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Difference between revisions of "2011 AMC 12A Problems/Problem 7"

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== Problem ==
 
== Problem ==
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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 <math>\</math>17.71. What was the cost of a pencil in cents?
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A majority of the <math>30</math> 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 <math>1</math>. 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 <math>17.71</math>. What was the cost of a pencil in cents?
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<math> \textbf{(A)}\ 7 \qquad
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<math>
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\textbf{(A)}\ 7 \qquad
 
\textbf{(B)}\ 11 \qquad
 
\textbf{(B)}\ 11 \qquad
 
\textbf{(C)}\ 17 \qquad
 
\textbf{(C)}\ 17 \qquad
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\textbf{(D)}\ 23 \qquad
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\textbf{(D)}\ 23 \qquad
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\textbf{(E)}\ 77
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\textbf{(E)}\ 77 </math>
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</math>
 
  
 
== Solution ==
 
== Solution ==
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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>.
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 +
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.
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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.
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==Video Solution==
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https://www.youtube.com/watch?v=6tlqpAcmbz4
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~Shreyas S
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== See also ==
 
== See also ==
 
{{AMC12 box|year=2011|num-b=6|num-a=8|ab=A}}
 
{{AMC12 box|year=2011|num-b=6|num-a=8|ab=A}}
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{{MAA Notice}}

Revision as of 03:40, 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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