Difference between revisions of "1959 AHSME Problems/Problem 30"
Brackie. . (talk | contribs) (Created page with "Solution : B (Can someone change it to latex) In 15 seconds, A will complete 3/8 of the track. This means that B will complete 5/8 of the track in 15 seconds, meaning that...") |
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| − | + | == Problem == | |
| − | ( | + | <math>A</math> can run around a circular track in <math>40</math> seconds. <math>B</math>, running in the opposite direction, meets <math>A</math> every <math>15</math> seconds. |
| + | What is <math>B</math>'s time to run around the track, expressed in seconds? | ||
| + | <math>\textbf{(A)}\ 12\frac12 \qquad\textbf{(B)}\ 24\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 27\frac12\qquad\textbf{(E)}\ 55 </math> | ||
| − | In 15 seconds, A will complete 3/ | + | == Solution == |
| + | |||
| + | In <math>15</math> seconds, A will complete <math>\frac{3}{8}</math> of the track. This means that B will complete <math>\frac{5}{8}</math> of the track in <math>15</math> seconds, meaning that to complete the whole track (making the fraction 1), it will take <math>\frac{8}{5} \cdot 15 = 24</math> seconds. So the answer is <math>\boxed{B}</math>. | ||
Revision as of 13:02, 16 July 2024
Problem
can run around a circular track in 
seconds. 
, running in the opposite direction, meets every 
seconds.
What is 's time to run around the track, expressed in seconds?
Solution
In seconds, A will complete 
of the track. This means that B will complete 
of the track in seconds, meaning that to complete the whole track (making the fraction 1), it will take 
seconds. So the answer is 
.
