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

Hint:

Set the scheduled time as distance/original speed, and the actual time (0.5 hour less) as distance/(original speed + 250), then solve the resulting quadratic.

Answer 1

The Correct Option is B

Step 1: Set up the relation between speed and time.
Let the plane's original (scheduled) speed be \(v\) km/h, so the scheduled flying time for the 1500 km trip is \(t = \dfrac{1500}{v}\) hours.
The plane actually leaves 30 minutes late, that is \(\dfrac{1}{2}\) hour late, but still reaches on time because it flies at the higher speed \(v+250\) km/h. So its actual flying time is only \(t - \dfrac{1}{2}\) hours, and this must equal the distance divided by the new speed.

Step 2: Form and solve the equation.
\[ \dfrac{1500}{v+250} = \dfrac{1500}{v} - \dfrac{1}{2} \]
Multiplying out and simplifying gives \(v^2 + 250v - 750000 = 0\).
Factoring, \((v-750)(v+1000) = 0\), so \(v = 750\) or \(v = -1000\).
Speed cannot be negative, so \(v = 750\) km/h.

Final Answer:
A quick check confirms this: at 750 km/h the scheduled time is 2 hours, and at the increased speed of 1000 km/h the actual time is 1.5 hours, a saving of exactly 30 minutes, which matches the given condition. Options like 1000 km/h or 800 km/h fail this check because they don't produce exactly a 30 minute time saving when the speed is increased by 250 km/h. \[ \boxed{v = 750 \text{ km/h}} \]