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Latest revision as of 16:32, 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 $11\cdot13$ and $3\cdot5\cdot13$. 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 $52/13 = \boxed{\textbf{(D)}\ 4}$ 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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