Water current

John swims upstream. After a while, he passes the bottle, from that moment he floats for 20 minutes in the same direction. He then turns around and swims back, and from the first meeting with the bottle, he sails 2 kilometers before he reaches the bottle. What is the current speed? (John is still swimming at the same speed regardless of the current.

Correct result:

f =  3 km/h

Solution:

$$\begin{gathered} s_1=(j-f)\cdot 20\mathrm{/}60 \\ {s}_2=2\text{ km } \\ {s}_2=f\cdot ({\displaystyle \frac{s_1+s_2}{f+j}}+20\mathrm{/}60) \\ 2=f\cdot ({\displaystyle \frac{(j-f)\cdot 20\mathrm{/}60+2}{f+j}}+20\mathrm{/}60) \\ 2=f\cdot ({\displaystyle \frac{(j-f)\mathrm{/}3+2}{f+j}}+1\mathrm{/}3) \\ (f-3)j\mathrm{/}(f+j)=0 \\ j,f>0 \\ j\mathrm{\ne }f \\ f-3=0 \\ f=3=3\text{ km/h} \end{gathered}$$

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