IB Mathematics SL 2.1 Different forms of the equation of a straight line AI SL Paper 1- Exam Style Questions- New Syllabus
Question
A community swimming pool is designed with a constant slope transitioning from the shallow end to the deep end. The water depth at the shallow end is \( 1.2 \text{ m} \), while the deep end reaches a depth of \( 2.0 \text{ m} \). A cross-sectional view of the pool is provided below (diagram not to scale).
The horizontal span of the sloped section between the two vertical ends is \( 3.0 \text{ m} \).
To ensure swimmer safety, local regulations dictate that the gradient (slope) of the pool floor must be no greater than \( \frac{1}{3} \).
(a) Demonstrate through calculation that the design of this pool complies with the safety regulation.
The time, \( t \), required to fill the pool is inversely proportional to the flow rate, \( r \), of the water source. Using a high-capacity hose with a flow rate of \( 300 \text{ litres per minute} \), the pool is filled from empty in \( 4.5 \text{ hours} \).
(b) Calculate the time, in hours, required to fill the same pool using a standard hose with a flow rate of \( 170 \text{ litres per minute} \).
Most appropriate topic codes (IB Mathematics: applications and interpretation):
• SL 2.1: Gradient of a straight line between two points — part (a)
• SL 2.5: Inverse variation \( y = \frac{k}{x} \) — part (b)
• SL 2.5: Inverse variation \( y = \frac{k}{x} \) — part (b)
▶️ Answer/Explanation
(a)
Vertical difference = \( 2.0 – 1.2 = 0.8 \text{ m} \)
Horizontal distance = \( 3.0 \text{ m} \)
Gradient = \( \frac{0.8}{3.0} = 0.2666\ldots \) or \( \frac{4}{15} \)
Since \( 0.2666 < \frac{1}{3} \) (or \( \frac{4}{15} < \frac{5}{15} \)), the pool satisfies the safety regulation.
(b)
Let \( t \) be time and \( r \) be flow rate. Since \( t \propto \frac{1}{r} \),
\( 300 \times 4.5 = 170 \times t \)
\( t = \frac{300 \times 4.5}{170} \approx 7.94 \text{ hours} \)
Thus, the filling time is approximately \( \boxed{7.94 \text{ hours}} \) (or \( 476 \text{ minutes} \)).