If a steamer and a motor-launch go down stream, then the steamer covers the distance from A to B $1.5$ times as fast as the motor-launch, the latter lagging behind the steamer $8$ km more each hour. Now if they go up stream, then the steamer covers the distance from B to A twice as fast as the motor-launch. Find the speeds of the steamer and the motor-launch in still water.
Direction A to B is downstream
Direction B to A is upstream
Let speed of steamer in still water $= u$ kmph
Let speed of motor-launch in still water $= v$ kmph
Let speed of stream $= s$ kmph
Speed of steamer downstream $= \left(u + s\right)$ kmph
Speed of motor-launch downstream $= \left(v + s\right)$ kmph
Given: Steamer downstream speed $=$ motor-launch downstream speed $\times 1.5$
i.e. $\;$ $u + s = \left(v + s\right) \times 1.5$
i.e. $\;$ $u + s = 1.5 v + 1.5s$
i.e. $\;$ $u - 1.5 v = 0.5 s$
i.e. $\;$ $s = 2u - 3v$ $\;\;\; \cdots \; (1)$
In view of equation $(1)$,
speed of steamer downstream $= u + 2u - 3v = 3u - 3v$ kmph $\;\;\; \cdots \; (2)$
speed ofmotor-launch downstream $= v + 2u - 3v = 2u - 2v$ kmph $\;\;\; \cdots \; (3)$
speed of steamer upstream $= \left(u - s\right) = u - 2u + 3v = 3v - u$ kmph $\;\;\; \cdots \; (4)$
speed of motor-launch upstream $= \left(v - s\right) = v - 2u + 3v = 4v - 2u$ kmph $\;\;\; \cdots \; (5)$
Given: Steamer upstream speed $=$ motor-launch upstream speed $\times 2$
i.e. $\;$ $3v - u = \left(4v - 2u\right) \times 2$
i.e. $\;$ $3v - u = 8v - 4u$
i.e. $\;$ $3u = 5v$
i.e. $\;$ $v = 0.6 u$ $\;\;\; \cdots \; (6)$
In view of equation $(6)$, we have
from equation $(2)$, speed of steamer downstream $= 3u - \left(3 \times 0.6u\right) = 1.2u$ kmph $\;\;\; \cdots \; (7)$
from equation $(3)$, speed of motor-launch downstream $= 2u - \left(2 \times 0.6u\right) = 0.8u$ kmph $\;\;\; \cdots \; (8)$
Let distance between points A and B be $= d$ km
While going downstream,
time taken by steamer to cover distance AB $= \dfrac{d}{1.2u}$ hr
time taken by motor-launch to cover distance AB $= \dfrac{d}{0.8u}$ hr
$\therefore \;$ Excess time taken by motor-launch to cover AB $= T = \dfrac{d}{0.8u} - \dfrac{d}{1.2u} = \dfrac{d}{2.4u}$ hr $\;\;\; \cdots \; (9)$
For every hour, the motor-launch lags the steamer by $8$ km
$\therefore \;$ For time $T$, the motor-launch lags the steamer by $= \dfrac{d}{2.4u} \times 8 = \dfrac{d}{0.3u}$ km $\;\;\; \cdots \; (10)$
Now, in time $T = \dfrac{d}{2.4u}$ hr,
the steamer covers downstream a distance $= d_1 = 1.2u \times \dfrac{d}{2.4u} = \dfrac{d}{2}$ km
and the motor-launch covers downstream a distance $= d_2 = 0.8u \times \dfrac{d}{2.4u} = \dfrac{d}{3}$ km
$\therefore \;$ In time $T$, the motor-launch lags the steamer by $= d_1 - d_2 = \dfrac{d}{2} - \dfrac{d}{3} = \dfrac{d}{6}$ km $\;\;\; \cdots \; (11)$
We have from equations $(10)$ and $(11)$
$\dfrac{d}{0.3u} = \dfrac{d}{6}$ $\implies$ $u = 20$ kmph
$\therefore \;$We have from equation $(6)$, $v = 0.6 \times 20 = 12$ kmph
Therefore, speed of steamer in still water $= 20$ kmph
and speed of motor-launch in still water $= 12$ kmph