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 $\fbox{(C) 2:3}$ .

1968 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 31	Followed by Problem 33
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1968 AHSME Problems/Problem 32

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