A man takes halftime in rowing a certain distance downstream than upstream. What is the ratio of the speed in still water to the speed of current?

Correct Option (4)

Let the speed of the man in still water be $u$u km/hr and the speed of the current be $v$v km/hr.

• The speed downstream (with the current) is $(u + v)$(u+v) km/hr.
• The speed upstream (against the current) is $(u - v)$(u−v) km/hr.

Let $d$d be the distance covered. The time taken to row upstream $\left( t_{1} \right)$(t1​) and downstream $\left( t_{2} \right)$(t2​) are given by:

• $t_{1} = \frac{d}{u - v}$t1​=u−vd​
• $t_{2} = \frac{d}{u + v}$t2​=u+vd​

According to the problem statement, the man takes half the time rowing downstream than upstream. This implies $t_{2} = \frac{1}{2} t_{1}$t2​=21​t1​, or equivalently, $\frac{t_{1}}{t_{2}} = \frac{2}{1}$t2​t1​​=12​.

Substituting the expressions for $t_{1}$t1​ and $t_{2}$t2​ into the ratio:

$$\frac{\frac{d}{(u - v)}}{\frac{d}{(u + v)}} = \frac{2}{1}$$(u+v)d​(u−v)d​​=12​

Simplifying the equation:

$$\frac{u + v}{u - v} = \frac{2}{1}$$u−vu+v​=12​

$$u + v = 2(u - v)$$u+v=2(u−v)

$$u + v = 2u - 2v$$u+v=2u−2v

Rearranging the terms to find the ratio of $u$u to $v$v:

$$v + 2v = 2u - u$$v+2v=2u−u

$$3v = u$$3v=u

Therefore, the ratio of the speed in still water to the speed of the current is $u:v = 3:1$u:v=3:1.

Incorrect Options:

Options 1 (1:2), 2 (2:1), and 3 (1:3) are incorrect. The mathematical derivation based on the given condition that the time taken downstream is half the time taken upstream unequivocally establishes the ratio of the speed in still water to the speed of the current as 3:1. These alternative ratios do not satisfy the problem's premise.