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.

Step-by-step explanation:

Since we have given that

Distance = 2500 meter

Outbound speed be x

Average speed be $\dfrac{100+20}{100}x=\dfrac{120}{100}x=1.2x$

Time = 9 hours 10 minutes = $9\dfrac{10}{60}=9\dfrac{1}{6}=\dfrac{55}{6}$

According to question, it becomes,

$\dfrac{2.5}{x}+\dfrac{2.5}{1.2x}=\dfrac{55}{6}\\\\\dfrac{1.2x+x}{1.2x^2}=\dfrac{55}{6\times 2.5}\\\\\dfrac{2.2x}{1.2x^2}=\dfrac{11}{3}\\\\\dfrac{2.2}{1.2x}=\dfrac{11}{3}\\\\2.2\times 3=1.2x\\\\6.6=1.2x\\\\\dfrac{6.6}{1.2}=x\\\\x=5.5\ km/hr$

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