A man rows a boat upstream a certain distance and then returns back to the same place. If the time taken by him in going upstream is twice the time taken in rowing downstream, find the ratio of the speed of the boat in still water and the speed of the stream.

Options

A. `2:1`
B. `3:2`
C. `5:3`
D. `3:1`

Answer

To solve the problem step by step, we will denote the following:

- Let the distance traveled upstream (and downstream) be \( x \).
- Let the speed of the boat in still water be \( s \).
- Let the speed of the stream be \( y \).

### Step 1: Determine the time taken for upstream and downstream
1. **Upstream**: The effective speed while rowing upstream is \( s - y \). The time taken to row upstream is given by:
\[ \text{Time upstream} = \frac{x}{s - y} \]

2. **Downstream**: The effective speed while rowing downstream is \( s + y \). The time taken to row downstream is given by:
\[ \text{Time downstream} = \frac{x}{s + y} \]

### Step 2: Set up the equation based on the given condition
According to the problem, the time taken to go upstream is twice the time taken to go downstream:
\[ \frac{x}{s - y} = 2 \times \frac{x}{s + y} \]

### Step 3: Simplify the equation
1. We can cancel \( x \) from both sides (assuming \( x \neq 0 \)):
\[ \frac{1}{s - y} = \frac{2}{s + y} \]

2. Cross-multiplying gives:
\[ s + y = 2(s - y) \]

### Step 4: Expand and rearrange the equation
1. Expanding the right side:
\[ s + y = 2s - 2y \]

2. Rearranging the terms to isolate \( s \) and \( y \):
\[ s + y + 2y = 2s \]
\[ s + 3y = 2s \]

3. Subtract \( s \) from both sides:
\[ 3y = 2s - s \]
\[ 3y = s \]

### Step 5: Find the ratio of the speed of the boat to the speed of the stream
1. From the equation \( s = 3y \), we can express the ratio of the speed of the boat in still water to the speed of the stream:
\[ \frac{s}{y} = \frac{3y}{y} = 3 \]

Thus, the ratio of the speed of the boat in still water to the speed of the stream is:
\[ \text{Ratio} = 3:1 \]