Difference between revisions of "1973 AHSME Problems/Problem 24"
Rockmanex3 (talk | contribs) (Solution to Problem 24) |
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==See Also== | ==See Also== | ||
| − | {{AHSME | + | {{AHSME 30p box|year=1973|num-b=23|num-a=25}} |
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
Latest revision as of 13:03, 20 February 2020
Problem
The check for a luncheon of 3 sandwiches, 7 cups of coffee and one piece of pie came to 
. The check for a luncheon consisting of 4 sandwiches, 10 cups of coffee and one piece of pie came to 
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
Solution
Let 
be the cost of one sandwich, 
be the cost of one cup of coffee, and 
be the price of one piece of pie. With the information,
![\[3s+7c+p=3.15\]](http://latex.artofproblemsolving.com/c/6/1/c61a84497e3502a890592987dc0b51407fe10196.png)
![\[4s+10c+p=4.20\]](http://latex.artofproblemsolving.com/7/4/6/7463636d5cd265b49eff34210864f5f6a81f7a82.png)
Subtract the first equation from the second to get
![\[s+3c=1.05\]](http://latex.artofproblemsolving.com/e/d/2/ed25e2d4b508a2f7c1078a0d81ad030153efe191.png)
That means 
. Substituting it back in the second equation results in.
![\[4.20-12c+10c+p=4.20\]](http://latex.artofproblemsolving.com/3/1/4/31469830ac8ca8d4b798d164b5117d5ef6cd6537.png)
Solving for yields 
. With the substitutions, the cost of one sandwich, one cup of coffee, and one slice of pie is 
.
See Also
| 1973 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 23 |
Followed by Problem 25 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 | ||
| All AHSME Problems and Solutions | ||
