<PRE><B>Question 1124065</B>
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+{{{10/60}}} = 9.167 hrs
:
Write a time equation; time = dist/speed
time out + time back = 9 hrs 10 min
{{{2500/s}}} + {{{2500/(1.2s)}}} = 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
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total time: 9.167 hr which is 9 hrs 10 min</PRE>