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> | + | 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 |
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