1968 AHSME Problems/Problem 25

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Problem

Ace runs with constant speed and Flash runs $x$ times as fast, $x>1$. Flash gives Ace a head start of $y$ yards, and, at a given signal, they start off in the same direction. Then the number of yards Flash must run to catch Ace is:

$\text{(A) } xy\quad \text{(B) } \frac{y}{x+y}\quad \text{(C) } \frac{xy}{x-1}\quad \text{(D) } \frac{x+y}{x+1}\quad \text{(E) } \frac{x+y}{x-1}$

Solution

$\fbox{C}$

Solution 2

Let $k$ denotes the distance Ace needs to run after the $y$ yard. Since the distance, they run with same amount of time is proportional to their speed, we have \[\frac{1}{x}=\frac{k}{y+k}\] \[k=\frac{y}{x-1}\] Thus the total distance ran by Flash is \[y+k=y+\frac{y}{x-1}=\frac{xy}{x-1}\boxed{C}\]

~ Nafer

See also

1968 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 24
Followed by
Problem 26
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