A man walks from his house at an average speed of 5km per hour and reaches his office 6 minutes late. If he walks at an average speed of 6km/h he reaches 2 minutes early. The distance of the office from his house is

Options

A. 6km
B. 9km
C. 12km
D. 4km

Answer

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

### Step 1: Define Variables
Let the distance from the man's house to his office be \( D \) kilometers.

### Step 2: Calculate Time Taken at Different Speeds
1. When the man walks at 5 km/h, he is 6 minutes late.
- Let the time he should take to reach the office on time be \( T \) hours.
- The time taken at 5 km/h is \( \frac{D}{5} \) hours.
- Since he is 6 minutes late, we can express this as:
\[ \frac{D}{5} = T + \frac{6}{60} \quad \text{(convert 6 minutes to hours)} \]
\[ \frac{D}{5} = T + 0.1 \]

2. When he walks at 6 km/h, he is 2 minutes early.
- The time taken at 6 km/h is \( \frac{D}{6} \) hours.
- Since he is 2 minutes early, we can express this as:
\[ \frac{D}{6} = T - \frac{2}{60} \quad \text{(convert 2 minutes to hours)} \]
\[ \frac{D}{6} = T - \frac{1}{30} \]

### Step 3: Set Up the Equations
Now we have two equations:
1. \( \frac{D}{5} = T + 0.1 \) (Equation 1)
2. \( \frac{D}{6} = T - \frac{1}{30} \) (Equation 2)

### Step 4: Solve for \( T \)
From Equation 1, we can express \( T \):
\[ T = \frac{D}{5} - 0.1 \]

Substituting \( T \) from Equation 1 into Equation 2:
\[ \frac{D}{6} = \left(\frac{D}{5} - 0.1\right) - \frac{1}{30} \]

### Step 5: Simplify the Equation
To simplify, we need a common denominator:
- The common denominator of 5, 6, and 30 is 30.

Rewriting the equation:
\[ \frac{5D}{30} = \frac{6D}{30} - \frac{3}{30} - \frac{1}{30} \]
\[ \frac{5D}{30} = \frac{6D - 4}{30} \]

### Step 6: Clear the Denominator
Multiplying through by 30:
\[ 5D = 6D - 4 \]

### Step 7: Solve for \( D \)
Rearranging gives:
\[ 6D - 5D = 4 \]
\[ D = 4 \text{ km} \]

### Final Answer
The distance from the man's house to his office is **4 km**.

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