A motor boat goes from A to B and comes back located at bank of river. If speed of motor boat becomes double in still water, it takes 20 % of its original time to travel from A to B and B to A. Find speed of motorboat is how many times of flow of river?

\(\sqrt {\frac{3}{2}}\)

Let boat's speed = A

River flow's speed = B

Speed of boat in downstream = A + B

Speed of boat in upstream = A - B

Distance = \(\frac{Product\;of\;speed}{sum\;of\;speed}\) × Time

ATQ, 

⇒ \(\frac{(A+B) (A-B)}{(A+B)+(A-B)}\)   × T = \(\frac{(2A+B) (2A-B)}{(2A+B)+(2A-B)}\) × 20% of T

⇒ \(\frac{A^2 - B^2}{2A}\) × 5T = \(\frac{4A^2 - B^2}{4B}\) × T

⇒ 10 A² - 10B² = 4A² - B²

⇒ 6A² = 9B²

⇒ A = \(\sqrt {\frac{3}{2}}\) × B