Question 646572
A search plane has a cruising speed of 250 mph and carries enough fuel
 for at most 5 hours of flying. 
If there is a wind that averages 30 mph and the direction of the search
 is with the wind one way and against it the other, how far can the search
 team travel before it has to turn back?
:
This is a "point of no return" problem
:
Let d = the one-way distance to the point of no return
:
250 - 30 = 220 mph ground speed against the wind
250 + 30 = 280 mph ground speed with the wind
:
Write a time equation; Time = dist/speed
:
outbound time + inbound time = 5 hrs
{{{d/220}}} + {{{d/280}}} = 5
find the least common multiple, prime factor
22: 2, 11
28: 2, 2, 7
2*2*7*11*10 = 3080
Multiply equation by 3080
3080*{{{d/220}}} + 3080*{{{d/280}}} = 3080*5
Cancel the denominators
14d + 11d = 15400
d = 15400/25
d = 616 miles to turn back point
:
:
Check this by finding the actual time each way
616/220 = 2.8 hrs outbound
616/280 = 2.2 hrs to return
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total time: 5 hrs