Question

A pilot flew a jet from City A to City B, a distance of 2300 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?

Answer 1

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

$20 \text{ minutes } = \frac{20}{60} \text{ hour } =\frac{1}{3} \text{ hour }$

Thus, we get,

Total time taken for both trips = $8 + \frac{1}{3} = \frac{24+1}{3} = \frac{25}{3} \text{ hour }$

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

$x + 20\% x = x + 0.2x = 1.2x$

Speed for return trip = 1.2x

Time taken is given by formula:

$\text{total time } = \frac{distance}{\text{speed of outbound trip}} + \frac{distance}{\text{speed of return trip}}$

$\frac{25}{3} = \frac{2300}{x} + \frac{2300}{1.2x}\\\\\frac{25x}{3} = \frac{2300 \times 1.2 + 2300}{1.2}\\\\10x = 5060\\\\x = 506$

Thus speed from City A to City B is 506 miles per hour