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== | + | ==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. | ||
| + | |||
| + | ==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>.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 12:48, 21 August 2021
Contents
Problem
Jamar bought some pencils costing more than a penny each at the school bookstore and paid 
. Sharona bought some of the same pencils and paid 
. How many more pencils did Sharona buy than Jamar?
Solution 1
We assume that the price of the pencils remains constant. Convert and 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 
and 
, which is 
. Therefore, Jamar bought 
pencils and Sharona bought 
pencils. Thus, Sharona bought 
more pencils than Jamar.
Solution 2
We find the difference between 
and 
is 
.It cost Sharona more 
to buy pencils. Because the difference between the numbers of peciles they bought must be divided evenly by . So the answer should be 
or 
, which gives us the cost of each pencile should be 
or 
. Then we find only can be divided evenly by and . So the answer is 
---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. 
