Difference between revisions of "2009 AMC 8 Problems/Problem 11"
m (→Solution) |
|||
| Line 10: | Line 10: | ||
==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> | + | 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. |
==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 14:15, 26 August 2017
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 
dollars. Some of the 
sixth graders each bought a pencil, and they paid a total of 
dollars. How many more sixth graders than seventh graders bought a pencil?
Solution
Because the pencil costs a whole number of cents, the cost must be a factor of both 
and 
. They can be factored into 
and 
. The common factor cannot be 
or there would have to be more than sixth graders, so the pencil costs 
cents. The difference in costs that the sixth and seventh graders paid is 
cents, which is equal to 
sixth graders.
See Also
| 2009 AMC 8 (Problems • Answer Key • Resources) | ||
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
