SOLUTION: a pilot flew a jet from point a to point b, a distance of 2500 mi. on the return trip, the average speed was 20% faster than the outbound speed. the round-trip took 9h and 10min. w

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Question 1124065: a pilot flew a jet from point a to point b, a distance of 2500 mi. on the return trip, the average speed was 20% faster than the outbound speed. the round-trip took 9h and 10min. what was the speed from point a to point b?
Found 3 solutions by ankor@dixie-net.com, josmiceli, MathTherapy:
Answer by ankor@dixie-net.com(22740)   (Show Source): You can put this solution on YOUR website!
a pilot flew a jet from point a to point b, a distance of 2500 mi.
on the return trip, the average speed was 20% faster than the outbound speed.
the round-trip took 9h and 10min.
what was the speed from point a to point b?
:
let s = the outbound speed of the plane (a to b)
then
1.2s = the return speed
:
Change 9 hr, 10 min to hrs. 9+ = 9.167 hrs
:
Write a time equation; time = dist/speed
time out + time back = 9 hrs 10 min
+ = 9.167
multiply equation by 1.2s, cancel the denominators
1.2(2500) + 2500 = 1.2s(9.167)
3000 + 2500 = 11s
5500 = 11s
s = 5500/11
s = 500 mph from a to b
;
:
Confirm this,find the actual time each way, return 1.2(500) = 600 mph)
2500/500 = 5.000 hrs
2500/600 = 4.167 hrs
----------------------
total time: 9.167 hr which is 9 hrs 10 min
Answer by josmiceli(19441)   (Show Source): You can put this solution on YOUR website!
Let = the jet’s speed going from a to b
= the jet’s speed going from b to a
Let = jet’s time going from a to b
= jet’s time from b to a
——————————-
Equation for a to b:
(1)
Equation for b to a:
(2)
——————————-
(1)
Plug this result into (2)
(2)
(2)
(2)
(2)
The speed from a to b was 500 mi/hr
———————-
Check:
(1)
(1) hrs
And
(2)
(2)
(2)
(2)
(2)
OK

Answer by MathTherapy(10552)   (Show Source): You can put this solution on YOUR website!

a pilot flew a jet from point a to point b, a distance of 2500 mi. on the return trip, the average speed was 20% faster than the outbound speed. the round-trip took 9h and 10min. what was the speed from point a to point b?
Let outbound speed be S

Then return speed  = 1.2S

Then we get the following TIME equation: _______

12(2,500) + 10(2,500) = 110S -------- Multiplying by LCD, 12S

22(2,500) = 110S

S, or outbound speed =  

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