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Difference between revisions of "1959 AHSME Problems/Problem 30"

(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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Solution : B
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== Problem ==
  
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(Can someone change it to latex)  
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<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.
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What is <math>B</math>'s time to run around the track, expressed in seconds?
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<math>\textbf{(A)}\ 12\frac12 \qquad\textbf{(B)}\ 24\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 27\frac12\qquad\textbf{(E)}\ 55    </math>
  
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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 to complete the whole track (making the fraction 1) , it will take 8/5x15 = 24 seconds.
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== Solution ==
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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>.

Latest revision as of 14:02, 16 July 2024

Problem

$A$ can run around a circular track in $40$ seconds. $B$, running in the opposite direction, meets $A$ every $15$ seconds. What is $B$'s time to run around the track, expressed in seconds? $\textbf{(A)}\ 12\frac12 \qquad\textbf{(B)}\ 24\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 27\frac12\qquad\textbf{(E)}\ 55$

Solution

In $15$ seconds, A will complete $\frac{3}{8}$ of the track. This means that B will complete $\frac{5}{8}$ of the track in $15$ seconds, meaning that to complete the whole track (making the fraction 1), it will take $\frac{8}{5} \cdot 15 = 24$ seconds. So the answer is $\boxed{B}$.

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