Difference between revisions of "2012 AMC 8 Problems/Problem 13"

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

Revision as of 12:57, 9 December 2012

Problem

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 (ProblemsAnswer KeyResources)
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