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, we need to analyze the motion of the swimmer and the float in the river. Let's denote:

- \( v \): Speed of the swimmer with respect to still water.
- \( u \): Speed of the river.

### Step-by-step Solution:

1. **Understanding the Scenario**:
The swimmer swims downstream and passes the float at point M. He swims a distance \( D \) downstream, then turns back and passes the float again at a distance \( \frac{D}{2} \) from point M.

2. **Defining the Distances**:
- Distance from M to the point where the swimmer turns back: \( D \).
- Distance from the turning point back to the float: \( \frac{D}{2} \).

3. **Time Taken by the Float**:
The float moves downstream with the speed of the river \( u \). The total distance covered by the float from point M to the new position (let's call it point N) is:
\[ \text{Distance covered by float} = D + \frac{D}{2} = \frac{3D}{2} \]
The time taken by the float to cover this distance is:
\[ t_{\text{float}} = \frac{\frac{3D}{2}}{u} = \frac{3D}{2u} \]

4. **Time Taken by the Swimmer**:
- Time taken to swim downstream to point O (distance \( D \)):
\[ t_{\text{down}} = \frac{D}{v + u} \]
- Time taken to swim back to point N (distance \( \frac{D}{2} \)):
\[ t_{\text{up}} = \frac{\frac{D}{2}}{v - u} \]
Therefore, the total time taken by the swimmer is:
\[ t_{\text{swimmer}} = t_{\text{down}} + t_{\text{up}} = \frac{D}{v + u} + \frac{\frac{D}{2}}{v - u} \]

5. **Setting the Times Equal**:
Since the time taken by the swimmer is equal to the time taken by the float, we can set the equations equal:
\[ \frac{D}{v + u} + \frac{\frac{D}{2}}{v - u} = \frac{3D}{2u} \]

6. **Cancelling \( D \)**:
We can cancel \( D \) from both sides (assuming \( D \neq 0 \)):
\[ \frac{1}{v + u} + \frac{1/2}{v - u} = \frac{3}{2u} \]

7. **Finding a Common Denominator**:
The common denominator for the left side is \( 2(v + u)(v - u) \):
\[ \frac{2(v - u) + (v + u)}{2(v + u)(v - u)} = \frac{3}{2u} \]
Simplifying the numerator:
\[ \frac{2v - 2u + v + u}{2(v + u)(v - u)} = \frac{3}{2u} \]
This simplifies to:
\[ \frac{3v - u}{2(v + u)(v - u)} = \frac{3}{2u} \]

8. **Cross Multiplying**:
Cross-multiplying gives:
\[ (3v - u) \cdot 2u = 3 \cdot 2(v + u)(v - u) \]
Simplifying this leads to:
\[ 6uv - 2u^2 = 6(v^2 - u^2) \]

9. **Rearranging the Equation**:
Rearranging gives us a quadratic equation in terms of \( v \):
\[ 6v^2 - 6uv - 2u^2 = 0 \]

10. **Using the Quadratic Formula**:
The roots of the quadratic can be found using the quadratic formula:
\[ v = \frac{6u \pm \sqrt{(6u)^2 + 4 \cdot 6 \cdot 2u^2}}{2 \cdot 6} \]
Simplifying gives:
\[ v = \frac{6u \pm \sqrt{36u^2 + 48u^2}}{12} = \frac{6u \pm 8u}{12} \]
This results in two possible values for \( v \):
\[ v = \frac{14u}{12} = \frac{7u}{6} \quad \text{and} \quad v = \frac{-2u}{12} = -\frac{u}{6} \quad (\text{not valid}) \]

11. **Finding the Ratio**:
The valid solution gives:
\[ \frac{v}{u} = \frac{7}{6} \]

### Final Answer:
The ratio of the speed of the swimmer with respect to still water to the speed of the river is \( \frac{7}{6} \).