Aman and Chirag decided to meet each other at point B. Aman started to point at 11:00 a.m. and at the same time Chirag started from point C. They move towards each other to meet at point B. Chirag take 1 hour more to reach point B relative to Aman. Speed of Chirag is 20% more than speed of Aman and distance covered by Aman `33(1)/(3)%` less than distance covered by Chirag.
At what time Chirag will reach at point B?

Options

A. `2:30` P.M.
B. `3:00` P.M.
C. `3:30` P.M.
D. `4:00` P.M.

Answer

To solve the problem step by step, let's define the variables and relationships based on the information given:

1. **Define Variables**:
- Let the speed of Aman be \( S_A \).
- Since Chirag's speed is 20% more than Aman's, we can express Chirag's speed as:
\[ S_C = S_A + 0.2 S_A = 1.2 S_A \]

2. **Define Distances**:
- Let the distance covered by Chirag be \( D_C \).
- According to the problem, the distance covered by Aman is \( 33 \frac{1}{3} \% \) less than that covered by Chirag. This can be expressed as:
\[ D_A = D_C - \frac{1}{3} D_C = \frac{2}{3} D_C \]

3. **Time Relationship**:
- Let the time taken by Aman to reach point B be \( T_A \).
- Since Chirag takes 1 hour more than Aman to reach point B, we have:
\[ T_C = T_A + 1 \]

4. **Using the formula for time**:
- Time is calculated as distance divided by speed. Therefore, we can express the times for Aman and Chirag as:
\[ T_A = \frac{D_A}{S_A} \quad \text{and} \quad T_C = \frac{D_C}{S_C} \]

5. **Substituting the distances and speeds**:
- For Aman:
\[ T_A = \frac{\frac{2}{3} D_C}{S_A} \]
- For Chirag:
\[ T_C = \frac{D_C}{1.2 S_A} \]

6. **Setting up the equation**:
- Since \( T_C = T_A + 1 \), we can substitute the expressions for \( T_A \) and \( T_C \):
\[ \frac{D_C}{1.2 S_A} = \frac{\frac{2}{3} D_C}{S_A} + 1 \]

7. **Eliminating \( D_C \) and \( S_A \)**:
- We can cancel \( D_C \) from both sides (assuming \( D_C \neq 0 \)):
\[ \frac{1}{1.2 S_A} = \frac{2/3}{S_A} + \frac{1}{D_C} \]
- Rearranging gives:
\[ \frac{1}{1.2} = \frac{2}{3} + \frac{S_A}{D_C} \]
- Solving for \( \frac{S_A}{D_C} \) gives:
\[ \frac{S_A}{D_C} = \frac{1}{1.2} - \frac{2}{3} \]

8. **Finding a common denominator**:
- The common denominator for \( 1.2 \) and \( 3 \) is \( 6 \):
\[ \frac{1}{1.2} = \frac{5}{6} \quad \text{and} \quad \frac{2}{3} = \frac{4}{6} \]
- Thus:
\[ \frac{S_A}{D_C} = \frac{5}{6} - \frac{4}{6} = \frac{1}{6} \]

9. **Finding the time taken by Aman**:
- Now, we can express \( D_C \) in terms of \( S_A \):
\[ D_C = 6 S_A \]
- Substituting back to find \( T_A \):
\[ T_A = \frac{\frac{2}{3} (6 S_A)}{S_A} = 4 \text{ hours} \]

10. **Finding Chirag's time**:
- Since Chirag takes 1 hour more than Aman:
\[ T_C = T_A + 1 = 4 + 1 = 5 \text{ hours} \]

11. **Final Time Calculation**:
- Aman starts at 11:00 AM, so:
\[ \text{Chirag will reach point B at } 11:00 \text{ AM} + 5 \text{ hours} = 4:00 \text{ PM} \]

### Final Answer:
Chirag will reach point B at **4:00 PM**.