A man went downstream for 28km in a motor boat and immediately returned. It took the man twice as long to make the return trip. If the speed of the river flow were twice as high, the trip downstream and back would take 672 minutes. Find the speed of the boat in still water and the speed of the river flow.

: প্রশ্নে বলা হচ্ছে, এক ব্যক্তি স্রোতের অনুকূলে 28km পথ যেয়ে পুনরায় ফিলে আসলো। যেতে যে সময় লেগেছিল। ফিরে আসতে তার দ্বিগুণ সময় লাগল। স্রোতের বেগ দ্বিগুণ হলে ঐ দূরত্ব স্রোতের অনুকূলে যেয়ে পুনরায় ফিরে আসতে মোট 672 মিনিট সময় লাগে । স্থির পানিতে নৌকার বেগ ও স্রোতের বেগ কত?

Let, the speed of boat in still water be x kmph and the speed of stream in still water be y kmph.

∴ The boat's downstream speed will be (x + y)

and the boat's upstream speed will be (x − y)

According to question,

$$\begin{gathered} 2 \left(\frac{28}{x+y}\right) = \frac{28}{x-y} \\ \Rightarrow \frac{2}{x+y} = \frac{1}{x-y} [Dividing both sides by 28] \\ \Rightarrow 2x -2y = x+y \\ \Rightarrow 2x - x = y + 2y = 3y \\ \therefore x = 3y \end{gathered}$$

Now, total time taken by the boat (if the stream gets 2 times high) to go and come back $= \frac{672}{60} = \frac{56}{5} hours$

So, according to the question

$$\begin{gathered} \frac{28}{x+2y} + \frac{28}{x-2y} = \frac{56}{5} [As, the speed of the stream is two times the speed of boat] \\ \Rightarrow \frac{28}{3y + 2y} + \frac{28}{3y - 2y} = \frac{56}{5} \\ \Rightarrow 28 \left(\frac{1}{5y} + \frac{1}{y}\right) = \frac{56}{5} \\ \Rightarrow \frac{1}{5y} + \frac{1}{y} = \frac{2}{5} \\ \Rightarrow \frac{1+5}{5y} = \frac{2}{5} \\ \Rightarrow \frac{6}{5y} = \frac{2}{5} \\ \Rightarrow 10y =30 \\ \therefore y = \frac{30}{10} = 3 kmph \\ Now, by putting the value of y in equation \left(i\right), we get \\ x=3y \\ \Rightarrow x = 3 \times 3 = 9 kmph \\ \therefore Speed of the boat in still water is 9 kmph and speed of the stream in still water is 3 kmph. \end{gathered}$$