Secondary 1 Mathematics Calculus Quiz

Secondary 1 Mathematics Quiz - Rates of Change and Linear Graphs (Answer Key)

Total Marks: 40

Section A: Gradients and Equations of Lines

1. Find the gradient of the straight line passing through the points $A(2, 5)$ and $B(6, 13)$. [2]

• Answer: $2$
• Working: $\text{Gradient } m = \frac{y_2 - y_1}{x_2 - x_1}$ $m = \frac{13 - 5}{6 - 2} = \frac{8}{4} = 2$
• Teaching Note: The gradient is the "rise over run". Ensure students subtract coordinates in the correct order (top minus bottom, bottom minus bottom).

2. Determine the gradient of the line given by the equation $3y + 2x = 12$. [2]

• Answer: $-\frac{2}{3}$
• Working: Rearrange into the form $y = mx + c$. $3y = -2x + 12$ $y = -\frac{2}{3}x + 4$ Comparing to $y = mx + c$, the gradient $m = -\frac{2}{3}$.
• Teaching Note: A common mistake is to identify the coefficient of $x$ in the original equation ($2$) as the gradient without isolating $y$ first.

3. A straight line has a gradient of $-4$ and passes through the point $(0, 7)$. Write down the equation of this line. [1]

• Answer: $y = -4x + 7$
• Working: Using $y = mx + c$, where $m = -4$ and the $y$-intercept $c = 7$ (since it passes through $(0,7)$).
• Teaching Note: The point $(0, 7)$ is the $y$-intercept, so $c$ is directly given.

4. Find the $y$-intercept of the line passing through $(1, 2)$ and $(3, 8)$. [2]

• Answer: $-1$ (or coordinate $(0, -1)$)
• Working: First, find gradient $m$: $m = \frac{8 - 2}{3 - 1} = \frac{6}{2} = 3$ Equation is $y = 3x + c$. Substitute $(1, 2)$: $2 = 3(1) + c \implies c = 2 - 3 = -1$ The $y$-intercept is $-1$.
• Teaching Note: Students must find the gradient first, then substitute a point to find $c$.

5. State whether the line with equation $y = -2x + 5$ is increasing or decreasing as $x$ increases. Give a reason for your answer. [1]

• Answer: Decreasing.
• Reason: The gradient ($m = -2$) is negative.
• Teaching Note: A negative gradient means the line goes down from left to right.

6. Two lines are parallel. One line has the equation $y = 3x - 1$. What is the gradient of the other line? [1]

• Answer: $3$
• Teaching Note: Parallel lines have equal gradients.

7. Find the coordinates of the point where the line $y = 2x - 4$ crosses the $x$-axis. [1]

• Answer: $(2, 0)$
• Working: At the $x$-axis, $y = 0$. $0 = 2x - 4 \implies 2x = 4 \implies x = 2$
• Teaching Note: Crossing the $x$-axis always means the $y$-coordinate is 0.

Section B: Rates of Change in Context

8. A car travels a distance of 150 km in 2 hours. Calculate the average speed of the car in km/h. [2]

• Answer: $75$ km/h
• Working: $\text{Speed} = \frac{\text{Distance}}{\text{Time}} = \frac{150}{2} = 75$
• Teaching Note: Ensure units are included in the final answer for full marks in context questions.

9. Calculate the rate at which water is being added to the tank in Litres per minute. [2]

• Answer: $6$ L/min
• Working: Change in Volume = $50 - 20 = 30$ L. Change in Time = $5 - 0 = 5$ min. $\text{Rate} = \frac{30}{5} = 6 \text{ L/min}$ (Check with other intervals: $\frac{80-50}{10-5} = \frac{30}{5} = 6$).
• Teaching Note: The rate is constant because the change is uniform. This is the gradient of the Volume-Time graph.

10. Distance-time graph analysis. (a) Describe the motion between $t = 1$ and $t = 2$ hours. [1] * Answer: The cyclist is stationary (at rest / not moving). * Reason: The distance from home does not change (horizontal line).

(b) Calculate the speed during the first hour. [2]
*   **Answer:** $10$ km/h
*   **Working:**
    $$ \text{Speed} = \frac{\text{Change in Distance}}{\text{Change in Time}} = \frac{10 - 0}{1 - 0} = 10 \text{ km/h} $$

11. Cooling coffee. (a) Average rate of cooling. [2] * Answer: $3^\circ\text{C/min}$ * Working: Change in Temp = $80 - 50 = 30^\circ\text{C}$. Time = $10$ min. Rate = $\frac{30}{10} = 3^\circ\text{C/min}$.

(b) Why might the rate not be constant? [1]
*   **Answer:** The rate of cooling depends on the temperature difference between the coffee and the room. As the coffee cools, the difference decreases, so it cools slower. (Or: Heat loss is faster when hotter).

12. T-shirt cost $C = 5n + 20$. (a) Fixed cost. [1] * Answer: \20$* **Reason:** This is the value of$C$when$n=0$(the$y$-intercept).

(b) Variable cost per T-shirt. [1]
*   **Answer:** $\$5$
*   **Reason:** This is the coefficient of $n$ (the gradient).

(c) Cost of 50 T-shirts. [2]
*   **Answer:** $\$270$
*   **Working:**
    $$ C = 5(50) + 20 = 250 + 20 = 270 $$

13. Water leakage rate. [3]

• Answer: $5$ Litres per hour.
• Working: Change in Volume = $25 - 10 = 15$ Litres. Change in Time = $5 - 2 = 3$ hours. $\text{Rate} = \frac{15}{3} = 5 \text{ L/h}$
• Teaching Note: Do not just divide 25 by 5. You must use the change in values ($\Delta y / \Delta x$).

Section C: Graph Interpretation and Problem Solving

14. Speed-time graph analysis. (a) Maximum speed. [1] * Answer: $30$ m/s.

(b) Motion from $t=10$ to $t=20$. [1]
*   **Answer:** The train is decelerating (slowing down) at a constant rate.

(c) Acceleration in first 10 seconds. [2]
*   **Answer:** $3$ m/s$^2$
*   **Working:**
    $$ \text{Acceleration} = \frac{\text{Change in Speed}}{\text{Time Taken}} = \frac{30 - 0}{10 - 0} = \frac{30}{10} = 3 \text{ m/s}^2 $$
*   **Teaching Note:** In a speed-time graph, the gradient represents acceleration.

15. What does the gradient of a distance-time graph represent? [1]

• Answer: Speed (or Velocity).

16. Line through $(2, k)$ and $(5, 11)$ with gradient $2$. Find $k$. [3]

• Answer: $5$
• Working: $m = \frac{y_2 - y_1}{x_2 - x_1}$ $2 = \frac{11 - k}{5 - 2}$ $2 = \frac{11 - k}{3}$ $6 = 11 - k$ $k = 11 - 6 = 5$

17. Equation $y = 3x - 2$. (a) Gradient $m$. [1] * Answer: $3$

(b) $y$-intercept $c$. [1]
*   **Answer:** $-2$

(c) Does $(4, 10)$ lie on the line? [2]
*   **Answer:** Yes.
*   **Working:**
    Substitute $x = 4$ into the equation:
    $$ y = 3(4) - 2 = 12 - 2 = 10 $$
    Since the calculated $y$ matches the point's $y$-coordinate ($10$), the point lies on the line.

18. Runners A and B. (a) Equation for Runner A. [1] * Answer: $d_A = 5t$

(b) Equation for Runner B. [1]
*   **Answer:** $d_B = 4t + 10$
*   **Reason:** Starts 10m ahead ($+10$), speed 4 m/s ($4t$).

(c) Time to catch up. [2]
*   **Answer:** $10$ seconds.
*   **Working:**
    Set $d_A = d_B$:
    $$ 5t = 4t + 10 $$
    $$ 5t - 4t = 10 $$
    $$ t = 10 $$

19. Graph passes through $(0, -3)$ and $(2, 1)$. Find $m$ and $c$. [3]

• Answer: $m = 2, c = -3$
• Working: Since it passes through $(0, -3)$, the $y$-intercept $c = -3$. Calculate gradient $m$: $m = \frac{1 - (-3)}{2 - 0} = \frac{4}{2} = 2$ So, $m = 2$ and $c = -3$.

20. Volume-Time graph gradient is 5. What does 5 represent? [2]

• Answer: The rate of flow of water into the tank is 5 Litres per second (assuming standard units, or whatever units are on axes, e.g., units$^3$/time).
• Teaching Note: The gradient of a Volume-Time graph is the flow rate. Units are crucial here (Volume units / Time units).