Difference between revisions of "2009 AMC 8 Problems/Problem 11"

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==Solution==
 
==Solution==
Because the pencil costs a whole number of cents, the cost must be a factor of both <math>143</math> and <math>195</math>. They can be factored into <math>11\cdot13</math> and <math>3\cdot5\cdot13</math>. The common factor cannot be <math>1</math> or there would have to be more than <math>30</math> sixth graders, so the pencil costs <math>13</math> cents. The difference in costs that the sixth and seventh graders paid is <math>195-143=52</math> cents, which is equal to <math>52/13 = \boxed{\textbf{(D)}\ 4}</math> sixth graders.
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Because the pencil costs a whole number of cents, the cost must be a factor of both 143 and 195 They can be factored into 11*13 and 3*5*13 The common factor cannot be 1 or there would have to be more than 30 sixth graders, so the pencil costs 13 cents. The difference in costs that the sixth and seventh graders paid is 195-143=52 cents which is equal to 52/13 = 4 sixth graders.
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2009|num-b=10|num-a=12}}
 
{{AMC8 box|year=2009|num-b=10|num-a=12}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 14:57, 14 August 2021

Problem

The Amaco Middle School bookstore sells pencils costing a whole number of cents. Some seventh graders each bought a pencil, paying a total of $1.43$ dollars. Some of the $30$ sixth graders each bought a pencil, and they paid a total of $1.95$ dollars. How many more sixth graders than seventh graders bought a pencil?

$\textbf{(A)}\  1  \qquad \textbf{(B)}\   2  \qquad \textbf{(C)}\   3  \qquad \textbf{(D)}\   4  \qquad \textbf{(E)}\   5$

Solution

Because the pencil costs a whole number of cents, the cost must be a factor of both 143 and 195 They can be factored into 11*13 and 3*5*13 The common factor cannot be 1 or there would have to be more than 30 sixth graders, so the pencil costs 13 cents. The difference in costs that the sixth and seventh graders paid is 195-143=52 cents which is equal to 52/13 = 4 sixth graders.

See Also

2009 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
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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