A motoboat whose speed in still …

Let the speed of stream be x km/h
Speed of boat in still stream = 9 km/h
Speed of boat down the stream = 9 + x km/h
Speed of boat upstream = 9 - x
{tex}\therefore{/tex} Time  taken by boat to go to downstream 15 km = {tex}\frac { 15 } { 9 + x }{/tex} hours.
{tex}\therefore{/tex} Time  taken by boat to go 15 km of stream = {tex}\frac { 15 } { 9 - x }{/tex} hours.
Total time {tex}= \frac { 15 } { 9 + x } + \frac { 15 } { 9 - x } = 3 \frac { 45 } { 60 } = 3 \frac { 3 } { 4 } = \frac { 15 } { 4 }{/tex}
Dividing by 15
{tex}\frac { 1 } { 9 + x } + \frac { 1 } { 9 - x } = \frac { 1 } { 4 } \text { or } \frac { 9 + x + 9 - x } { ( 9 + x ) ( 9 - x ) } = \frac { 1 } { 4 }{/tex}
{tex}\Rightarrow \frac { 18 } { 81 - x ^ { 2 } } = \frac { 1 } { 4 }{/tex} or 81 - x² = 72
{tex}\therefore{/tex} x² = 81 - 72 = 9 {tex}\therefore{/tex} x = +3 and x = -3
But x {tex}\neq{/tex} -3
{tex}\therefore{/tex} Speed of stream = 3 km/h