A train leaves the station 1 hour before the scheduled time. The driver decreases its speed by 50 km/hr. At the next station 300 km away, the train reached on time. Find the original speed of the train.

Options

A. 100 km/hr
B. 150 km/hr
C. 125 km/hr
D. 200 km/hr

Answer

To solve the problem step by step, we will follow a structured approach:

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

### Step 2: Calculate the normal time to cover 300 km
The normal time taken to cover 300 km at speed \( x \) km/hr is given by the formula:
\[ \text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{300}{x} \text{ hours} \]

### Step 3: Determine the time taken with the new speed
Since the train leaves the station 1 hour early, it has 1 hour more than the normal time to reach the next station. Therefore, the time taken with the new speed is:
\[ \text{New Time} = 1 + \frac{300}{x} \text{ hours} \]

### Step 4: Calculate the new speed
The new speed of the train after decreasing the speed by 50 km/hr is:
\[ \text{New Speed} = x - 50 \text{ km/hr} \]

### Step 5: Set up the equation
Since the distance remains the same (300 km), we can equate the distance covered with the new speed and the new time:
\[ \text{Distance} = \text{New Speed} \times \text{New Time} \]
This gives us the equation:
\[ 300 = (x - 50) \left(1 + \frac{300}{x}\right) \]

### Step 6: Expand and simplify the equation
Expanding the right-hand side:
\[ 300 = (x - 50) + \frac{300(x - 50)}{x} \]
This simplifies to:
\[ 300 = x - 50 + 300 - \frac{15000}{x} \]
Combining like terms:
\[ 300 = x + 250 - \frac{15000}{x} \]
Rearranging gives:
\[ x - \frac{15000}{x} = 50 \]

### Step 7: Multiply through by \( x \) to eliminate the fraction
Multiplying both sides by \( x \):
\[ x^2 - 15000 = 50x \]
Rearranging gives us a quadratic equation:
\[ x^2 - 50x - 15000 = 0 \]

### Step 8: Factor the quadratic equation
To factor the quadratic equation, we look for two numbers that multiply to \(-15000\) and add to \(-50\). The factors are:
\[ (x - 150)(x + 100) = 0 \]

### Step 9: Solve for \( x \)
Setting each factor to zero gives:
\[ x - 150 = 0 \quad \text{or} \quad x + 100 = 0 \]
Thus, we have:
\[ x = 150 \quad \text{or} \quad x = -100 \]

### Step 10: Determine the valid solution
Since speed cannot be negative, we reject \( x = -100 \) and accept:
\[ x = 150 \text{ km/hr} \]

### Conclusion
The original speed of the train is \( 150 \) km/hr.

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