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An air rescue plane averages 300 miles per hour in still air. It carries enough fuel for 5 hours of flying time. If, upon takeoff it encounters a head wind of 30 mi/hr and the wind remains constant, how far can the plane fly and then return safely?

Let $t =$ flying time of the leg with the wind.

Let $(5 - t) = $ flying time of the leg against the wind

Let $d = $ distance with $d = rate \times time$:

against the wind $d = (300 - 30)(5 - t)$

with the wind $d = (300 + 30) t$

Thus: $$270(5 - t) = 330t$$ which yields the correct answer for $t$. However if you switch the time around and

Let $t =$ flying time of the leg against the wind.

Let $(5 - t) =$ flying time of the leg with the wind.

so that:$$270t = 330(5 - t)$$ You don't get the correct answer. Why? It doesn't seem that it should matter. Thanks

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    $\begingroup$ Both methods give a distance of $\frac {1485}2$, there is no significant difference between them. In each case, you fly out for a total time of $\frac {11}4$. $\endgroup$
    – lulu
    Oct 1, 2022 at 23:05
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    $\begingroup$ Note: of course $t$ changes... but that's to be expected. In both cases, the flying time out is $\frac {11}4$. What variable you assign to that is up to you. $\endgroup$
    – lulu
    Oct 1, 2022 at 23:06
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    $\begingroup$ @JohnDouma: The question asks "how far can the plane fly and then return safely?" This does not remain the same with wind, so idk what you mean by saying it is a "trick question" $\endgroup$ Oct 2, 2022 at 7:30
  • $\begingroup$ @lulu yes, that's what I was missing. The definition of $t$ changes so the results are different. Time with the wind is $2.25$ hours while against the wind it's $2.75$ hours. For some reason I was thinking the results had to be the same. $\endgroup$
    – Rolomoto
    Oct 3, 2022 at 21:13

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You should frame the equations as a fraction of total time to get a better insight as to what is happening.

Let $f$ = fraction of total flying time of the leg with the wind.

then $(1-f)$ = fraction of total time flying time of the leg against the wind

Then $330(f) = 270(1-f) \Rightarrow f = \frac{9}{20}$

Total distance $5\times \left(\frac{9}{20}\cdot 330 +\frac{11}{20}\cdot 270\right) = 1485$ miles

Distance out = distance in $= 742.5$ miles

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