2012 AMC 8 Problems/Problem 13

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 1

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.

Solution 2

We find the difference between $1.43$ and $1.87$ is $1.87-1.43 = 0.44$ , which is the extra cost of Sharona's pencils than Jamar's pencils. Because the difference between the amounts of the pencils they bought must be divided evenly by $0.44$ , looking into the answers, $2$ or $4$ is possibly correct. It gives us the price of each pencil should be $0.44/2=0.22$ or $0.44/4=0.11$ , respectively. Then we find only $0.11$ can be divided evenly by $1.43$ and $1.87$ . So the answer is $\boxed{\textbf{(C)}\ 4}$ ---LarryFlora

2012 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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2012 AMC 8 Problems/Problem 13

Contents

Problem

Solution 1

Solution 2

See Also