Question

A plane left 30 min later than its scheduled time to reach its destination 1500 km away. In order to reach in time it increases its speed by 250 km/h. What is its original speed ?

• A. 1000km/h
• B. 750km/h
• C. 600km/h
• D. 800km/h

Answer 1

The Correct Option is B

Let the original speed of the plane be \( x \) km/h. The plane covers a distance of 1500 km. If the plane is late by 30 minutes (0.5 hours) and increases its speed by 250 km/h to reach on time, the increased speed is \( x + 250 \) km/h.
Time taken at original speed: \(\frac{1500}{x}\) hours.
Time taken at increased speed: \(\frac{1500}{x+250}\) hours.
Since the plane is 30 minutes late but still reaches on time, these times differ by 0.5 hours:
\(\frac{1500}{x} - \frac{1500}{x+250} = 0.5\)
Solving the equation for \( x \):
1. Multiply through by \( x(x+250) \) to clear the denominators:
\(1500(x+250) - 1500x = 0.5x(x+250)\)
2. Simplify and distribute:
\(1500x + 375000 - 1500x = 0.5x^2 + 125x\)
Remove like terms:
\(375000 = 0.5x^2 + 125x\)
3. Multiply through by 2 to eliminate the fraction:
\(750000 = x^2 + 250x\)
4. Rearrange terms:
\(x^2 + 250x - 750000 = 0\)
5. Factor the quadratic equation:
\((x-750)(x+1000) = 0\)
Solve for \( x \):
\(x = 750\) or \(x = -1000\)
Since speed cannot be negative, the original speed of the plane is \(750\) km/h.