1973 AMC 12 Problem 24

Question

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

\$1.70

\$1.65

\$1.20

\$1.05

\$0.95

Let

s,

c,

p

be the three item prices; only

s+c+p

is required

A suitable linear combination of

3s+7c+p=3.15

and

4s+10c+p=4.20

isolates the desired sum

Let

s,

c,

p

be the prices of a sandwich, coffee, and pie. The checks give

\begin{aligned} 3s+7c+p&=3.15,\\ 4s+10c+p&=4.20. \end{aligned}

Three times the first equation minus twice the second gives

\begin{aligned} s+c+p &=3(3.15)-2(4.20)\\ &=1.05. \end{aligned}

The requested luncheon costs

\$1.05.

Therefore, the correct answer is D.

Problem 24 in other years