If Seeta walks from her home to school at a speed of 4 km/h, she is late by 9 min. If she walks at a speed of 6 km/h, she reaches the school 6 min early. What is the distance between her scholl and her home?

Options

A. 4km
B. 7km
C. 6km
D. 3km

Answer

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

### Step 1: Define Variables
Let \( T \) be the actual time (in minutes) it takes Seeta to walk from her home to school.

### Step 2: Set Up Equations for Distances
1. When Seeta walks at a speed of 4 km/h, she is late by 9 minutes. Therefore, the time taken is \( T + 9 \) minutes.
- The distance \( D \) can be expressed as:
\[ D = \text{Speed} \times \text{Time} = 4 \times \left(\frac{T + 9}{60}\right) \]
Simplifying this gives:
\[ D = \frac{4(T + 9)}{60} = \frac{T + 9}{15} \]

2. When Seeta walks at a speed of 6 km/h, she reaches 6 minutes early. Therefore, the time taken is \( T - 6 \) minutes.
- The distance \( D \) can also be expressed as:
\[ D = 6 \times \left(\frac{T - 6}{60}\right) \]
Simplifying this gives:
\[ D = \frac{6(T - 6)}{60} = \frac{T - 6}{10} \]

### Step 3: Set the Two Distance Equations Equal
Since the distance \( D \) is the same in both cases, we can set the two equations equal to each other:
\[ \frac{T + 9}{15} = \frac{T - 6}{10} \]

### Step 4: Cross Multiply to Solve for \( T \)
Cross multiplying gives:
\[ 10(T + 9) = 15(T - 6) \]
Expanding both sides:
\[ 10T + 90 = 15T - 90 \]

### Step 5: Rearranging the Equation
Rearranging the equation to isolate \( T \):
\[ 90 + 90 = 15T - 10T \]
\[ 180 = 5T \]
\[ T = \frac{180}{5} = 36 \text{ minutes} \]

### Step 6: Calculate the Distance
Now, we can find the distance \( D \) using either of the distance equations. Using the first equation:
\[ D = \frac{T + 9}{15} = \frac{36 + 9}{15} = \frac{45}{15} = 3 \text{ km} \]

### Final Answer
The distance between Seeta's home and school is **3 km**.

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