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Difference between revisions of "2011 AMC 12A Problems/Problem 7"
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== Solution ==
 
== Solution ==
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The total cost of the pencils can be found by <math>(\text{students}*\text{pencils purchased by each}*\text{price of each pencil})</math>.
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The total cost of the pencils can be found by <math>(\text{students}\cdot\text{pencils purchased by each}\cdot\text{price of each pencil})</math>.
  
 
Since <math>1771</math> is the product of three sets of values, we can begin with prime factorization, since it gives some insight into the values: <math>7, 11, 23</math>. Since neither <math>(C)</math> nor <math>(E)</math> are any of these factors, they can be eliminated immediately, leaving <math>(A)</math>, <math>(B)</math>, and <math>(D)</math>.
 
Since <math>1771</math> is the product of three sets of values, we can begin with prime factorization, since it gives some insight into the values: <math>7, 11, 23</math>. Since neither <math>(C)</math> nor <math>(E)</math> are any of these factors, they can be eliminated immediately, leaving <math>(A)</math>, <math>(B)</math>, and <math>(D)</math>.
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Beginning with <math>(A) 7</math>, we see that the number of pencils purchased by each student must be either <math>11</math> or <math>23</math>. However, the problem states that the price of each pencil must exceed the number of pencils purchased, so we can eliminate this.
 
Beginning with <math>(A) 7</math>, we see that the number of pencils purchased by each student must be either <math>11</math> or <math>23</math>. However, the problem states that the price of each pencil must exceed the number of pencils purchased, so we can eliminate this.
  
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Continuing with <math>(B) 11</math>, we can conclude that the only case that fulfills the restrictions are that there are <math>23</math> students who each purchased <math>7</math> such pencils, so the answer is <math>\boxed{B}</math>. We can apply the same logic to <math>(E)</math> as we applied to <math>(A)</math>, if one wants to make doubly sure.
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Continuing with <math>(B) 11</math>, we can conclude that the only case that fulfils the restrictions are that there are <math>23</math> students who each purchased <math>7</math> such pencils, so the answer is <math>\boxed{B}</math>. We can apply the same logic to <math>(E)</math> as we applied to <math>(A)</math> if one wants to make doubly sure.
  
 
==Video Solution==
 
==Video Solution==

Revision as of 02:40, 28 October 2020

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 $(\text{students}\cdot\text{pencils purchased by each}\cdot\text{price of each pencil})$.

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 fulfils 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.

Video Solution

https://www.youtube.com/watch?v=6tlqpAcmbz4 ~Shreyas S

See also

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

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