2009 AMC 8 Problems/Problem 11

Revision as of 14:57, 14 August 2021 by Raina0708 (talk | contribs) (Solution)

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
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All AJHSME/AMC 8 Problems and Solutions

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