Question

A pilot flew a jet from location A to B, a distance of 2500 meter,on the return trip the average speed was 20 percent faster than the outbound speed,the round-trip took 9 hours 10 minutes . What is the jet's speed from A to B.

Answer 1

Answer: The jet's speed from A to B would be 5.5 km/hr.

Explanation:

Since we have given that

Distance = 2500 meter

Outbound speed be x

Average speed be
`(100+20)/(100)x=(120)/(100)x=1.2x`

Time = 9 hours 10 minutes =
`9(10)/(60)=9(1)/(6)=(55)/(6)`

According to question, it becomes,

`(2.5)/(x)+(2.5)/(1.2x)=(55)/(6)\\\\(1.2x+x)/(1.2x^2)=(55)/(6* 2.5)\\\\(2.2x)/(1.2x^2)=(11)/(3)\\\\(2.2)/(1.2x)=(11)/(3)\\\\2.2* 3=1.2x\\\\6.6=1.2x\\\\(6.6)/(1.2)=x\\\\x=5.5\ km/hr`

Hence, the jet's speed from A to B would be 5.5 km/hr.