2011 AMC 12A Problems/Problem 7

Problem

A majority of the $30$ students in Ms. Demeanor's class bought pencils at the school bookstore. Each of these students bought the same number of pencils, and this number was greater than $1$ . The cost of a pencil in cents was greater than the number of pencils each student bought, and the total cost of all the pencils was $17.71$ . What was the cost of a pencil in cents?

$\textbf{(A)}\ 7 \qquad \textbf{(B)}\ 11 \qquad \textbf{(C)}\ 17 \qquad \textbf{(D)}\ 23 \qquad \textbf{(E)}\ 77$

Solution

The total cost of the pencils can be found by $(\text{students}*\text{pencils purchased by each}*\text{price of each pencil})$ .

Since $1771$ is the product of three sets of values, we can begin with prime factorization, since it gives some insight into the values: $7, 11, 23$ . Since neither $(C)$ nor $(E)$ are any of these factors, they can be eliminated immediately, leaving $(A)$ , $(B)$ , and $(D)$ .

Beginning with $(A) 7$ , we see that the number of pencils purchased by each student must be either $11$ or $23$ . However, the problem states that the price of each pencil must exceed the number of pencils purchased, so we can eliminate this.

Continuing with $(B) 11$ , we can conclude that the only case that fulfills the restrictions are that there are $23$ students who each purchased $7$ such pencils, so the answer is $\boxed{B}$ . We can apply the same logic to $(E)$ as we applied to $(A)$ , if one wants to make doubly sure.

2011 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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 AMC 12 Problems and Solutions

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

Page

Toolbox

Search

2011 AMC 12A Problems/Problem 7

Problem

Solution

See also