A motor boat whose speed is …

Speed of the motorboat in still water = 24 km/hr.
Let the speed of the stream be x km/hr.
Then, speed upstream = (24 - x) km/hr.
Speed downstream = (24 + x) km/hr.
Time taken to go 32 km upstream = {tex}\frac { 32 } { ( 24 - x ) }{/tex} hours.
Time taken to return 32 km downstream = {tex}\frac { 32 } { ( 24 + x ) }{/tex} hours.

Now, according to question, we have
{tex}\frac { 32 } { ( 24 - x ) } - \frac { 32 } { ( 24 + x ) } = 1{/tex}
{tex}\Rightarrow \frac { 1 } { ( 24 - x ) } - \frac { 1 } { ( 24 + x ) } = \frac { 1 } { 32 } \Rightarrow \frac { ( 24 + x ) - ( 24 - x ) } { ( 24 - x ) ( 24 + x ) } = \frac { 1 } { 32 }{/tex}

{tex}\Rightarrow{/tex} {tex}\frac{{2x}}{576 + 24x - 24x -x^2} =\frac{{1}}{32}{/tex}
{tex}\Rightarrow \frac { 2 x } { \left( 576 - x ^ { 2 } \right) } = \frac { 1 } { 32 } \Rightarrow{/tex}576 - x² = 64x
{tex}\Rightarrow{/tex} x² + 64x - 576 = 0 {tex}\Rightarrow{/tex} x² + 72x - 8x - 576 = 0
{tex}\Rightarrow{/tex} x(x + 72) - 8(x + 72) = 0 {tex}\Rightarrow{/tex} (x + 72)(x - 8) = 0
{tex}\Rightarrow{/tex} x + 72 = 0 or x - 8 = 0
{tex}\Rightarrow{/tex} x = -72 or x = 8
{tex}\Rightarrow{/tex} x = 8 [{tex}{/tex}x {tex} \neq{/tex} -72, because  speed of the stream cannot be negative]
Hence, the speed of the stream is 8 km/hr.