# 1973 AHSME Problems/Problem 24

## Problem

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 $\textbf{(A)}\ 1.70 \qquad \textbf{(B)}\ 1.65 \qquad \textbf{(C)}\ 1.20 \qquad \textbf{(D)}\ 1.05 \qquad \textbf{(E)}\ 0.95$

## Solution

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 $(1.05-3c)+c+(2c) = \boxed{\textbf{(D)}\ 1.05}$.