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

Latest revision as of 00:25, 10 November 2024

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 1

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 $(C)$ has none of these factors, it can be eliminated immediately, leaving $(A)$ , $(B)$ , $(D)$ , and $(E)$ .

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 is 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.

~PerseverePlayer (minor edits, changed some grammar, E can't be eliminated at first, fixed that)

Solution 2

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

Using prime factorization like in the solution above, we see that there is only one combination of three whole numbers whose product is equal to $1771$ : $7, 11, 23$ (without using $1$ ). So we know that $7, 11$ , and $23$ must be the number of students, the number of pencils purchased by each student and the price of each pencil in cents.

We know that $23$ must be the number of students, as it is the only number that makes up the majority of 30.

We pick the greater of the remaining numbers for the price of each pencil in cents, which is $11$ .

Therefore, our answer is $\boxed{B}$ .

-JinhoK

Solution 3 (essentially Solutions 1 and 2 but cleaned with variables)

We let $s$ be the number of students that bought pencils at the bookstore, $c$ be the cost of each pencil in cents, and $n$ the number of pencils each student bought. Thus, we are looking for $c.$ Since a majority of the students in the class bought pencils at the bookstore, $s>\dfrac{30}2=15.$ (s>15) We also know that $n>1$ and $c>n>1.$ Finally, $s\cdot c\cdot n=1771.$ We can factor $1771$ as $7\cdot11\cdot23.$ Since $n>15$ and $c>n>1,$ none of them can be $1,$ and therefore $c,n,$ and $s$ are $7,11,$ and $23$ in some order. We know that $n>15,$ so $n$ must be $23.$ $c>n,$ so $c=11$ and $n=7.$ Thus, $c=\boxed{\text{(B)}~11}$ cents. ~Technodoggo

Video Solution

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

https://youtu.be/E5UoNyFjSO0

~savannahsolver

@@ Line 1: / Line 1: @@
 == 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 <math>\</math>17.71. What was the cost of a pencil in cents?
+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?
- <math> \textbf{(A)}\ 7 \qquad
+<math>
+\textbf{(A)}\ 7 \qquad
 \textbf{(B)}\ 11 \qquad
 \textbf{(C)}\ 17 \qquad
-\textbf{(D)}\  23 \qquad
+\textbf{(D)}\ 23 \qquad
-\textbf{(E)}\ 77
+\textbf{(E)}\ 77 </math>
-</math>
+== Solution 1 ==
+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 <math>(C)</math> has none of these factors, it can be eliminated immediately, leaving <math>(A)</math>, <math>(B)</math>, <math>(D)</math>, and <math>(E)</math>.
+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.
+Continuing with <math>(B) 11</math>, we can conclude that the only case that fulfills the restrictions is 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.
+~PerseverePlayer (minor edits, changed some grammar, E can't be eliminated at first, fixed that)
+== Solution 2 ==
+We know 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>.
+Using prime factorization like in the solution above, we see that there is only one combination of three whole numbers whose product is equal to <math>1771</math>: <math>7, 11, 23</math> (without using <math>1</math>). So we know that <math>7, 11</math>, and <math>23</math> must be the number of students, the number of pencils purchased by each student and the price of each pencil in cents.
+We know that <math>23</math> must be the number of students, as it is the only number that makes up the majority of 30.
+We pick the greater of the remaining numbers for the price of each pencil in cents, which is <math>11</math>.
+Therefore, our answer is <math>\boxed{B}</math>.
+-JinhoK
+==Solution 3 (essentially Solutions 1 and 2 but cleaned with variables)==
+We let <math>s</math> be the number of students that bought pencils at the bookstore, <math>c</math> be the cost of each pencil in cents, and <math>n</math> the number of pencils each student bought. Thus, we are looking for <math>c.</math>
+Since a majority of the students in the class bought pencils at the bookstore, <math>s>\dfrac{30}2=15.</math> (s>15) We also know that <math>n>1</math> and <math>c>n>1.</math> Finally, <math>s\cdot c\cdot n=1771.</math> We can factor <math>1771</math> as <math>7\cdot11\cdot23.</math> Since <math>n>15</math> and <math>c>n>1,</math> none of them can be <math>1,</math> and therefore <math>c,n,</math> and <math>s</math> are <math>7,11,</math> and <math>23</math> in some order. We know that <math>n>15,</math> so <math>n</math> must be <math>23.</math> <math>c>n,</math> so <math>c=11</math> and <math>n=7.</math> Thus, <math>c=\boxed{\text{(B)}~11}</math> cents.
+~Technodoggo
+==Video Solution==
+https://www.youtube.com/watch?v=6tlqpAcmbz4
+~Shreyas S
+https://youtu.be/E5UoNyFjSO0
+~savannahsolver
-== Solution ==
 == See also ==
 {{AMC12 box|year=2011|num-b=6|num-a=8|ab=A}}
+{{AMC10 box|year=2011|num-b=9|num-a=11|ab=A}}
+{{MAA Notice}}

2011 AMC 12A (Problems • Answer Key • Resources)
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2011 AMC 10A (Problems • Answer Key • Resources)
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