1952 AHSME Problems/Problem 10

Problem

An automobile went up a hill at a speed of $10$ miles an hour and down the same distance at a speed of $20$ miles an hour. The average speed for the round trip was:

$\textbf{(A) \ }12\frac{1}{2}\text{mph} \qquad \textbf{(B) \ }13\frac{1}{3}\text{mph} \qquad \textbf{(C) \ }14\frac{1}{2}\text{mph} \qquad \textbf{(D) \ }15\text{mph} \qquad \textbf{(E) \ }\text{none of these}$

Solution

Let $r$ , $t$ , and $d$ represent the rate, time, and distance (respectively) of each trip. We know that $rt=d$ , or $r_1t_1=r_2t_2$ . Because it is given that $2r_1=r_2$ , $t_1=2t_2$ . Furthermore, $d=10t_1=20t_2$ . The average speed can be expressed as $\frac{2d}{t_1+t_2}=\frac{2(20t_2)}{2t_2+t_2}=\frac{40}{3}=\boxed{\textbf{(B)}\ 13\frac{1}{3}\text{ mph}}$ .

1952 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

1952 AHSME Problems/Problem 10

Problem

Solution

See also