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Home » A motorboat whose speed is in still water, goes downstream and comes back in a total time of 3 hours 45 minutes. Find the speed of the stream.
A motorboat whose speed is 9 \mathrm{~km} / \mathrm{hr} in still water, goes 15 \mathrm{~km} downstream and comes back in a total time of 3 hours 45 minutes. Find the speed of the stream.
CBSE | Class 10 | Exercise 10e | Maths | Quadratic Equations | RS Aggarwal
A motorboat whose speed is 9 \mathrm{~km} / \mathrm{hr} in still water, goes 15 \mathrm{~km} downstream and comes back in a total time of 3 hours 45 minutes. Find the speed of the stream.

Let the speed of the stream be x \mathrm{~km} / \mathrm{hr}.

\therefore Downstream speed =(9+x) \mathrm{km} / \mathrm{hr}

Upstream speed =(9-x) \mathrm{km} / \mathrm{hr}

Distance covered downstream = Distance covered upstream =15 \mathrm{~km}

\begin{array}{l} \text { Total time taken }=3 \text { hours } 45 \text { minutes }=\left(3+\frac{45}{60}\right) \text { minutes }=\frac{225}{60} \text { minutes }=\frac{15}{4} \text { 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 \end{array}
\begin{array}{l} \Rightarrow-x^{2}+9=0 \\ \Rightarrow x^{2}=9 \\ \Rightarrow x=3 \text { or } x=-3 \end{array}

The value of x cannot be negative; therefore, the speed of the stream is 3 \mathrm{~km} / \mathrm{hr}.

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