Choose the correct alternative. Ratish and Satish start walking from a point A to point B, a distance of 45 km at the speed of 4 and $$ 5 \mathrm{~km} / \mathrm{hr} $$ respectively. Satish reaches B, turns around and starts walking towards A. He meets Ratish at a point X. Find the distance from A to X?

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To determine the distance from point $$ A $$ to the meeting point $$ X $$, let's analyze the movements of Ratish and Satish step by step.

1. **Initial Movement:**
   - **Ratish** walks from $$ A $$ to $$ B $$ at $$ 4 \, \text{km/hr} $$.
   - **Satish** walks from $$ A $$ to $$ B $$ at $$ 5 \, \text{km/hr} $$.
   - The distance between $$ A $$ and $$ B $$ is $$ 45 \, \text{km} $$.

2. **Time to Reach $$ B $$:**
   - **Ratish** takes $$ \frac{45}{4} = 11.25 $$ hours to reach $$ B $$.
   - **Satish** takes $$ \frac{45}{5} = 9 $$ hours to reach $$ B $$.

3. **After Reaching $$ B $$:**
   - **Satish** reaches $$ B $$ at $$ t = 9 $$ hours and immediately starts walking back towards $$ A $$ at $$ 5 \, \text{km/hr} $$.

4. **Positions at $$ t = 9 $$ Hours:**
   - **Ratish** has walked $$ 9 \times 4 = 36 \, \text{km} $$ from $$ A $$, so he is $$ 45 - 36 = 9 \, \text{km} $$ away from $$ B $$.
   - **Satish** starts returning towards $$ A $$ from $$ B $$.

5. **Meeting Point $$ X $$:**
   - Let $$ t' $$ be the time after $$ t = 9 $$ hours when they meet.
   - **Ratish** walks an additional $$ 4t' $$ km towards $$ B $$.
   - **Satish** walks $$ 5t' $$ km back towards $$ A $$.

At the meeting point:
   $$
   36 + 4t' = 45 - 5t'
   $$
   Solving for $$ t' $$:
   $$
   9t' = 9 \implies t' = 1 \, \text{hour}
   $$

Therefore, the meeting point $$ X $$ is:
   $$
   36 \, \text{km} + 4 \times 1 \, \text{km} = 40 \, \text{km} \text{ from } A
   $$

**Answer:** The distance from $$ A $$ to $$ X $$ is **40 km**.
The distance from $$ A $$ to $$ X $$ is **40 km**.

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