1950 AHSME Problems/Problem 28
Contents
[hide]Problem
Two boys and
start at the same time to ride from Port Jervis to Poughkeepsie,
miles away.
travels
miles an hour slower than
.
reaches Poughkeepsie and at once turns back meeting
miles from Poughkeepsie. The rate of
was:
Solution
Let the speed of boy be
, and the speed of boy
be
. Notice that
travels
miles per hour slower than boy
, so we can replace
with
.
Now let us see the distances that the boys each travel. Boy travels
miles, and boy
travels
miles. Now, we can use
to make an equation, where we set the time to be equal:
Cross-multiplying gives
. Isolating the variable
, we get the equation
, so
.
Alternate Solution
Note that travels
miles in the time it takes
to travel
miles. Thus,
travels
more miles in the given time, meaning $\frac{24\text{miles}}{4\text{miles}\{hour}} = 6 \text{hours}$ (Error compiling LaTeX. Unknown error_msg) have passed, as
goes
miles per hour faster. Thus,
travels
miles per
hours, or
miles per hour.
See Also
1950 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 27 |
Followed by Problem 29 | |
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.