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

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== Solution ==
 
== 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>.
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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{\textbf{(B) }24}</math>.
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== See also ==
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{{AHSME 50p box|year=1959|num-b=29|num-a=31}}
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{{MAA Notice}}
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[[Category:Introductory Algebra Problems]]

Latest revision as of 14:33, 21 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 $\boxed{\textbf{(B) }24}$.

See also

1959 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 29 		Followed by
Problem 31
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

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