Difference between revisions of "1973 AHSME Problems/Problem 34"

Latest revision as of 13:05, 20 February 2020

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 $\boxed{\textbf{(C)}\ 63 \text{ or } 12}$ .

1973 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 33	Followed by Problem 35
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35
All AHSME Problems and Solutions

@@ Line 26: / Line 26: @@
 ==See Also==
-{{AHSME 35p box|year=1973|num-b=33|num-a=35}}
+{{AHSME 30p box|year=1973|num-b=33|num-a=35}}
 [[Category:Intermediate Algebra Problems]]

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Difference between revisions of "1973 AHSME Problems/Problem 34"

Latest revision as of 13:05, 20 February 2020

Problem

Solution

See Also