Paula the painter and her two helpers each paint at constant, but different, rates. They always start at 8:00 AM, and all three always take the same amount of time to eat lunch. On Monday the three of them painted 50% of a house, quitting at 4:00 PM. On Tuesday, when Paula wasn't there, the two helpers painted only 24% of the house and quit at 2:12 PM. On Wednesday Paula worked by herself and finished the house by working until 7:12 P.M. How long, in minutes, was each day's lunch break?
$\textbf{(A)}\ 30\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 60$
Let Paula work at a rate of $p$, the two helpers work at a combined rate of $h$, and the time it takes to eat lunch be $L$, where $p$ and $h$ are in house/hours and L is in hours. Then the labor on Monday, Tuesday, and Wednesday can be represented by the three following equations:
\[(8-L)(p+h)=50\]
\[(6.2-L)h=24\]
\[(11.2-L)p=26\]
With three equations and three variables, we need to find the value of $L$.
Adding the second and third equations together gives us $6.2h+11.2p-L(p+h)=50$. Subtracting the first equation from this new one gives us $-1.8h+3.2p=0$, so we get $h=\frac{16}{9}p$. 
Plugging into the second equation:
\[(6.2-L)\frac{16}{9}p=24\]
\[(6.2-L)p=\frac{27}{2}\]
We can then subtract this from the third equation:
\[5p=26-\frac{27}{2}\]
\[p=\frac{5}{2}\]
Plugging $p$ into our third equation gives: \[L=\frac{4}{5}\]
Converting $L$ from hours to minutes gives us $L=48$ minutes, which is $\boxed{\textbf{(D)}\ 48}$.