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

(Solution)
m (Problem)
 
(5 intermediate revisions by 3 users not shown)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
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?
+
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>
 
<math>
Line 21: Line 21:
 
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>.
 
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, and 23</math> must be the number of students, the number of pencils purchased by each student and the price of each pencil in cents.
+
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 know that <math>23</math> must be the number of students, as it is the only number that makes up the majority of 30.
Line 35: Line 35:
 
https://www.youtube.com/watch?v=6tlqpAcmbz4
 
https://www.youtube.com/watch?v=6tlqpAcmbz4
 
~Shreyas S
 
~Shreyas S
 +
 +
https://youtu.be/E5UoNyFjSO0
 +
 +
~savannahsolver
  
 
== 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}}
 +
{{AMC10 box|year=2011|num-b=9|num-a=11|ab=A}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 23:58, 18 July 2021

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 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.

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

Video Solution

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

https://youtu.be/E5UoNyFjSO0

~savannahsolver

See also

2011 AMC 12A (ProblemsAnswer KeyResources)
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
2011 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
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 10 Problems and Solutions

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

Invalid username
Login to AoPS