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A motor boat whose speed in still water is 178 \mathrm{~km} / \mathrm{hr}, takes 1 hour more to go 24 \mathrm{~km} upstream than to return to the same spot. Find the speed of the stream.
CBSE | Class 10 | Exercise 10e | Maths | Quadratic Equations | RS Aggarwal
A motor boat whose speed in still water is 178 \mathrm{~km} / \mathrm{hr}, takes 1 hour more to go 24 \mathrm{~km} upstream than to return to the same spot. Find the speed of the stream.

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

Given:

Speed of the boat =18 \mathrm{~km} / \mathrm{hr}

\therefore Speed downstream =(18+x) \mathrm{km} / h r

Speed upstream =(18-x) \mathrm{km} / \mathrm{hr}

\begin{array}{l} \therefore \frac{24}{(18-x)}-\frac{24}{(18-x)}=1 \\ \Rightarrow \frac{1}{(18-x)}-\frac{1}{(18+x)}=\frac{1}{24} \\ \Rightarrow \frac{18+x-18+x}{(18-x)(18+x)}=\frac{1}{24} \\ \Rightarrow \frac{2 x}{18^{2}-x^{2}}=\frac{1}{24} \\ \Rightarrow 324-x^{2}=48 x \\ \Rightarrow 324-x^{2}-48 x=0 \\ \Rightarrow x^{2}+48 x-324=0 \\ \Rightarrow x^{2}+(54-6) x-324=0 \\ \Rightarrow x^{2}+54 x-6 x-324=0 \\ \Rightarrow x(x+54)-6(x+54)=0 \\ \Rightarrow(x+54)(x-6)=0 \\ \Rightarrow x=-54 \text { or } x=6 \end{array}

The value of x cannot be negative; therefore, the speed of the stream is 6 \mathrm{~km} / \mathrm{hr}.

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