The check for a luncheon of 3 sandwiches, 7 cups of coffee and one piece of pie came to $$3.15$. The check for a luncheon consisting of 4 sandwiches, 10 cups of coffee and one piece of pie came to $$4.20$ at the same place. The cost of a luncheon consisting of one sandwich, one cup of coffee, and one piece of pie at the same place will come to 
$\textbf{(A)}\ $1.70 \qquad \textbf{(B)}\ $1.65 \qquad \textbf{(C)}\ $1.20 \qquad \textbf{(D)}\ $1.05 \qquad \textbf{(E)}\ $0.95$
Let $s$ be the cost of one sandwich, $c$ be the cost of one cup of coffee, and $p$ be the price of one piece of pie.  With the information,
\[3s+7c+p=3.15\]
\[4s+10c+p=4.20\]
Subtract the first equation from the second to get
\[s+3c=1.05\]
That means $s=1.05-3c$.  Substituting it back in the second equation results in.
\[4.20-12c+10c+p=4.20\]
Solving for $p$ yields $p=2c$.  With the substitutions, the cost of one sandwich, one cup of coffee, and one slice of pie is $(1.05-3c)+c+(2c) = \boxed{\textbf{(D)}\ $1.05}$.