Question

A boat goes 30 km upstream and 44 km downstream in 10 hours. It can go 40 km upstream and 55 km downstream in 13 hours. Find the speed of the stream and that of the boat in still water.

Answer 1

Let speed of boat in still water be x km/h and speed of stream be y km/h.
Speed Upstream = (x − y) km/h
Speed downstream = (x + y) km/h

According to the question,
$$\begin{gathered} \frac{30}{x-y}+\frac{44}{x+y}=10 \\ \mathrm{and} \\ \frac{40}{x-y}+\frac{55}{x+y}=13 \\ \mathrm{Let}\frac{1}{x-y}=p\mathrm{and}\frac{1}{x+y}=q...\left(1\right) \\ Therefore,theequationbecomes \\ 30p+44q=10...\left(2\right) \\ 40p+55q=13...\left(3\right) \\ 40p+55q=13 \\ \Rightarrow 40p=13-55q \\ \Rightarrow p=\frac{13-55q}{40}...\left(4\right) \\ \mathrm{Substituting}\mathrm{the}\mathrm{value}\mathrm{of}p\mathrm{in}\left(2\right),\mathrm{we}\mathrm{get} \\ 30\left(\frac{13-55q}{40}\right)+44q=10 \\ \Rightarrow 30\left(\frac{13-55q}{40}\right)+44q=10 \\ \Rightarrow \frac{3\left(13-55q\right)+4\left(44q\right)}{4}=10 \\ \Rightarrow 39-165q+176q=40 \\ \Rightarrow 11q=40-39 \\ \Rightarrow 11q=1 \\ \Rightarrow q=\frac{1}{11}...\left(5\right) \\ \mathrm{Substituting}\mathrm{the}\mathrm{value}\mathrm{of}q\mathrm{in}\left(4\right),\mathrm{we}\mathrm{get} \\ p=\frac{13-55{\displaystyle \frac{1}{11}}}{40} \\ \Rightarrow p=\frac{13-5}{40} \\ \Rightarrow p=\frac{8}{40} \\ \Rightarrow p=\frac{1}{5}...\left(6\right) \\ \mathrm{From}\left(1\right),\left(5\right)\mathrm{and}\left(6\right),\mathrm{we}\mathrm{get} \\ x-y=5\mathrm{and}x+y=11 \\ \mathrm{Solving}\mathrm{both},\mathrm{we}\mathrm{get} \\ x=8\mathrm{and}y=3 \end{gathered}$$

Hence, the speed of the stream and that of the boat in still water is 3 km/h and 8 km/h, respectively.