Question

MCO
G Select Language
02
What is the expected result?
A plane is flying 2400 kilometers (km) from source A to destination B. The airspeed of the plane is 600
km/hr, and the speed of the tailwind is 40 km/hr. How long after takeoff will it be strictly faster to go to B
than to return to A?

Answer 1

Answer:

Right Answer is: D

From A to B speed of flight = 640 km/hr , while for the reverse it is = 560 km/hr. ⇒x≥1120 km.

Answer 2

The time taken for how long after takeoff will plane be strictly faster to go to B than to return to A is 2 hours.

Given:-

Total Distance traveled by the plane = 2400km

Air speed of plane = 600km/hr

Speed of tail wind = 40km/hr

To Find:-

The time taken for how long after takeoff will plane be strictly faster to go to B than to return to A.

Solution:-

We can easily calculate the time taken for how long after takeoff will plane be strictly faster to go to B than to return to A by using these simple steps.

As

Total Distance traveled by the plane (s) = 2400km

Air speed of plane (v) = 600km/hr

Speed of tail wind (v') = 40km/hr

If the distance traveled by plane from A to B is x, then for return

distance = (2400-x) as 2400km is the total distance

Also, total speed from A to B

$v(AB) = v + v'$

$v(AB) = 600 + 40$

$v(AB) = 640$

Similarly, total speed from B to A

$v(BA) = v - v'$

$v(BA) = 600 - 40$

$v(BA) = 560$

Now, According to the formula,

$speed = \frac{distance}{time}$

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

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

here, time for both the paths will be equal. So

$t1 \geqslant t2$

$\frac{x}{560} \geqslant \frac{2400 - x}{640}$

$x \geqslant \frac{560(2400 - x)}{640}$

$x \geqslant \frac{56(2400 - x)}{64}$

$x \geqslant \frac{ 7(2400 - x) }{8}$

$x \geqslant \frac{16800 - 7x}{8}$

On solving we get,

$x \geqslant 1120km$

So,

Minimum distance traveled from A to B = 1120km

Now,

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

$t = \frac{1120}{560}$

$t = 2hr$

Hence, The time taken for how long after takeoff will plane be strictly faster to go to B than to return to A is 2 hours.

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