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) About Me  (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+10%2F60 = 9.167 hrs
:
Write a time equation; time = dist/speed
time out + time back = 9 hrs 10 min
2500%2Fs + 2500%2F%281.2s%29 = 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) About Me  (Show Source):
You can put this solution on YOUR website!
Let +s+ = the jet’s speed going from a to b
+1.2s+ = the jet’s speed going from b to a
Let +t+ = jet’s time going from a to b
+9+%2B+1%2F6+-+t+ = jet’s time from b to a
——————————-
Equation for a to b:
(1) +2500+=+s%2At+
Equation for b to a:
(2) +2500+=+1.2s%2A%28+9+%2B+1%2F6+-+t+%29+
——————————-
(1) +t+=+2500%2Fs+
Plug this result into (2)
(2) +2500+=+1.2s%2A%28+9+%2B+1%2F6+-+2500%2Fs+%29+
(2) +2500+=+10.8s+%2B+.2s+-+3000+
(2) +11s+=+5500+
(2) +s+=+500+
The speed from a to b was 500 mi/hr
———————-
Check:
(1) +t+=+2500%2F500+
(1) +t+=+5+ hrs
And
(2) +2500+=+1.2s%2A%28+9+%2B+1%2F6+-+5+%29+
(2) +2500+=+1.2%2As%2A%28+4+%2B+1%2F6+%29+
(2) +2500+=+s%2A%28+4.8+%2B+.2+%29+
(2) +s+=+2500%2F5+
(2) +s+=+500+
OK

Answer by MathTherapy(10552) About Me  (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 =