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Difference between revisions of "2012 AMC 8 Problems/Problem 13"
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<math> \textbf{(A)}\hspace{.05in}2\qquad\textbf{(B)}\hspace{.05in}3\qquad\textbf{(C)}\hspace{.05in}4\qquad\textbf{(D)}\hspace{.05in}5\qquad\textbf{(E)}\hspace{.05in}6 </math>
 
<math> \textbf{(A)}\hspace{.05in}2\qquad\textbf{(B)}\hspace{.05in}3\qquad\textbf{(C)}\hspace{.05in}4\qquad\textbf{(D)}\hspace{.05in}5\qquad\textbf{(E)}\hspace{.05in}6 </math>
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==Solution==
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We assume that the price of the pencils remains constant. Convert <math> \textdollar 1.43 </math> and <math> \textdollar 1.87 </math> to cents.  Since the price of the pencils is more than one penny, we can find the price of one pencil (in cents) by taking the greatest common divisor of <math> 143 </math> and <math> 187 </math>, which is <math> 11 </math>. Therefore, Jamar bought <math> \frac{143}{11} \implies 13 </math> pencils and Sharona bought <math> \frac{187}{11} \implies 17 </math> pencils. Thus, Sharona bought <math> 17-13 = \boxed{\textbf{(C)}\ 4} </math>  more pencils than Jamar.
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2012|num-b=12|num-a=14}}
 
{{AMC8 box|year=2012|num-b=12|num-a=14}}

Revision as of 11:57, 24 November 2012

Jamar bought some pencils costing more than a penny each at the school bookstore and paid $\textdollar 1.43$. Sharona bought some of the same pencils and paid $\textdollar 1.87$. How many more pencils did Sharona buy than Jamar?

$\textbf{(A)}\hspace{.05in}2\qquad\textbf{(B)}\hspace{.05in}3\qquad\textbf{(C)}\hspace{.05in}4\qquad\textbf{(D)}\hspace{.05in}5\qquad\textbf{(E)}\hspace{.05in}6$

Solution

We assume that the price of the pencils remains constant. Convert $\textdollar 1.43$ and $\textdollar 1.87$ to cents. Since the price of the pencils is more than one penny, we can find the price of one pencil (in cents) by taking the greatest common divisor of $143$ and $187$, which is $11$. Therefore, Jamar bought $\frac{143}{11} \implies 13$ pencils and Sharona bought $\frac{187}{11} \implies 17$ pencils. Thus, Sharona bought $17-13 = \boxed{\textbf{(C)}\ 4}$ more pencils than Jamar.

See Also

2012 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
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 AJHSME/AMC 8 Problems and Solutions
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