# 1973 AHSME Problems/Problem 34

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

## Problem

A plane flew straight against a wind between two towns in 84 minutes and returned with that wind in 9 minutes less than it would take in still air. The number of minutes (2 answers) for the return trip was

$\textbf{(A)}\ 54 \text{ or } 18 \qquad \textbf{(B)}\ 60 \text{ or } 15 \qquad \textbf{(C)}\ 63 \text{ or } 12 \qquad \textbf{(D)}\ 72 \text{ or } 36 \qquad \textbf{(E)}\ 75 \text{ or } 20$

## Solution

Let $d$ be the distance between the two towns, $p$ be the speed of the plane, and $w$ be the speed of the wind. Since the time it took to fly against the wind towards the other town is $84$ minutes, $$\frac{d}{p-w} = 84$$ $$d = 84p-84w$$ Since flying with the wind on the return trip takes $9$ minutes less compared to the return trip without wind, $$\frac{d}{p+w} + 9 = \frac{d}{p}$$ Substitute $d$ and substitute $p$ and $w$ $$\frac{84p-84w}{p+w} + 9 = \frac{84p-84w}{p}$$ $$\frac{93p-75w}{p+w} = \frac{84p-84w}{p}$$ $$93p^2 - 75pw = 84p^2 + 84pw - 84pw - 84w^2$$ $$9p^2 - 75pw + 84w^2 = 0$$ $$3p^2 - 25pw + 28w^2 = 0$$ $$(3p-4w)(p-7w)=0$$ That means $p = \tfrac{4}{3}w$ or $p = 7w$. If $p = \tfrac{4}{3}w$, then $\frac{112w-84w}{\tfrac{7}{3}w} = 12$. If $p=7w$, then $\frac{84(7w)-84w}{8w} = 63$. The number of minutes for the return trip can be $\boxed{\textbf{(C)}\ 63 \text{ or } 12}$.