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 | + | 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 13:57, 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 
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 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 (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. 
