Difference between revisions of "2011 AMC 12A Problems/Problem 7"
(→Solution) |
(→Solution) |
||
Line 12: | Line 12: | ||
<cmath>1771 = 7 \times 11 \times 23</cmath> | <cmath>1771 = 7 \times 11 \times 23</cmath> | ||
− | The only factor of 1771 between 16 and 30 (16 because it has to be a majority of her students) is 23. Therefore, we can conclude that there are 23 students, 11 cents per pencil, and each student bought 7 pencils. <math>\Rightarrow</math> | + | The only factor of 1771 between 16 and 30 (16 because it has to be a majority of her students) is 23. Therefore, we can conclude that there are 23 students, 11 cents per pencil, and each student bought 7 pencils. <math>\Rightarrow \box{B}</math> |
== 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}} |
Revision as of 18:56, 10 February 2011
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
The only factor of 1771 between 16 and 30 (16 because it has to be a majority of her students) is 23. Therefore, we can conclude that there are 23 students, 11 cents per pencil, and each student bought 7 pencils. $\Rightarrow \box{B}$ (Error compiling LaTeX. ! Missing number, treated as zero.)
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 |