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