A train has been stopped for 6 minutes. Next stations is 36 kms away and to reach on scheduled time its speed is increased by 4 km/h. Find its initial speed.

Answer

To solve the problem step by step, we will first define the variables and then set up the equations based on the information given.

### Step 1: Define the Variables
Let the initial speed of the train be \( x \) km/h.

### Step 2: Calculate the Time Taken at Initial Speed
The distance to the next station is 36 km. The time taken to cover this distance at the initial speed \( x \) is given by the formula:
\[ \text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{36}{x} \text{ hours} \]

### Step 3: Calculate the Time Taken at Increased Speed
When the speed is increased by 4 km/h, the new speed becomes \( x + 4 \) km/h. The time taken to cover the same distance at this new speed is:
\[ \text{Time} = \frac{36}{x + 4} \text{ hours} \]

### Step 4: Account for the Stoppage Time
Since the train has stopped for 6 minutes, which is equivalent to \( \frac{6}{60} = \frac{1}{10} \) hours, the time taken at the increased speed plus the stoppage time must equal the time taken at the initial speed:
\[ \frac{36}{x + 4} + \frac{1}{10} = \frac{36}{x} \]

### Step 5: Rearranging the Equation
To eliminate the fractions, we can multiply through by \( 10x(x + 4) \):
\[ 10x(36) + x(x + 4) = 10(36)(x + 4) \]
This simplifies to:
\[ 360x + x^2 + 4x = 360x + 1440 \]

### Step 6: Simplifying the Equation
Subtract \( 360x \) from both sides:
\[ x^2 + 4x = 1440 \]
Rearranging gives us:
\[ x^2 + 4x - 1440 = 0 \]

### Step 7: Factoring the Quadratic Equation
Now we will factor the quadratic equation:
\[ (x + 40)(x - 36) = 0 \]
Setting each factor to zero gives us:
\[ x + 40 = 0 \quad \text{or} \quad x - 36 = 0 \]
Thus, \( x = -40 \) (not valid since speed cannot be negative) or \( x = 36 \).

### Step 8: Conclusion
The initial speed of the train is:
\[ \boxed{36 \text{ km/h}} \]