A boy walking at a speed of 20 km/h reaches his school 30 min late. Next time he in- creases his speed by 4 km/h but still he is late by 10 min. Find the distance of the school from his home.

Answer

To solve the problem step by step, we will follow the information given in the question and use the formulas for speed, time, and distance.

### Step 1: Understand the Problem
The boy walks to school at two different speeds and is late both times. We need to find the distance from his home to the school.

### Step 2: Define Variables
Let the distance from the boy's home to the school be \( x \) kilometers.

### Step 3: Set Up the Equations
1. **First Journey:**
- Speed = 20 km/h
- Time taken = \( \frac{x}{20} \) hours
- He is 30 minutes late, which is \( \frac{30}{60} = \frac{1}{2} \) hours.
- Let the time he should take to reach school on time be \( t \) hours.
- Therefore, we have:
\[ \frac{x}{20} = t + \frac{1}{2} \]

2. **Second Journey:**
- Speed = 24 km/h (20 km/h + 4 km/h)
- Time taken = \( \frac{x}{24} \) hours
- He is 10 minutes late, which is \( \frac{10}{60} = \frac{1}{6} \) hours.
- Therefore, we have:
\[ \frac{x}{24} = t + \frac{1}{6} \]

### Step 4: Equate the Two Expressions for \( t \)
From the first journey:
\[ t = \frac{x}{20} - \frac{1}{2} \]
From the second journey:
\[ t = \frac{x}{24} - \frac{1}{6} \]

### Step 5: Set the Two Expressions for \( t \) Equal
\[ \frac{x}{20} - \frac{1}{2} = \frac{x}{24} - \frac{1}{6} \]

### Step 6: Solve for \( x \)
1. To eliminate the fractions, find a common denominator. The least common multiple of 20 and 24 is 120.
2. Multiply the entire equation by 120:
\[ 120 \left( \frac{x}{20} \right) - 120 \left( \frac{1}{2} \right) = 120 \left( \frac{x}{24} \right) - 120 \left( \frac{1}{6} \right) \]
This simplifies to:
\[ 6x - 60 = 5x - 20 \]

3. Rearranging gives:
\[ 6x - 5x = 60 - 20 \]
\[ x = 40 \]

### Step 7: Conclusion
The distance from the boy's home to the school is \( 40 \) kilometers.