Walking with 3/4 of his usual speed ,Sachin covers a certain distance in 2 h more than the time he takes to cover the distance at his usual speed. Find the time taken by him to cover this distance with usual speed.

Options

A. 5H
B. 6H
C. 9H
D. 4H

Answer

To solve the problem step by step, we can use the relationship between speed, distance, and time. Let's denote:

- \( d \) = distance to be covered
- \( s \) = Sachin's usual speed
- \( t \) = time taken by Sachin to cover the distance at his usual speed

1. **Express the distance in terms of speed and time:**
The distance can be expressed as:
\[ d = s \times t \]

2. **Determine the reduced speed:**
Sachin walks at \( \frac{3}{4} \) of his usual speed, which is:
\[ \text{Reduced speed} = \frac{3}{4}s \]

3. **Express the time taken at the reduced speed:**
The time taken to cover the same distance at the reduced speed can be expressed as:
\[ \text{Time at reduced speed} = \frac{d}{\text{Reduced speed}} = \frac{d}{\frac{3}{4}s} = \frac{4d}{3s} \]

4. **Set up the equation based on the problem statement:**
According to the problem, the time taken at the reduced speed is 2 hours more than the time taken at the usual speed:
\[ \frac{4d}{3s} = t + 2 \]

5. **Substitute \( d \) from step 1 into the equation:**
Replace \( d \) with \( s \times t \):
\[ \frac{4(s \times t)}{3s} = t + 2 \]
Simplifying this gives:
\[ \frac{4t}{3} = t + 2 \]

6. **Clear the fraction by multiplying through by 3:**
\[ 4t = 3t + 6 \]

7. **Solve for \( t \):**
Subtract \( 3t \) from both sides:
\[ 4t - 3t = 6 \]
\[ t = 6 \]

Thus, the time taken by Sachin to cover the distance at his usual speed is **6 hours**.