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

Let the speed of the man in still water be S_m and the speed of the current be S_c.

1. Define speeds:

   • Downstream speed = S_m + S_c (speed with the current)
   • Upstream speed = S_m - S_c (speed against the current)

2. Define time and distance:

   • Let the certain distance be D.
   • Time taken downstream (T_d) = D / (S_m + S_c)
   • Time taken upstream (T_u) = D / (S_m - S_c)

3. Apply the given condition: The problem states that the man takes half time downstream than upstream: T_d = (1/2) * T_u

4. Substitute and solve: D / (S_m + S_c) = (1/2) * [D / (S_m - S_c)] Cancel D from both sides: 1 / (S_m + S_c) = 1 / [2 * (S_m - S_c)] Cross-multiply: 2 * (S_m - S_c) = S_m + S_c 2S_m - 2S_c = S_m + S_c Rearrange the terms to group S_m and S_c: 2S_m - S_m = S_c + 2S_c S_m = 3*S_c

5. Determine the ratio: The question asks for the ratio of the speed in still water to the speed of current, which is S_m : S_c. From S_m = 3*S_c, we can write: S_m / S_c = 3 / 1 So, the ratio S_m : S_c = 3:1.

Analysis of Options:

• A) 1:2: Incorrect. This would imply the current is twice as fast as the man in still water, which is physically impossible for the man to move upstream.
• B) 2:1: Incorrect. This would mean S_m = 2*S_c.
• C) 1:3: Incorrect. This would mean S_m = (1/3)*S_c, again implying current is much faster.
• D) 3:1: Correct. This matches our derived ratio S_m : S_c = 3:1.

The final answer is D.