When a person goes to his office from his house with a speed of 10 km/hr, he is late by 20 minutes. When he goes with speed of 15 km/hr, he is late by 5 minutes. What is the distance (in km between his office and house?

Options

A. 8
B. 7.5
C. 7
D. 8.5

Answer

To solve the problem step by step, we can follow these steps:

### Step 1: Understand the Problem
We have two scenarios based on the speed of the person traveling to his office:
- At a speed of 10 km/hr, he is late by 20 minutes.
- At a speed of 15 km/hr, he is late by 5 minutes.

### Step 2: Convert Late Time to Hours
Since the speeds are given in km/hr, we need to convert the late times from minutes to hours:
- 20 minutes = 20/60 hours = 1/3 hours
- 5 minutes = 5/60 hours = 1/12 hours

### Step 3: Set Up the Equations
Let the distance between the house and the office be \( D \) km.

Using the formula for time, we can express the time taken for each speed:
- Time taken at 10 km/hr: \( \frac{D}{10} \)
- Time taken at 15 km/hr: \( \frac{D}{15} \)

Since he is late, we can set up the following equations based on the late times:
1. \( \frac{D}{10} = T + \frac{1}{3} \) (where \( T \) is the actual time he should take)
2. \( \frac{D}{15} = T + \frac{1}{12} \)

### Step 4: Rearranging the Equations
From the first equation:
\[ T = \frac{D}{10} - \frac{1}{3} \]

From the second equation:
\[ T = \frac{D}{15} - \frac{1}{12} \]

### Step 5: Set the Two Expressions for T Equal
Now we can set the two expressions for \( T \) equal to each other:
\[ \frac{D}{10} - \frac{1}{3} = \frac{D}{15} - \frac{1}{12} \]

### Step 6: Solve for D
To eliminate the fractions, we can find a common denominator (which is 60 in this case):
- Multiply the entire equation by 60:
\[ 6D - 20 = 4D - 5 \]

Now, simplify:
\[ 6D - 4D = 20 - 5 \]
\[ 2D = 15 \]
\[ D = 7.5 \]

### Step 7: Conclusion
The distance between his house and office is **7.5 km**.

---