Question

A flight, traveling to a destination 11,200 kms away, was supposed to take off at 6:30 AM. Due to bad weather, the departure of the flight got delayed by three hours.\ The pilot increased the average speed of the airplane by 100 km/hr from the initially planned average speed, to reduce the overall delay to one hour.
Had the pilot increased the average speed by 350 km/hr from the initially planned average speed, when would have the flight reached its destination?

• A. 0.979166666666667
• B. 0.826388888888889
• C. 0.715277777777778
• D. 0.840277777777778
• E. 0.941666666666667

Answer 1

The Correct Option is D

Let the initially planned speed of the flight be v km/hr. The original time to travel 11200 km would have been:

$t = \frac{11200}{v}$

Step 1: Using the information about speed increase by 100 km/hr. With the new speed of v + 100, the time becomes:

$t' = \frac{11200}{v + 100}$

The flight delay is reduced to 1 hour, meaning the time difference is 2 hours (3 − 1):

t − t' = 2

Substitute t and t':

$\frac{11200}{v} - \frac{11200}{v + 100} = 2$

Simplify:

$11200 \left( \frac{1}{v} - \frac{1}{v + 100} \right) = 2$

$\frac{11200}{v(v + 100)} = 2$

$v(v + 100) = 560000$

Solve for v:

v² + 100v − 560000 = 0

Use the quadratic formula:

$v = \frac{-100 \pm \sqrt{100^2 + 4 \times 560000}}{2}$

$v = \frac{-100 \pm \sqrt{2250000}}{2}$

v = −100 ± 750

v = 700 (ignoring the negative root as speed must be positive).

Step 2: Calculate the time with 350 km/hr increase. If the pilot increased the speed by 350 km/hr, the new speed would be:

v + 350 = 1050 km/hr.

The time taken to travel 11,200 km is:

$t'' = \frac{11200}{1050} = 10.66 \text{ hours (10 hours and 40 minutes).}$

Step 3: Determine the arrival time. The flight took off at 9:30 AM (6:30 AM + 3-hour delay). Adding 10 hours and 40 minutes to this:

9:30 AM + 10:40 = 8:10 PM.

Answer: 8:10 PM