A train leaves the station 1 hour before the schedules time. The driver decreases its speed by 2km/h. At the next station 60 km away, the train reached on time. Find the original speed of the train ?

Answer

To solve the problem step by step, let's define the variables and use the information provided in the question.

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

### Step 2: Determine the New Speed
Since the driver decreases the speed by 2 km/h, the new speed of the train becomes \( (x - 2) \) km/h.

### Step 3: Calculate Time Taken to Travel 60 km
The distance to the next station is 60 km. The time taken to travel this distance at the new speed is given by the formula:
\[ \text{Time} = \frac{\text{Distance}}{\text{Speed}} \]
Thus, the time taken at the new speed is:
\[ \text{Time} = \frac{60}{x - 2} \text{ hours} \]

### Step 4: Determine the Scheduled Time
Since the train leaves 1 hour early, the scheduled time to reach the next station would be:
\[ \text{Scheduled Time} = \frac{60}{x} \text{ hours} \]
However, since the train leaves 1 hour early, the time taken to reach the station at the new speed must equal the scheduled time minus 1 hour:
\[ \frac{60}{x - 2} = \frac{60}{x} - 1 \]

### Step 5: Solve the Equation
Now, we can solve the equation:
\[ \frac{60}{x - 2} = \frac{60}{x} - 1 \]

Multiply through by \( x(x - 2) \) to eliminate the denominators:
\[ 60x = 60(x - 2) - x(x - 2) \]
Expanding both sides:
\[ 60x = 60x - 120 - x^2 + 2x \]
Rearranging gives:
\[ 0 = -x^2 + 2x - 120 \]
This simplifies to:
\[ x^2 - 2x + 120 = 0 \]

### Step 6: Use the Quadratic Formula
Now we can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = -2, c = -120 \):
\[ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-120)}}{2 \cdot 1} \]
Calculating the discriminant:
\[ x = \frac{2 \pm \sqrt{4 + 480}}{2} \]
\[ x = \frac{2 \pm \sqrt{484}}{2} \]
\[ x = \frac{2 \pm 22}{2} \]
Calculating the two possible values:
1. \( x = \frac{24}{2} = 12 \)
2. \( x = \frac{-20}{2} = -10 \) (not a valid speed)

Thus, the original speed of the train is \( 12 \) km/h.

### Final Answer
The original speed of the train is **12 km/h**.

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