Difference between revisions of "2006 AMC 12A Problems/Problem 1"

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{{duplicate|[[2006 AMC 12A Problems|2006 AMC 12A #1]] and [[2006 AMC 10A Problems|2006 AMC 10A #1]]}}
 
{{duplicate|[[2006 AMC 12A Problems|2006 AMC 12A #1]] and [[2006 AMC 10A Problems|2006 AMC 10A #1]]}}
 
== Problem ==
 
== Problem ==
Sandwiches at Joe's Fast Food cost <math>\$3 each and sodas cost </math>\$2 each. How many dollars will it cost to purchase <math>5</math> sandwiches and <math>8</math> sodas?
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Sandwiches at Joe's Fast Food cost <math>\$3</math> each and sodas cost <math>\$2</math> each. How many dollars will it cost to purchase <math>5</math> sandwiches and <math>8</math> sodas?
  
 
<math>\mathrm{(A)}\ 31\qquad\mathrm{(B)}\ 32\qquad\mathrm{(C)}\ 33\qquad\mathrm{(D)}\ 34\qquad\mathrm{(E)}\ 35</math>
 
<math>\mathrm{(A)}\ 31\qquad\mathrm{(B)}\ 32\qquad\mathrm{(C)}\ 33\qquad\mathrm{(D)}\ 34\qquad\mathrm{(E)}\ 35</math>

Revision as of 18:35, 5 March 2016

The following problem is from both the 2006 AMC 12A #1 and 2006 AMC 10A #1, so both problems redirect to this page.

Problem

Sandwiches at Joe's Fast Food cost $$3$ each and sodas cost $$2$ each. How many dollars will it cost to purchase $5$ sandwiches and $8$ sodas?

$\mathrm{(A)}\ 31\qquad\mathrm{(B)}\ 32\qquad\mathrm{(C)}\ 33\qquad\mathrm{(D)}\ 34\qquad\mathrm{(E)}\ 35$

Solution

The $5$ sandwiches cost $5\cdot 3=15$ dollars. The $8$ sodas cost $8\cdot 2=16$ dollars. In total, the purchase costs $15+16=31$ dollars. The answer is $\mathrm{(A)}$.

See also

2006 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
First Question
Followed by
Problem 2
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
All AMC 12 Problems and Solutions
2006 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
First Question
Followed by
Problem 2
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
All AMC 10 Problems and Solutions

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