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

The correct option is 3 : 1.

Step 1: Understand the given information and define variables.
Let the speed of the man in still water be v and the speed of the stream/current be u.
Let the distance covered be d.
When traveling downstream (in the direction of the stream), the effective speed is:

$v_{\text{down}}=v+u$

When traveling upstream (against the stream), the effective speed is:

$v_{\text{up}}=v-u$

Step 2: Relate the time taken for upstream and downstream journeys.
We know that $\text{Time}=\frac{\text{Distance}}{\text{Speed}}$.
Time taken downstream, $t_{\text{down}}=\frac{d}{v+u}$
Time taken upstream, $t_{\text{up}}=\frac{d}{v-u}$

According to the problem, the time taken to row downstream is half the time taken to row upstream:

$t_{\text{down}}=\frac{1}{2}{t}_{\text{up}}$

Step 3: Solve for the ratio of speed in still water to the speed of the current (v : u).
Substitute the time formulas into the equation:

$\frac{d}{v+u}=\frac{1}{2}\times \frac{d}{v-u}$

Divide both sides by d:

$\frac{1}{v+u}=\frac{1}{2(v-u)}$

Cross-multiplying gives:

$2(v-u)=v+u$

Expand the left side:

$2v-2u=v+u$

Rearrange terms to group v on one side and u on the other side:

$2v-v=u+2u$

$v=3u$

Now, find the ratio of v to u:

$\frac{v}{u}=\frac{3}{1}$

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