A) 2.0 km/hr
B) 2.5 km/hr
C) 3 km/hr
D) 3.5 km/hr
Correct Answer: B
Solution :
Let \[{{v}_{1}}\] be the flow velocity of the be ship and \[{{v}_{2}}\] be the velocity of the river flow In downward stream, \[{{t}_{1}}=\frac{s}{{{v}_{1}}+{{v}_{2}}}\] \[\therefore \] \[10=\frac{300}{{{v}_{1}}+{{v}_{2}}}\] or \[{{v}_{1}}+{{v}_{2}}=30\] ? (i) In upstream, \[12=\frac{300}{{{v}_{1}}-{{v}_{2}}}\] \[{{v}_{1}}-{{v}_{2}}=25\] ? (ii) Solving equations (i) and (ii), we get \[{{v}_{2}}=\mathbf{2}\mathbf{.5}\,\,\mathbf{km/hr}\]
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