# 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 $\boxed{\textbf{(C)}}$.

## Solution 2

Let $s_a$, $s_b$, $s_c$ be the speeds of $A$, $B$, $C$, respectively and let $t_a$, $t_b$, $t_c$ be their respective times needed, such that $$d=s_at_a=s_bt_b=s_ct_b$$ Putting each pair of them into a race gets the $3$ equations $$s_at_a-s_bt_a=20$$ $$s_bt_b-s_ct_b=10$$ $$s_at_a-s_ct_a=28$$ Subtract the first equation from the third yields $$s_bt_a-s_ct_a=t_a(s_b-s_c)=8$$ Divide the equation by the second original equation yields $$\frac{t_a}{t_b}=\frac{4}{5}$$ Since $s_at_a=s_bt_b$ we have $$\frac{s_a}{s_b}=\frac{5}{4}$$ Thus from the first equation we have $$s_at_a-s_bt_a=s_at_a-\frac{4}{5}s_at_a=20\Longrightarrow d=s_at_a=100 \boxed{C}$$

~ Nafer