Difference between revisions of "1964 AHSME Problems/Problem 26"
Talkinaway (talk | contribs) (Created page with "==Problem== In a ten-mile race <math>\textit{First}</math> beats <math>\textit{Second}</math> by <math>2</math> miles and <math>\textit{First}</math> beats <math>\textit{Thi...") |
Talkinaway (talk | contribs) (→Solution) |
||
Line 16: | Line 16: | ||
Let the speeds of the runners in miles per hour be <math>a, b, c</math>, with <math>a>b>c</math>. If person <math>a</math> reaches the finish line in <math>h</math> hours, then we have <math>10 = ah, 8 = bh, 6 = bh</math>. | Let the speeds of the runners in miles per hour be <math>a, b, c</math>, with <math>a>b>c</math>. If person <math>a</math> reaches the finish line in <math>h</math> hours, then we have <math>10 = ah, 8 = bh, 6 = bh</math>. | ||
− | Thus, the speeds of the second and third place people are <math>\frac{8}{h}</math> and <math>|frac{6}{h}</math>. The ratio of those speeds is <math>\frac{\frac{8}{h}}{\frac{6}{h}} = \frac{4}{3}</math>, meaning person <math>b</math> runs <math>\frac{4}{3}</math> times as fast as person <math>c</math>, or conversely that person <math>c</math> is <math>\frac | + | Thus, the speeds of the second and third place people are <math>\frac{8}{h}</math> and <math>|frac{6}{h}</math>. The ratio of those speeds is <math>\frac{\frac{8}{h}}{\frac{6}{h}} = \frac{4}{3}</math>, meaning person <math>b</math> runs <math>\frac{4}{3}</math> times as fast as person <math>c</math>, or conversely that person <math>c</math> is <math>\frac{3}{4}</math> times slower than person <math>b</math>. |
If <math>c</math> runs <math>0.75</math> times as fast as <math>b</math>, then when <math>b</math> has finished the race at <math>10</math> miles, <math>c</math> will run <math>10 \cdot 0.75 = 7.5</math> miles. This is a difference of <math>10 - 75. = 2.5</math> miles, which is answer <math>\boxed{\textbf{(C)}}</math> | If <math>c</math> runs <math>0.75</math> times as fast as <math>b</math>, then when <math>b</math> has finished the race at <math>10</math> miles, <math>c</math> will run <math>10 \cdot 0.75 = 7.5</math> miles. This is a difference of <math>10 - 75. = 2.5</math> miles, which is answer <math>\boxed{\textbf{(C)}}</math> |
Revision as of 21:43, 24 July 2019
Problem
In a ten-mile race beats
by
miles and
beats
by
miles.
If the runners maintain constant speeds throughout the race,
by how many miles does
beat
?
Solution
Let the speeds of the runners in miles per hour be , with
. If person
reaches the finish line in
hours, then we have
.
Thus, the speeds of the second and third place people are and
. The ratio of those speeds is
, meaning person
runs
times as fast as person
, or conversely that person
is
times slower than person
.
If runs
times as fast as
, then when
has finished the race at
miles,
will run
miles. This is a difference of
miles, which is answer
See Also
1964 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 25 |
Followed by Problem 27 | |
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 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.