If a child goes to his school at the speed of 40 km/h. He reaches 2 hours early and if the travels at the speed of 30 km/h then the reaches 1 hour early. Find out his actual speed and distance actual time taken in order to reach at time.

Answer

To solve the problem, we need to find the actual speed of the child and the distance to the school based on the information provided. Let's denote the following variables:

- Let \( D \) be the distance to the school.
- Let \( T \) be the actual time taken to reach the school on time.

### Step 1: Set up the equations based on the given speeds and time differences.

1. When the child travels at 40 km/h and reaches 2 hours early:
\[ \frac{D}{40} = T - 2 \]

2. When the child travels at 30 km/h and reaches 1 hour early:
\[ \frac{D}{30} = T - 1 \]

### Step 2: Rearrange both equations to express \( D \) in terms of \( T \).

From the first equation:
\[ D = 40(T - 2) \]

From the second equation:
\[ D = 30(T - 1) \]

### Step 3: Set the two expressions for \( D \) equal to each other.

Since both expressions represent the same distance \( D \):
\[ 40(T - 2) = 30(T - 1) \]

### Step 4: Expand and simplify the equation.

Expanding both sides:
\[ 40T - 80 = 30T - 30 \]

Now, rearranging the equation:
\[ 40T - 30T = 80 - 30 \]
\[ 10T = 50 \]
\[ T = 5 \]

### Step 5: Substitute \( T \) back to find \( D \).

Now that we have \( T \), we can substitute it back into either equation for \( D \). Using the first equation:
\[ D = 40(T - 2) = 40(5 - 2) = 40 \times 3 = 120 \text{ km} \]

### Step 6: Find the actual speed.

The actual speed is calculated as:
\[ \text{Actual Speed} = \frac{D}{T} = \frac{120 \text{ km}}{5 \text{ hours}} = 24 \text{ km/h} \]

### Final Results:
- Actual Speed: \( 24 \text{ km/h} \)
- Distance: \( 120 \text{ km} \)
- Actual Time Taken: \( 5 \text{ hours} \)