Walking at a speed of 10 km/hr ,Madhu reaches her university 15 minutes late . Next time ,she increases her speed by 2 km/hr ,but is still late by 5 minutes .The distance of the university from her house is :

Options

A. 15 km
B. 30 km
C. 20 km
D. 10 km

Answer

To solve the problem step by step, let's denote the distance from Madhu's house to her university as \( x \) kilometers.

### Step 1: Establish the time taken at each speed
1. **Speed 1**: When Madhu walks at 10 km/hr, she is 15 minutes late.
- Time taken = \( \frac{x}{10} \) hours.
- Let the scheduled time to reach the university be \( T \) hours.
- Since she is 15 minutes late, we can write:
\[ \frac{x}{10} = T + \frac{15}{60} \]
\[ \frac{x}{10} = T + \frac{1}{4} \]

2. **Speed 2**: When Madhu increases her speed to 12 km/hr, she is still late by 5 minutes.
- Time taken = \( \frac{x}{12} \) hours.
- Since she is 5 minutes late, we can write:
\[ \frac{x}{12} = T + \frac{5}{60} \]
\[ \frac{x}{12} = T + \frac{1}{12} \]

### Step 2: Set up the equations
Now we have two equations:
1. \( \frac{x}{10} = T + \frac{1}{4} \) (Equation 1)
2. \( \frac{x}{12} = T + \frac{1}{12} \) (Equation 2)

### Step 3: Eliminate \( T \)
From Equation 1, we can express \( T \):
\[ T = \frac{x}{10} - \frac{1}{4} \]

Substituting \( T \) into Equation 2:
\[ \frac{x}{12} = \left(\frac{x}{10} - \frac{1}{4}\right) + \frac{1}{12} \]

### Step 4: Solve for \( x \)
1. Rearranging the equation:
\[ \frac{x}{12} = \frac{x}{10} - \frac{1}{4} + \frac{1}{12} \]

2. To eliminate the fractions, find a common denominator, which is 60:
\[ \frac{5x}{60} = \frac{6x}{60} - \frac{15}{60} + \frac{5}{60} \]
\[ \frac{5x}{60} = \frac{6x - 10}{60} \]

3. Cross-multiply:
\[ 5x = 6x - 10 \]
\[ 10 = 6x - 5x \]
\[ x = 10 \]

### Final Answer
The distance of the university from her house is \( \mathbf{10 \text{ km}} \).

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