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1968 AHSME Problems/Problem 32

Problem

$A$ and $B$ move uniformly along two straight paths intersecting at right angles in point $O$. When $A$ is at $O$, $B$ is $500$ yards short of $O$. In two minutes they are equidistant from $O$, and in $8$ minutes more they are again equidistant from $O$. Then the ratio of $A$'s speed to $B$'s speed is:

$\text{(A) } 4:5\quad \text{(B) } 5:6\quad \text{(C) } 2:3\quad \text{(D) } 5:8\quad \text{(E) } 1:2$


Solution

Let the speed of $A$ be $a$ and the speed of $B$ be $b$. The first time that $A$ and $B$ will be equidistant from $O$, $B$ will have not yet reached $O$. Thus, after two minutes, $B$'s distance from $O$ will be $500-2b$, and $A$'s distance from $O$ will be $2a$. Setting these expressions equal to each other and dividing by 2, we see that $a=250-b$.

After another eight minutes (or after a total of ten minutes since $A$ was at $O$), $A$ and $B$ will again be equidistant from $O$, but this time $B$ will have passed $O$. The distance $A$ will be from $O$ is $10a$, and the distance $B$ will be from $O$ is $10b-500$. Setting these expressions equal to each other and dividing by 10, we see that $a=b-50$.

Adding the two equations that we have obtained above, we see that $2a=250-b+b-50$, and so $a=100$. Substituting this value of $a$ into the second equation, we see that $100=b-50$, or $b=150$. Then, $\frac{a}{b}=\frac{100}{150}=\frac{2}{3}$, so the ratio of $A$'s speed to that of $B$ is $\fbox{(C) 2:3}$.

See also

1968 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 31 		Followed by
Problem 33
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
All AHSME Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: Unable to save thumbnail to destination

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