Question

A motorboat whose speed is 9 km/hr in still water, goes 15 km downstream and comes back in a total time of 3 hours 45 minutes. Find the speed of the stream.

Answer 1

$$\begin{gathered} \begin{array}{l}\\ \text{Let the speed of the stream be}x\mathrm{km}/\mathrm{hr}.\end{array} \\ \therefore \mathrm{D}\text{ownstream speed=}(9+x)\text{km/hr} \\ \text{Upstream speed=}(9-x)\text{km/hr} \\ \mathrm{Distance}\mathrm{covered}\mathrm{downstream}=\mathrm{Distance}\mathrm{covered}\mathrm{upstream}=15\mathrm{km} \\ \mathrm{Total}\mathrm{time}\mathrm{taken}=3\mathrm{hours}45\mathrm{minutes}=\left(3+\frac{45}{60}\right)\mathrm{minutes}=\frac{225}{60}\mathrm{minutes}=\frac{15}{4}\mathrm{minutes} \\ \therefore \frac{15}{(9+x)}+\frac{15}{(9-x)}=\frac{15}{4} \\ \Rightarrow \frac{1}{(9+x)}+\frac{1}{(9-x)}=\frac{1}{4} \\ \Rightarrow \frac{9-x+9+x}{(9+x)(9-x)}=\frac{1}{4} \\ \Rightarrow \frac{18}{9^2-x^2}=\frac{1}{4} \\ \Rightarrow \frac{18}{81-x^2}=\frac{1}{4} \\ \Rightarrow 81-x^2=72 \\ \Rightarrow 81-x^2-72=0 \\ \Rightarrow -x^2+9=0 \\ \Rightarrow x^2=9 \\ \Rightarrow x=3\text{or}x=-3 \\ \text{The value of}x\text{cannot be negative; therefore, the speed of the stream is 3 km/hr.} \end{gathered}$$