A person can row a boat in still water at the rate of 5 km/hr. It takes him 4 times as long to row upstream of a river as to row downstream to cover same distance in the same river. The speed of flow of the stream is
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For problems where the ratio of upstream and downstream times is given for the same distance, you can use the direct formula: \( \frac{u}{v} = \frac{T_u + T_d}{T_u - T_d} \). Here, \( \frac{T_u}{T_d} = 4 \), so \( \frac{5}{v} = \frac{4+1}{4-1} = \frac{5}{3} \). This gives \(v = 3\) km/hr.
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.