Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 2)
(→‎Solution 2)
Line 9: Line 9:
  
 
==Solution 2==
 
==Solution 2==
−
We find the difference between <math>1.43</math> and <math>1.87</math> is <math>1.87-1.43 = 0.44</math>, 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 <math>0.44</math>, looking into the answers, <math>2</math> or <math>4</math> is possibly correct. It gives us the price of each pencil should be <math>0.44/2=0.22</math> or <math>0.44/4=0.11</math>, respectively. 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
+
We find the difference between <math>1.43</math> and <math>1.87</math> is <math> 1.87-1.43 = 0.44 </math>, 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 <math>0.44</math>, looking into the answers, <math>2</math> or <math>4</math> is possibly correct. It gives us the price of each pencil should be <math> 0.44/2=0.22 </math> or <math> 0.44/4=0.11 </math>, respectively. 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 12:56, 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 $1.87-1.43 = 0.44$, 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

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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_8_Problems/Problem_13&oldid=160708"