Step 1: Conceptualization:
This is a standard 'Boats and Streams' problem. The boat's velocity is influenced by the river's current. Downstream, velocities are additive. Upstream, the current's velocity is subtracted from the boat's velocity.
Step 2: Core Formula/Methodology:
Let \(u\) represent the boat's speed in still water.
Let \(v\) represent the stream's speed.
Speed downstream (\(S_d\)) = \(u + v\).
Speed upstream (\(S_u\)) = \(u - v\).
Formula: Time = Distance / Speed.
The problem states that for an equal distance, upstream travel time is four times downstream travel time.
\[ T_{\text{upstream}} = 4 \times T_{\text{downstream}} \]
Step 3: Detailed Calculation:
Given data:
Boat speed in still water, \(u = 5\) km/hr.
Stream speed = \(v\) km/hr.
Distance = 'd' km.
Applying the time-distance-speed relationship:
\[ T_{\text{downstream}} = \frac{d}{S_d} = \frac{d}{5 + v} \]
\[ T_{\text{upstream}} = \frac{d}{S_u} = \frac{d}{5 - v} \]
Utilizing the provided condition \( T_{\text{upstream}} = 4 \times T_{\text{downstream}} \):
\[ \frac{d}{5 - v} = 4 \times \left(\frac{d}{5 + v}\right) \]
Since 'd' is constant on both sides, it can be canceled:
\[ \frac{1}{5 - v} = \frac{4}{5 + v} \]
Cross-multiplication to solve for \(v\):
\[ 1 \times (5 + v) = 4 \times (5 - v) \]
\[ 5 + v = 20 - 4v \]
Rearranging terms to group \(v\) and constants:
\[ v + 4v = 20 - 5 \]
\[ 5v = 15 \]
\[ v = \frac{15}{5} \]
\[ v = 3 \]
Step 4: Conclusive Result:
The stream's flow speed is 3 km/hr.