Point C is at a distance of $12$ km from point B down the river. A fisher left point A, which is somewhat farther up the river than point B, for point C in a boat. He arrived at C four hours later and covered the return trip in six hours. Fixing a motor to the boat and thus trebling its speed relative to the water, the fisher covered the distance from A to B in $45$ minutes. Determine the speed of the river flow, considering it to be constant.
Direction A to B to C is downstream
Direction C to B to A is upstream
Let speed of boat in still water $= u$ kmph
Let speed of river flow $= v$ kmph
Downstream speed of boat $= u + v$ kmph $\;\;\; \cdots \; (1)$
Upstream speed of boat $= u - v$ kmph $\;\;\; \cdots \; (2)$
Let distance AB $= x$ km
Given: Distance BC $= 12$ km
Then, distance AC $= x + 12$ km
Time taken by the boat to arrive from point A to point C (downstream) $= 4$ hours
Therefore, downstream speed of boat $= \dfrac{x + 12}{4}$ kmph $\;\;\; \cdots \; (3)$
Therefore, we have from equations $(1)$ and $(3)$
$u + v = \dfrac{x + 12}{4}$
i.e. $\;$ $4u + 4v = x + 12$ $\implies$ $x = 4u + 4v - 12$ $\;\;\; \cdots \; (4)$
Time taken by the boat to arrive from point C to point A (upstream) $= 6$ hours
Therefore, upstream speed of boat $= \dfrac{x + 12}{6}$ kmph $\;\;\; \cdots \; (5)$
Therefore, we have from equations $(2)$ and $(5)$
$u - v = \dfrac{x + 12}{6}$
i.e. $\;$ $6u - 6v = x + 12$ $\implies$ $x = 6u - 6v - 12$ $\;\;\; \cdots \; (6)$
Therefore, we have from equations $(4)$ and $(6)$
$4u + 4v - 12 = 6u - 6v - 12$
i.e. $\;$ $2u = 10 v$ $\implies$ $u = 5v$ $\;\;\; \cdots \; (7)$
After fixing motor, new speed of boat $= 3u = 3 \times 5v = 15v$ kmph [in view of equation $(7)$]
Therefore, speed of boat downstream from A to B $= 15v + v = 16v$ kmph
Time taken by the boat to cover distance AB $= \dfrac{x}{16v}$ hr $\;\;\; \cdots \; (8)$
Given: Time now taken to cover distance AB $= 45$ minutes $= \dfrac{45}{60} = \dfrac{3}{4}$ hr $\;\;\; \cdots \; (9)$
We have from equations $(8)$ and $(9)$
$\dfrac{x}{16v} = \dfrac{3}{4}$ $\implies$ $x = \dfrac{3}{4} \times 16v = 12v$ km $\;\;\; \cdots \; (10)$
In view of equations $(7)$ and $(10)$, equation $(4)$ becomes
$12v = 20v + 4v - 12$
i.e. $\;$ $12v = 12$ $\implies$ $v = 1$ kmph
i.e. Speed of river flow $= v = 1$ kmph