Difference between revisions of "2010 AMC 8 Problems/Problem 1"
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| + | ==Problem== | ||
| + | At Euclid Middle School the mathematics teachers are Miss Germain, Mr. Newton, and Mrs. Young. There are <math>11</math> students in Mrs. Germain's class, <math>8</math> students in Mr. Newton's class, and <math>9</math> students in Mrs. Young's class taking the AMC 8 this year. How many mathematics students at Euclid Middle School are taking the contest? | ||
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| + | <math> \textbf{(A)}\ 26 \qquad\textbf{(B)}\ 27\qquad\textbf{(C)}\ 28\qquad\textbf{(D)}\ 29\qquad\textbf{(E)}\ 30 </math> | ||
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==Solution== | ==Solution== | ||
Given that these are the only math teacher's at Euclid Middle School and we are told how many from each class are taking the AMC 8, we simply add the three numbers to find the total. <math>11+8+9=\boxed{\textbf{(C)}\ 28}</math> | Given that these are the only math teacher's at Euclid Middle School and we are told how many from each class are taking the AMC 8, we simply add the three numbers to find the total. <math>11+8+9=\boxed{\textbf{(C)}\ 28}</math> | ||
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| + | ==See Also== | ||
| + | {{AMC8 box|year=2010|before=First Problem|num-a=2}} | ||
Revision as of 12:05, 4 March 2012
Problem
At Euclid Middle School the mathematics teachers are Miss Germain, Mr. Newton, and Mrs. Young. There are 
students in Mrs. Germain's class, 
students in Mr. Newton's class, and 
students in Mrs. Young's class taking the AMC 8 this year. How many mathematics students at Euclid Middle School are taking the contest?
Solution
Given that these are the only math teacher's at Euclid Middle School and we are told how many from each class are taking the AMC 8, we simply add the three numbers to find the total. 
See Also
| 2010 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by First Problem |
Followed by Problem 2 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
