Shri X goes to his office by scooter at a speed of 30km/h and reaches 6 minutes earlier. If he goes at a speed of 24 km/h, he reaches 5 minutes late. The distance of his office is

Options

A. 20 km
B. 21 km
C. 22 km
D. 24 km

Answer

To solve the problem step by step, let's denote the distance to Shri X's office as \( D \) kilometers and the normal time taken to reach the office at the normal speed as \( t \) hours.

### Step 1: Set up the equations based on the information given.

1. When Shri X travels at 30 km/h, he reaches 6 minutes early. This means he takes \( t - \frac{6}{60} \) hours to reach the office.
2. When he travels at 24 km/h, he reaches 5 minutes late. This means he takes \( t + \frac{5}{60} \) hours to reach the office.

### Step 2: Write the equations for the distance.

Using the formula \( \text{Distance} = \text{Speed} \times \text{Time} \):

1. For the first case (30 km/h):
\[ D = 30 \left(t - \frac{6}{60}\right) \]
Simplifying gives:
\[ D = 30 \left(t - 0.1\right) = 30t - 3 \]

2. For the second case (24 km/h):
\[ D = 24 \left(t + \frac{5}{60}\right) \]
Simplifying gives:
\[ D = 24 \left(t + \frac{1}{12}\right) = 24t + 2 \]

### Step 3: Set the two expressions for distance equal to each other.

Since both expressions represent the same distance \( D \):
\[ 30t - 3 = 24t + 2 \]

### Step 4: Solve for \( t \).

Rearranging the equation:
\[ 30t - 24t = 2 + 3 \]
\[ 6t = 5 \]
\[ t = \frac{5}{6} \text{ hours} \]

### Step 5: Calculate the distance \( D \).

Substituting \( t \) back into either distance equation. Let's use the first one:
\[ D = 30 \left(\frac{5}{6} - \frac{6}{60}\right) \]
Calculating \( \frac{6}{60} = \frac{1}{10} \):
\[ D = 30 \left(\frac{5}{6} - \frac{1}{10}\right) \]
Finding a common denominator (30):
\[ D = 30 \left(\frac{25}{30} - \frac{3}{30}\right) = 30 \left(\frac{22}{30}\right) = 22 \text{ km} \]

### Final Answer:
The distance of Shri X's office is \( \boxed{22} \) km.