Difference between revisions of "2011 AMC 10A Problems/Problem 10"
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==Problem 10== | ==Problem 10== | ||
| − | 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> | + | 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>$</math>17.71<math>. What was the cost of a pencil in cents? |
| − | <math>\text{(A)}\,7 \qquad\text{(B)}\,11 \qquad\text{(C)}\,17 \qquad\text{(D)}\,23 \qquad\text{(E)}\,77< | + | </math>\text{(A)}\,7 \qquad\text{(B)}\,11 \qquad\text{(C)}\,17 \qquad\text{(D)}\,23 \qquad\text{(E)}\,77<math> |
== Solution == | == Solution == | ||
| − | Let <math>x>15< | + | Let </math>x>15<math> be the number of students at the bookstore, </math>p>1<math> be the number of pencils each student bought, and </math>c>p<math> be the price of one pencil. We see that </math>xcp = 1771 = 7 \cdot 11 \cdot 23<math>. Then we must have </math>x=23, c=11, p=7<math> by the original conditions and the answer is </math>c=\boxed{11 \ \mathbf{(B)}}$. |
== See Also == | == See Also == | ||
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{{AMC10 box|year=2011|ab=A|num-b=9|num-a=11}} | {{AMC10 box|year=2011|ab=A|num-b=9|num-a=11}} | ||
| + | {{MAA Notice}} | ||
Revision as of 12:00, 4 July 2013
Problem 10
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
\text{(A)}\,7 \qquad\text{(B)}\,11 \qquad\text{(C)}\,17 \qquad\text{(D)}\,23 \qquad\text{(E)}\,77$== Solution ==
Let$ (Error compiling LaTeX. Unknown error_msg)x>15
p>1
c>p
xcp = 1771 = 7 \cdot 11 \cdot 23
x=23, c=11, p=7
c=\boxed{11 \ \mathbf{(B)}}$.
See Also
| 2011 AMC 10A (Problems • Answer Key • Resources) | ||
| 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 | ||
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