A child while going to school reduces his speed to 4/5th of his actual speed and reaches 15 minutes late. Find his actual time.

Answer

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

### Step 1: Define Variables
Let the actual speed of the child be \( S \) and the actual time taken to reach school be \( T \). The distance to school can be expressed as:
\[ D = S \times T \]

### Step 2: Determine Reduced Speed
The child reduces his speed to \( \frac{4}{5} \) of his actual speed. Thus, the reduced speed \( S' \) is:
\[ S' = \frac{4}{5} S \]

### Step 3: Calculate Time with Reduced Speed
When the child travels at the reduced speed, the time taken \( T' \) can be calculated using the same distance:
\[ D = S' \times T' \]
Substituting for \( D \):
\[ S \times T = \left(\frac{4}{5} S\right) \times T' \]
From this, we can simplify and find \( T' \):
\[ T' = \frac{5}{4} T \]

### Step 4: Relate Time Difference
According to the problem, the child reaches school 15 minutes late. Therefore, we can express this relationship as:
\[ T' - T = 15 \text{ minutes} \]
Substituting for \( T' \):
\[ \frac{5}{4} T - T = 15 \]
This simplifies to:
\[ \frac{5}{4} T - \frac{4}{4} T = 15 \]
\[ \frac{1}{4} T = 15 \]

### Step 5: Solve for Actual Time
Now, we can solve for \( T \):
\[ T = 15 \times 4 = 60 \text{ minutes} \]

### Conclusion
Thus, the actual time taken by the child to reach school is:
\[ \boxed{60 \text{ minutes}} \]