Answer :
The speed from City A to City B is 506 miles per hour
Solution:
A pilot flew a jet from City A to City B, a distance of 2300 miles
Distance for trip = 2300 miles
Distance for return trip = 2300 miles
The round-trip took 8 h 20 min
Total time taken for both trips = 8 hour 20 minutes
We know that,
1 hour = 60 minutes
[tex]20 \text{ minutes } = \frac{20}{60} \text{ hour } =\frac{1}{3} \text{ hour }[/tex]
Thus, we get,
Total time taken for both trips = [tex]8 + \frac{1}{3} = \frac{24+1}{3} = \frac{25}{3} \text{ hour }[/tex]
On the return trip, the average speed was 20% faster than the outbound speed
Let "x" be the speed for out bound trip
Then, x + 20 % of x is the average speed of return trip
[tex]x + 20\% x = x + 0.2x = 1.2x[/tex]
Speed for return trip = 1.2x
Time taken is given by formula:
[tex]\text{total time } = \frac{distance}{\text{speed of outbound trip}} + \frac{distance}{\text{speed of return trip}}[/tex]
[tex]\frac{25}{3} = \frac{2300}{x} + \frac{2300}{1.2x}\\\\\frac{25x}{3} = \frac{2300 \times 1.2 + 2300}{1.2}\\\\10x = 5060\\\\x = 506[/tex]
Thus speed from City A to City B is 506 miles per hour