A train travels a distance of 300km at constant speed. If the speed of the train is increased by 5 km/kr, the journey would have taken 2 hours less. Find the original speed of the train.

Answer

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

### Step 1: Define the variables
Let the original speed of the train be \( x \) km/h. The distance traveled by the train is 300 km.

### Step 2: Write the equation for time taken at original speed
The time taken to travel 300 km at speed \( x \) is given by:
\[ t = \frac{300}{x} \]

### Step 3: Write the equation for time taken at increased speed
If the speed is increased by 5 km/h, the new speed becomes \( x + 5 \) km/h. The time taken at this new speed is:
\[ t' = \frac{300}{x + 5} \]

### Step 4: Set up the equation based on the time difference
According to the problem, the journey would have taken 2 hours less at the increased speed. Therefore, we can write the equation:
\[ t - t' = 2 \]
Substituting the expressions for \( t \) and \( t' \):
\[ \frac{300}{x} - \frac{300}{x + 5} = 2 \]

### Step 5: Clear the fractions
To eliminate the fractions, we can multiply through by \( x(x + 5) \):
\[ 300(x + 5) - 300x = 2x(x + 5) \]
This simplifies to:
\[ 300x + 1500 - 300x = 2x^2 + 10x \]
Thus, we have:
\[ 1500 = 2x^2 + 10x \]

### Step 6: Rearrange the equation
Rearranging gives us a standard quadratic equation:
\[ 2x^2 + 10x - 1500 = 0 \]

### Step 7: Simplify the quadratic equation
We can divide the entire equation by 2 to simplify:
\[ x^2 + 5x - 750 = 0 \]

### Step 8: Factor or use the quadratic formula
To solve for \( x \), we can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = 5 \), and \( c = -750 \).

### Step 9: Calculate the discriminant
First, calculate the discriminant:
\[ b^2 - 4ac = 5^2 - 4 \cdot 1 \cdot (-750) = 25 + 3000 = 3025 \]

### Step 10: Solve for \( x \)
Now substituting back into the quadratic formula:
\[ x = \frac{-5 \pm \sqrt{3025}}{2 \cdot 1} = \frac{-5 \pm 55}{2} \]
Calculating the two possible values:
1. \( x = \frac{50}{2} = 25 \)
2. \( x = \frac{-60}{2} = -30 \) (not valid since speed cannot be negative)

### Conclusion
The original speed of the train is:
\[ \boxed{25} \text{ km/h} \]