# Difference between revisions of "2009 AMC 8 Problems/Problem 11"

## 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.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 