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>
  
==Solution==
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==Solution 1==
 
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.
 
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.
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==Solution 2==
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We find the difference between <math>1.43</math> and <math>1.87</math> is <math>1.87-1.43 = 0.44</math>.It cost Sharona more <math>0.44</math> to buy pencils. Because the difference between the numbers of peciles they bought must be divided evenly by <math>0.44</math>. So the answer should be <math>2</math> or <math>4</math>, which gives us the cost of each pencile should be <math>0.22</math> or <math>0.11</math>. Then we find only <math>0.11</math> can be divided evenly by <math>1.43</math> and <math>1.87</math>. So the answer is <math> \boxed{\textbf{(C)}\ 4} </math>  ---LarryFlora
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2012|num-b=12|num-a=14}}
 
{{AMC8 box|year=2012|num-b=12|num-a=14}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 13:48, 21 August 2021

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 $1.87-1.43 = 0.44$.It cost Sharona more $0.44$ to buy pencils. Because the difference between the numbers of peciles they bought must be divided evenly by $0.44$. So the answer should be $2$ or $4$, which gives us the cost of each pencile should be $0.22$ or $0.11$. 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

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

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