Two boys P and Q are playing on a river bank. P plans to swim across the river directly and come back. Q plans to swim downstream by a length equal to the width of the river and then come back. Both of them bet each other, claiming that the boy succeeding in less time will win. Assuming the swimming rate of both P and Q to the same, it can be concluded that

Options

A. P wins
B. Q wins
C. A draw takes place
D. Nothing certain can be stated.

Explanation

To solve the problem, we need to analyze the swimming paths and times taken by both boys, P and Q.

### Step-by-Step Solution:

1. **Define the Variables:**
- Let \( D \) be the width of the river.
- Let \( v \) be the swimming speed of both boys.
- Let \( u \) be the speed of the river current.

2. **Path of Boy P:**
- Boy P swims directly across the river and then returns.
- The distance he swims in one direction is \( D \).
- The effective velocity component across the river is \( v \cos(\theta) \), where \( \theta \) is the angle made with the river bank.
- The time taken for P to swim across the river is given by:
\[ t_P = \frac{D}{v \cos(\theta)} \]
- Since he returns the same distance, the total time for P is:
\[ T_P = 2t_P = 2 \frac{D}{v \cos(\theta)} \]

3. **Path of Boy Q:**
- Boy Q swims downstream a distance equal to the width of the river and then returns.
- When swimming downstream, his effective speed is \( v + u \) and when swimming upstream, it is \( v - u \).
- The time taken for Q to swim downstream is:
\[ t_{Q1} = \frac{D}{v + u} \]
- The time taken for Q to swim upstream is:
\[ t_{Q2} = \frac{D}{v - u} \]
- Therefore, the total time for Q is:
\[ T_Q = t_{Q1} + t_{Q2} = \frac{D}{v + u} + \frac{D}{v - u} \]

4. **Combine the Times for Q:**
- To combine the times, we find a common denominator:
\[ T_Q = D \left( \frac{(v - u) + (v + u)}{(v + u)(v - u)} \right) = D \left( \frac{2v}{v^2 - u^2} \right) \]
- Thus, we have:
\[ T_Q = \frac{2Dv}{v^2 - u^2} \]

5. **Compare the Times:**
- Now we compare \( T_P \) and \( T_Q \):
\[ T_P = \frac{2D}{v \cos(\theta)} \quad \text{and} \quad T_Q = \frac{2Dv}{v^2 - u^2} \]
- To find out who wins, we take the ratio:
\[ \frac{T_P}{T_Q} = \frac{\frac{2D}{v \cos(\theta)}}{\frac{2Dv}{v^2 - u^2}} = \frac{v^2 - u^2}{v \cos(\theta)} \]

6. **Analyze the Ratio:**
- Since \( u < v \), \( v^2 - u^2 > 0 \).
- The term \( \cos(\theta) \) is always less than or equal to 1, thus:
\[ \frac{T_P}{T_Q} < 1 \]
- This means \( T_P < T_Q \), indicating that P takes less time than Q.

### Conclusion:
Boy P wins the bet as he swims across and back in less time than Boy Q who swims downstream and then back upstream.