A train covers a distance of `108` km at a certain speed. If its speed is increased by `18` km/hr , it would take `18` minutes less to cover the same distance. What is the original speed (in km/hr) of the train ?

Options

A. 72
B. 80
C. 63
D. 54

Answer

To find the original speed of the train, we can follow these steps:

### Step 1: Define the Variables
Let the original speed of the train be \( V \) km/hr.

### Step 2: Calculate the Time Taken at Original Speed
The distance covered by the train is \( 108 \) km. The time taken to cover this distance at the original speed \( V \) is given by:
\[ T = \frac{108}{V} \]

### Step 3: Calculate the Time Taken at Increased Speed
If the speed is increased by \( 18 \) km/hr, the new speed becomes \( V + 18 \) km/hr. The time taken to cover the same distance at this new speed is:
\[ T' = \frac{108}{V + 18} \]

### Step 4: Set Up the Equation for Time Difference
According to the problem, the time taken at the increased speed is \( 18 \) minutes less than the time taken at the original speed. We need to convert \( 18 \) minutes into hours, which is:
\[ \frac{18}{60} = \frac{3}{10} \text{ hours} \]
Thus, we can write the equation:
\[ T - T' = \frac{3}{10} \]
Substituting the expressions for \( T \) and \( T' \):
\[ \frac{108}{V} - \frac{108}{V + 18} = \frac{3}{10} \]

### Step 5: Solve the Equation
To solve the equation, we can find a common denominator:
\[ \frac{108(V + 18) - 108V}{V(V + 18)} = \frac{3}{10} \]
This simplifies to:
\[ \frac{108 \cdot 18}{V(V + 18)} = \frac{3}{10} \]
Cross-multiplying gives:
\[ 108 \cdot 18 \cdot 10 = 3V(V + 18) \]
\[ 1080 \cdot 18 = 3V^2 + 54V \]
Calculating \( 1080 \cdot 18 \):
\[ 19440 = 3V^2 + 54V \]

### Step 6: Rearranging the Equation
Rearranging gives:
\[ 3V^2 + 54V - 19440 = 0 \]
Dividing the entire equation by \( 3 \):
\[ V^2 + 18V - 6480 = 0 \]

### Step 7: Factor the Quadratic Equation
To factor the quadratic equation, we need two numbers that multiply to \( -6480 \) and add to \( 18 \). The numbers are \( 90 \) and \( -72 \):
\[ (V + 90)(V - 72) = 0 \]

### Step 8: Find the Possible Values of \( V \)
Setting each factor to zero gives:
\[ V + 90 = 0 \quad \Rightarrow \quad V = -90 \quad (\text{not valid, speed cannot be negative}) \]
\[ V - 72 = 0 \quad \Rightarrow \quad V = 72 \]

### Conclusion
The original speed of the train is:
\[ \boxed{72} \text{ km/hr} \]