A man riding a bicycle from his house at 10km/h and reaches his office late by 6min. He increases his speed by 2 km/h and reaches 6 min before. How far is the office from his house?

Options

A. 6 km
B. 7 km
C. 12 km
D. 16 km

Answer

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

### Step 1: Understand the situation
The man rides his bicycle at a speed of 10 km/h and arrives 6 minutes late. When he increases his speed to 12 km/h, he arrives 6 minutes early. We need to find the distance from his house to his office.

### Step 2: Define variables
Let:
- \( D \) = Distance from the house to the office (in km)
- \( T_1 \) = Time taken at 10 km/h (in hours)
- \( T_2 \) = Time taken at 12 km/h (in hours)

### Step 3: Convert minutes to hours
Since the man is late by 6 minutes and early by 6 minutes, the total time difference when he changes his speed is:
- Total time difference = 6 minutes late + 6 minutes early = 12 minutes
- Convert 12 minutes to hours: \( \frac{12}{60} = 0.2 \) hours

### Step 4: Set up the equations
Using the relationship between speed, distance, and time:
1. At 10 km/h:
\[ D = 10 \times T_1 \]
Since he is 6 minutes late, we can express the actual time to reach on time as:
\[ T_1 = T + \frac{6}{60} = T + 0.1 \]
where \( T \) is the actual time he should take.

2. At 12 km/h:
\[ D = 12 \times T_2 \]
Since he is 6 minutes early, we can express this time as:
\[ T_2 = T - \frac{6}{60} = T - 0.1 \]

### Step 5: Equate the distances
Since both expressions equal \( D \):
\[ 10(T + 0.1) = 12(T - 0.1) \]

### Step 6: Expand and simplify
Expanding both sides:
\[ 10T + 1 = 12T - 1.2 \]
Rearranging gives:
\[ 1 + 1.2 = 12T - 10T \]
\[ 2.2 = 2T \]
\[ T = 1.1 \text{ hours} \]

### Step 7: Calculate \( D \)
Now, substitute \( T \) back into either expression for \( D \):
Using \( D = 10(T + 0.1) \):
\[ D = 10(1.1 + 0.1) = 10 \times 1.2 = 12 \text{ km} \]

### Final Answer:
The distance from the man's house to his office is **12 km**.

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