Difference between revisions of "2011 AMC 12A Problems/Problem 7"
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\textbf{(E)}\ 77 </math> | \textbf{(E)}\ 77 </math> | ||
− | == Solution == | + | == 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>. | 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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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. | 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. | ||
+ | |||
+ | == 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, 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. | ||
+ | |||
+ | 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 | ||
==Video Solution== | ==Video Solution== |
Revision as of 04:43, 9 January 2021
Problem
A majority of the 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 . 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 . What was the cost of a pencil in cents?
Solution 1
The total cost of the pencils can be found by .
Since is the product of three sets of values, we can begin with prime factorization, since it gives some insight into the values: . Since neither nor are any of these factors, they can be eliminated immediately, leaving , , and .
Beginning with , we see that the number of pencils purchased by each student must be either or . However, the problem states that the price of each pencil must exceed the number of pencils purchased, so we can eliminate this.
Continuing with , we can conclude that the only case that fulfils the restrictions are that there are students who each purchased such pencils, so the answer is . We can apply the same logic to as we applied to if one wants to make doubly sure.
Solution 2
We know the total cost of the pencils can be found by .
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 : (without using ). So we know that 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 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 .
Therefore, our answer is .
-JinhoK
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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