | 201.1 |
(a) Find the average rate of change of area from t = 2 to t = 6.
(b) Estimate the rate of change of area when t = 4 by finding the average rate of change from t = 2 to t = 6.
10. The cost C (in dollars) of producing x items is given by C = x² − 10x + 50.
(a) Find the cost of producing 3 items.
(b) Find the average rate of change of cost when production increases from 3 items to 6 items.
Section C: Applications of Rate of Change (Questions 11–15)
Questions 11 to 15 carry 3 marks each.
11. A taxi company charges a flag-down fare of 0.25 for every 100 metres travelled.
(a) Write an equation connecting the total cost C (in dollars) and the distance travelled d (in metres).
(b) Find the rate of change of cost with respect to distance. State its units.
(c) Calculate the total cost of a journey of 2.5 km.
12. The population P of a town t years after 2020 is modelled by P = 5000 + 200t.
(a) Find the population in 2020.
(b) Find the rate of change of population per year.
(c) Use your answer to (b) to find the population in 2025.
13. Water is leaking from a tank. The volume of water V (in litres) remaining in the tank after t hours is shown in the table.
| t (hours) | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| V (litres) | 200 | 185 | 170 | 155 | 140 |
(a) Find the rate at which water is leaking from the tank. State your answer with units.
(b) Write an equation connecting V and t.
(c) How long will it take for the tank to be completely empty?
14. A car accelerates from rest. Its speed v (in m/s) at time t (in seconds) is given in the table.
| t (s) | 0 | 2 | 4 | 6 | 8 |
|---|---|---|---|---|---|
| v (m/s) | 0 | 8 | 16 | 24 | 32 |
(a) Find the rate of change of speed. State its units.
(b) Write an equation connecting v and t.
(c) Find the speed of the car after 15 seconds.
15. The profit P (in dollars) from selling x boxes of cookies is given by P = −x² + 30x − 100.
(a) Find the profit when 10 boxes are sold.
(b) Find the average rate of change of profit when sales increase from 10 to 15 boxes.
(c) What does a negative average rate of change mean in this context?
Section D: Gradient and Real-World Problems (Questions 16–20)
Questions 16 to 20 carry 3 marks each.
16. A straight road passes through two towns located at coordinates P(2, 5) and Q(8, 17) on a map, where each unit represents 1 km.
(a) Find the gradient of the road.
(b) A second road is perpendicular to the first road and passes through P. Find the gradient of the second road.
(c) A rest stop lies on the first road at a horizontal distance of 5 km from P. Find the coordinates of the rest stop.
17. The temperature T (°C) of a cooling object at time t (minutes) is given by T = 80 − 6t.
(a) Find the initial temperature of the object.
(b) Find the rate of change of temperature per minute. State what this value means.
(c) Find the time when the temperature reaches 20°C.
18. A company's revenue R (in thousands of dollars) from advertising spending x (in thousands of dollars) is given by R = −2x² + 24x + 10.
(a) Find the revenue when no money is spent on advertising.
(b) Find the average rate of change of revenue when advertising spending increases from 3000.
(c) Find the average rate of change of revenue when advertising spending increases from 6000.
19. The table shows the height h (in cm) of a plant d days after it was planted.
| d (days) | 0 | 5 | 10 | 15 | 20 |
|---|---|---|---|---|---|
| h (cm) | 2 | 5 | 11 | 20 | 32 |
(a) Find the average rate of growth from day 5 to day 15.
(b) Is the plant growing at a constant rate? Explain your answer using calculations.
(c) Estimate the rate of growth on day 10 by finding the average rate of change from day 5 to day 15.
20. A motorist drives from Town A to Town B, a distance of 120 km. The journey takes 2 hours.
(a) Find the average speed for the entire journey.
(b) For the first 60 km, the motorist drives at 50 km/h. Find the time taken for the first 60 km.
(c) Using your answers to (a) and (b), find the average speed for the remaining 60 km.
End of Quiz
Answers
Secondary 1 Mathematics Quiz - Calculus
Answer Key
Section A: Rate of Change and Gradient (Questions 1–5)
1. (a) Average rate of change = (45 − 5) ÷ (3 − 1) = 40 ÷ 2 = 20 m/s ✓ [1 mark]
(b) No. From t = 0 to t = 1: rate = 5 m/s. From t = 1 to t = 2: rate = 15 m/s. The rates are different, so the rate of change is not constant. ✓ [1 mark]
Common mistake: Students may assume constant rate because the data looks patterned. They must calculate at least two intervals to compare.
2. (a) When t = 0: V = 15(0) + 10 = 10 cm³ ✓ [1 mark]
(b) Rate of change = coefficient of t = 15 cm³/s ✓ [1 mark]
Marking note: Accept "15" with correct units. Award 1 mark for correct value even if units are missing.
3. (a) Gradient = 4 ✓ [1 mark]
(b) Change in y = gradient × change in x = 4 × 3 = 12 ✓ [1 mark]
Common mistake: Students may confuse gradient with the y-intercept (−7).
4. (a) Gradient = (11 − 3) ÷ (5 − 1) = 8 ÷ 4 = 2 ✓ [1 mark]
(b) Using y = mx + c and point A(1, 3): 3 = 2(1) + c, so c = 1. Equation: y = 2x + 1 ✓ [1 mark]
Marking note: Award 1 mark for correct gradient, 1 mark for correct equation.
5. (a) Rate of change = (7.50 − 1.50) ÷ (5 − 1) = 6.00 ÷ 4 = $1.50 per apple ✓ [1 mark]
(b) This represents the cost of one apple (unit price). ✓ [1 mark]
Common mistake: Students may give the answer as "1.5" without units or context.
Section B: Gradient of Curves and Tangents (Questions 6–10)
6. (a) Completed table:
| t | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| d | 0 | 4 | 10 | 18 | 28 |
d(0) = 0 + 0 = 0; d(1) = 1 + 3 = 4; d(2) = 4 + 6 = 10; d(3) = 9 + 9 = 18; d(4) = 16 + 12 = 28 ✓ [1 mark] for all correct
(b) Average rate of change = (18 − 4) ÷ (3 − 1) = 14 ÷ 2 = 7 m/s ✓ [1 mark]
7. (a) When t = 1: h = 20(1) − 5(1)² = 20 − 5 = 15 m ✓ [1 mark]
(b) h(0) = 0; h(2) = 40 − 20 = 20. Average rate of change = (20 − 0) ÷ (2 − 0) = 20 ÷ 2 = 10 m/s ✓ [1 mark]
Common mistake: Students may substitute incorrectly, e.g., forgetting to square t in the 5t² term.
8. (a) When x = 3: y = 9 − 6 + 1 = 4. Coordinates: (3, 4) ✓ [1 mark]
(b) When x = 2: y = 4 − 4 + 1 = 1. When x = 4: y = 16 − 8 + 1 = 9. Gradient of chord = (9 − 1) ÷ (4 − 2) = 8 ÷ 2 = 4 ✓ [1 mark]
Marking note: This is an estimate of the gradient of the tangent at x = 3 using the chord method.
9. (a) Average rate of change = (113.1 − 12.6) ÷ (6 − 2) = 100.5 ÷ 4 = 25.13 cm²/min ✓ [1 mark]
(b) The estimate is the same calculation: 25.13 cm²/min ✓ [1 mark]
Marking note: Accept answers correct to 2 decimal places. Award 1 mark for correct method, 1 mark for correct answer.
10. (a) C = 9 − 30 + 50 = $29 ✓ [1 mark]
(b) C(3) = 29; C(6) = 36 − 60 + 50 = 26. Average rate of change = (26 − 29) ÷ (6 − 3) = −3 ÷ 3 = −$1 per item ✓ [1 mark]
Common mistake: Students may forget the negative sign, indicating cost is decreasing.
Section C: Applications of Rate of Change (Questions 11–15)
11. (a) C = 0.0025d + 3.50 (or C = 0.25d/100 + 3.50) ✓ [1 mark]
(b) Rate of change = **0.25 per 100 m) ✓ [1 mark]
(c) 2.5 km = 2500 m. C = 0.0025(2500) + 3.50 = 6.25 + 3.50 = $9.75 ✓ [1 mark]
Common mistake: Students may forget to convert km to metres.
12. (a) When t = 0: P = 5000 + 0 = 5000 ✓ [1 mark]
(b) Rate of change = 200 people per year ✓ [1 mark]
(c) 2025 is 5 years after 2020, so t = 5. P = 5000 + 200(5) = 5000 + 1000 = 6000 ✓ [1 mark]
Marking note: Award 1 mark for each correct part.
13. (a) Rate of change = (140 − 200) ÷ (4 − 0) = −60 ÷ 4 = −15 litres per hour ✓ [1 mark]
(b) V = 200 − 15t ✓ [1 mark]
(c) When V = 0: 0 = 200 − 15t, so t = 200 ÷ 15 = 13.33 hours (or 13 hours 20 minutes) ✓ [1 mark]
Common mistake: Students may give the rate as positive 15 instead of −15.
14. (a) Rate of change = (32 − 0) ÷ (8 − 0) = 32 ÷ 8 = 4 m/s² ✓ [1 mark]
(b) v = 4t ✓ [1 mark]
(c) v = 4(15) = 60 m/s ✓ [1 mark]
Marking note: The rate of change of speed is acceleration.
15. (a) P = −100 + 300 − 100 = $100 ✓ [1 mark]
(b) P(10) = 100; P(15) = −225 + 450 − 100 = 125. Average rate of change = (125 − 100) ÷ (15 − 10) = 25 ÷ 5 = $5 per box ✓ [1 mark]
(c) A negative average rate of change would mean that profit is decreasing as more boxes are sold (i.e., the additional boxes sold are reducing overall profit). ✓ [1 mark]
Note: In this case the rate is positive ($5/box), but part (c) asks students to interpret what a negative rate would mean.
Section D: Gradient and Real-World Problems (Questions 16–20)
16. (a) Gradient = (17 − 5) ÷ (8 − 2) = 12 ÷ 6 = 2 ✓ [1 mark]
(b) Gradient of perpendicular line = −1/2 ✓ [1 mark]
(c) Horizontal distance from P = 5 km, so x = 2 + 5 = 7. y = 5 + 2(5) = 15. Coordinates: (7, 15) ✓ [1 mark]
Common mistake: Students may use the product of gradients = −1 incorrectly, e.g., giving −2 instead of −1/2.
17. (a) When t = 0: T = 80°C ✓ [1 mark]
(b) Rate of change = −6°C per minute. This means the temperature decreases by 6°C every minute. ✓ [1 mark]
(c) 20 = 80 − 6t, so 6t = 60, t = 10 minutes ✓ [1 mark]
18. (a) When x = 0: R = 10 (i.e., $10,000) ✓ [1 mark]
(b) R(0) = 10; R(3) = −18 + 72 + 10 = 64. Average rate of change = (64 − 10) ÷ (3 − 0) = 54 ÷ 3 = 18 (i.e., 1000 spent) ✓ [1 mark]
(c) R(3) = 64; R(6) = −72 + 144 + 10 = 82. Average rate of change = (82 − 64) ÷ (6 − 3) = 18 ÷ 3 = 6 (i.e., 1000 spent) ✓ [1 mark]
Marking note: This shows diminishing returns — the rate of revenue increase slows down.
19. (a) Average rate of growth = (20 − 5) ÷ (15 − 5) = 15 ÷ 10 = 1.5 cm/day ✓ [1 mark]
(b) No. From day 0 to 5: rate = (5 − 2) ÷ 5 = 0.6 cm/day. From day 5 to 10: rate = (11 − 5) ÷ 5 = 1.2 cm/day. From day 10 to 15: rate = (20 − 11) ÷ 5 = 1.8 cm/day. The rates are increasing, so the plant is not growing at a constant rate — it is growing faster over time. ✓ [1 mark]
(c) Estimate of rate of growth on day 10 = average rate from day 5 to day 15 = 1.5 cm/day ✓ [1 mark]
20. (a) Average speed = 120 ÷ 2 = 60 km/h ✓ [1 mark]
(b) Time = 60 ÷ 50 = 1.2 hours (or 1 hour 12 minutes) ✓ [1 mark]
(c) Remaining time = 2 − 1.2 = 0.8 hours. Average speed for remaining 60 km = 60 ÷ 0.8 = 75 km/h ✓ [1 mark]
Common mistake: Students may simply average 50 and another value, rather than using total distance ÷ total time for each section.
Total: 40 marks
| Section | Questions | Marks |
|---|---|---|
| A: Rate of Change and Gradient | 1–5 | 10 |
| B: Gradient of Curves and Tangents | 6–10 | 10 |
| C: Applications of Rate of Change | 11–15 | 15 |
| D: Gradient and Real-World Problems | 16–20 | 15 |
| Total | 40 |