1961 AHSME Problems/Problem 37

Problem

In racing over a distance $d$ at uniform speed, $A$ can beat $B$ by $20$ yards, $B$ can beat $C$ by $10$ yards, and $A$ can beat $C$ by $28$ yards. Then $d$ , in yards, equals:

$\textbf{(A)}\ \text{Not determined by the given information}\qquad \textbf{(B)}\ 58\qquad \textbf{(C)}\ 100\qquad \textbf{(D)}\ 116\qquad \textbf{(E)}\ 120$

Solution

Let $a$ be speed of $A$ , $b$ be speed of $B$ , and $c$ be speed of $C$ .

Person $A$ finished the track in $\frac{d}{a}$ minutes, so $B$ traveled $\frac{db}{a}$ yards at the same time. Since $B$ is $20$ yards from the finish line, the first equation is $\frac{db}{a} + 20 = d$ Using similar steps, the second and third equation are, respectively, $\frac{dc}{b} + 10 = d$ $\frac{dc}{a} + 28 = d$ Get rid of the denominator in all three equations by multiplying both sides by the denominator in each equation. $db + 20a = da$ $dc + 10b = db$ $dc + 28a = da$ These equations can be rearranged to get the following. $(d-20)a = db$ $(d-10)b = dc$ $(d-28)a = dc$

Solving for $b$ in the first equation yields $\frac{(d-20)a}{d} = b$ . Substituting for $b$ and solving for $c$ in the second equation yields $\frac{(d-10)(d-20)a}{d^2} = c$ . Substituting for $c$ in the third equation yields $(d-28)a = \frac{(d-10)(d-20)a}{d}$ $d^2 - 28d = d^2 - 30d + 200$ Solve the equation to get $d = 100$ . Thus, the track is $100$ yards long, which is answer choice $\boxed{\textbf{(C)}}$ .

1961 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 36	Followed by Problem 38
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1961 AHSME Problems/Problem 37

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