A pilot flew a jet from city A to city B, a distance of 2400 mi. On the return trip, the average speed was 20% faster than the outbound speed. The round-trip took 8 h 20 min. What was the speed from city A to city B?

Step-by-step explanation:

Given that,

A pilot flew from point A to point B

Distance travelled (d)=2400miles

On return from B to A

The pilot increases his speed by 20%

The total time for the trip is 8hour 20mins= 8+20/60= 8.33hours

We want to find speed from A to B

Let the speed from A to B be V,

Then the speed from B to A is

20% increased

20/100 ×V +V

0.2V+V=1.2V

Now,

Speed is given as

Speed=distance /time

So, time =distance /speed

The time he spend from A to B is

t1=d/s

Not that the distance is constant

t1=2400/V

The time he spend from B to A is

t2=d/s

Not that the distance is constant

t2=2400/(1.2V)

t2=2000/V

Now, total time is 8.33hours

Therefore, t1+t2=total time for that trip

2400/V + 2000/V = 8.33

(2400+2000)/V= 8.33

4400/V=8.33

Cross multiply

4400=8.33V

Then, divided both side by 8.33

V=4400/8.33

V=528miles/hour

Since we represent V as the speed from Point A to B, the required speed is

V=528miles/hour

according to egenuity the answer is

step-by-step explanation:

d. 344.22