A man swimming downstream overcomes a float at a point M. After travelling distance D,he turned back and passed the float at a distance of `D//2` from the point M. Then the ratio of speed of swimmer with respect to still water to the speed of the river will be

Options

A. 1
B. 2
C. 4
D. 3

Explanation

To solve the problem, let's denote the following variables:

- \( U \) = speed of the swimmer with respect to still water
- \( V \) = speed of the river
- \( D \) = distance traveled downstream before turning back
- \( D/2 \) = distance from point M where the swimmer passes the float on his way back

### Step 1: Determine the time taken to swim downstream

When the swimmer swims downstream, his effective speed is the sum of his speed and the speed of the river:

\[ \text{Effective speed downstream} = U + V \]

The time taken to swim the distance \( D \) downstream is given by:

\[ T_1 = \frac{D}{U + V} \]

### Step 2: Determine the time taken to swim upstream

When the swimmer turns back and swims upstream, his effective speed is the difference between his speed and the speed of the river:

\[ \text{Effective speed upstream} = U - V \]

The distance he swims upstream to pass the float is \( D/2 \). The time taken to swim this distance is:

\[ T_2 = \frac{D/2}{U - V} \]

### Step 3: Determine the total time taken

Since the float is also moving downstream with the river, we need to account for the time it takes the float to move downstream during the time \( T_1 + T_2 \). The float moves a distance of \( V(T_1 + T_2) \).

### Step 4: Set up the equation

The total distance traveled by the float in the time \( T_1 + T_2 \) should equal the distance \( D + D/2 \):

\[ D + \frac{D}{2} = V \left( T_1 + T_2 \right) \]

Substituting the expressions for \( T_1 \) and \( T_2 \):

\[ D + \frac{D}{2} = V \left( \frac{D}{U + V} + \frac{D/2}{U - V} \right) \]

### Step 5: Simplify the equation

Combine the left side:

\[ \frac{3D}{2} = V \left( \frac{D}{U + V} + \frac{D/2}{U - V} \right) \]

We can cancel \( D \) from both sides (assuming \( D \neq 0 \)):

\[ \frac{3}{2} = V \left( \frac{1}{U + V} + \frac{1/2}{U - V} \right) \]

### Step 6: Solve for the ratio \( \frac{U}{V} \)

Multiply both sides by \( 2(U + V)(U - V) \):

\[ 3(U + V)(U - V) = 2V \left( (U - V) + \frac{1}{2}(U + V) \right) \]

Expanding and simplifying leads to a quadratic equation in terms of \( \frac{U}{V} \). Let \( x = \frac{U}{V} \):

\[ 3(x + 1)(x - 1) = 2x \left( (x - 1) + \frac{1}{2}(x + 1) \right) \]

### Step 7: Solve the quadratic equation

After simplifying, we will find two solutions for \( x \). Since speed cannot be negative, we take the positive solution.

### Final Result

The ratio of the speed of the swimmer with respect to still water to the speed of the river is:

\[ \frac{U}{V} = 3 \]