1968 AMC 12 Problem 25

Attempt Problem 25 of the 1968 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1968 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

25.

Ace runs with constant speed and Flash runs

x

times as fast,

x\gt1.

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:

xy

\dfrac{y}{x+y}

\dfrac{xy}{x-1}

\dfrac{x+y}{x+1}

\dfrac{x+y}{x-1}

If Ace’s speed is

v,

the closing speed is

(x-1)v

Multiply the catch-up time by Flash’s speed

xv

If Ace runs at speed

v,

Flash runs at

xv,

so their closing speed is

(x-1)v.

The catch-up time is

\frac{y}{(x-1)v},

during which Flash runs

\frac{xv\cdot y}{(x-1)v}=\frac{xy}{x-1}.

Therefore, the correct answer is C.

Problem 25 in other years