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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]]}}
 
== Problem ==
 
== Problem ==
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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?
  
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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?
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<math>\textbf{(A)}\ 31\qquad\textbf{(B)}\ 32\qquad\textbf{(C)}\ 33\qquad\textbf{(D)}\ 34\qquad\textbf{(E)}\ 35</math>
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<math> \mathrm{(A) \ } 31\qquad \mathrm{(B) \ } 32\qquad \mathrm{(C) \ } 33\qquad \mathrm{(D) \ } 34</math>
 
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<math>\mathrm{(E) \ } 35 </math>
 
  
 
== Solution ==
 
== Solution ==
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The <math>5</math> sandwiches cost <math>5\cdot 3=15</math> dollars. The <math>8</math> sodas cost <math>8\cdot 2=16</math> dollars. In total, the purchase costs <math>15+16=\boxed{\textbf{(A) }31}</math> dollars.
  
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The <math>5</math> sandwiches cost <math>5\cdot 3=15</math> dollars.  The <math>8</math> sodas cost <math>8\cdot 2=16</math> dollars.  In total, the purchase costs <math>15+16=31</math> dollars.
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== See also ==
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{{AMC12 box|year=2006|ab=A|before=First Question|num-a=2}}
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{{AMC10 box|year=2006|ab=A|before=First Question|num-a=2}}
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{{MAA Notice}}
  
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== See also ==
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[[Category:Introductory Algebra Problems]]
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* [[2006 AMC 12A Problems]]
 

Latest revision as of 16:20, 16 December 2021

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?

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

Solution

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

See also

2006 AMC 12A (Problems • Answer Key • Resources)
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 (Problems • Answer Key • Resources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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