A train left 1 hour later than the scheduled time but in order to reach its destination 200 km away in time, it had to increase its usual speed by 10 km/hr. What is the usual speed (in km/hr) of the train?

Options

A. 38
B. 42
C. 45
D. 40

Answer

To find the usual speed of the train, we can set up the problem using the information given.

### Step-by-Step Solution:

1. **Define Variables:**
Let the usual speed of the train be \( s \) km/hr.

2. **Calculate Time Taken at Usual Speed:**
The time taken to cover 200 km at the usual speed \( s \) is given by:
\[ \text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{200}{s} \text{ hours} \]

3. **Calculate Time Taken at Increased Speed:**
If the train increases its speed by 10 km/hr, the new speed becomes \( s + 10 \) km/hr. The time taken to cover the same distance at this new speed is:
\[ \text{Time} = \frac{200}{s + 10} \text{ hours} \]

4. **Set Up the Equation:**
Since the train left 1 hour late but still arrived on time, the time taken at the increased speed is 1 hour less than the time taken at the usual speed. Therefore, we can set up the equation:
\[ \frac{200}{s} - \frac{200}{s + 10} = 1 \]

5. **Clear the Fractions:**
To eliminate the fractions, we can multiply through by \( s(s + 10) \):
\[ 200(s + 10) - 200s = s(s + 10) \]
Simplifying this gives:
\[ 2000 = s^2 + 10s \]

6. **Rearrange the Equation:**
Rearranging the equation, we have:
\[ s^2 + 10s - 2000 = 0 \]

7. **Factor the Quadratic Equation:**
We need to factor the quadratic equation. We look for two numbers that multiply to \(-2000\) and add to \(10\). The numbers are \(50\) and \(-40\):
\[ (s + 50)(s - 40) = 0 \]

8. **Solve for \( s \):**
Setting each factor to zero gives us:
\[ s + 50 = 0 \quad \text{or} \quad s - 40 = 0 \]
Thus, \( s = -50 \) (not valid since speed cannot be negative) or \( s = 40 \).

9. **Conclusion:**
The usual speed of the train is \( 40 \) km/hr.