A child while going to school increases his speed to 7/6th of his actual speed. He reaches his school 5 minutes early. Find his initial time.

Answer

To solve the problem step by step, let's denote the initial speed of the child as \( S \) and the initial time taken to reach school as \( T \).

### Step 1: Understand the relationship between speed, distance, and time
The distance to school can be expressed as:
\[ \text{Distance} = \text{Speed} \times \text{Time} \]
So, the distance to school can be represented as:
\[ D = S \times T \]

### Step 2: Determine the new speed and time
When the child increases his speed to \( \frac{7}{6}S \), he reaches school 5 minutes early. The new time taken to reach school can be represented as:
\[ \text{New Time} = T - 5 \text{ minutes} \]

### Step 3: Express the distance with the new speed and time
The distance to school remains the same, so we can write:
\[ D = \left(\frac{7}{6}S\right) \times (T - 5) \]

### Step 4: Set up the equation
Since both expressions for distance \( D \) are equal, we can set them equal to each other:
\[ S \times T = \left(\frac{7}{6}S\right) \times (T - 5) \]

### Step 5: Simplify the equation
We can cancel \( S \) from both sides (assuming \( S \neq 0 \)):
\[ T = \frac{7}{6} \times (T - 5) \]

### Step 6: Solve for \( T \)
Now, multiply both sides by 6 to eliminate the fraction:
\[ 6T = 7(T - 5) \]
Distributing the 7 on the right side gives:
\[ 6T = 7T - 35 \]

Now, rearranging the equation:
\[ 6T - 7T = -35 \]
\[ -T = -35 \]
Thus,
\[ T = 35 \text{ minutes} \]

### Final Answer
The initial time taken by the child to reach school is **35 minutes**.

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