A swimmer takes 4 second in crossing some distance in downstream and taken 6 second in upstream for same distance then find the time taken by him to cover same distance in still water.

Answer

To solve the problem of finding the time taken by a swimmer to cover the same distance in still water, we can follow these steps:

### Step 1: Define Variables
Let:
- \( v \) = velocity of the swimmer in still water
- \( u \) = velocity of the river
- \( s \) = distance covered by the swimmer

### Step 2: Write Equations for Downstream and Upstream
1. **Downstream**: The swimmer's effective velocity is \( v + u \).
\[ s = (v + u) \times t_{down} \]
Given \( t_{down} = 4 \) seconds, we can write:
\[ s = (v + u) \times 4 \quad \text{(1)} \]

2. **Upstream**: The swimmer's effective velocity is \( v - u \).
\[ s = (v - u) \times t_{up} \]
Given \( t_{up} = 6 \) seconds, we can write:
\[ s = (v - u) \times 6 \quad \text{(2)} \]

### Step 3: Set the Two Equations Equal
Since both equations equal \( s \), we can set them equal to each other:
\[ (v + u) \times 4 = (v - u) \times 6 \]

### Step 4: Expand and Rearrange the Equation
Expanding both sides gives:
\[ 4v + 4u = 6v - 6u \]
Rearranging the terms:
\[ 4u + 6u = 6v - 4v \]
\[ 10u = 2v \]
From this, we can express \( u \) in terms of \( v \):
\[ u = \frac{2v}{10} = 0.2v \]

### Step 5: Substitute \( u \) Back into One of the Original Equations
We can substitute \( u = 0.2v \) into equation (1):
\[ s = (v + 0.2v) \times 4 \]
\[ s = (1.2v) \times 4 = 4.8v \quad \text{(3)} \]

### Step 6: Substitute \( u \) Back into the Other Equation
Now substitute \( u = 0.2v \) into equation (2):
\[ s = (v - 0.2v) \times 6 \]
\[ s = (0.8v) \times 6 = 4.8v \quad \text{(4)} \]

### Step 7: Calculate Time Taken in Still Water
In still water, the swimmer's speed is \( v \) and the distance is \( s \):
\[ \text{Time} = \frac{s}{v} \]
Substituting \( s \) from equation (3):
\[ \text{Time} = \frac{4.8v}{v} = 4.8 \text{ seconds} \]

### Final Answer
The time taken by the swimmer to cover the same distance in still water is **4.8 seconds**.

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