1973 AHSME Problems/Problem 24

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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)}\ $1.05}$.

See Also

1973 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 23
Followed by
Problem 25
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