Difference between revisions of "2010 AMC 8 Problems/Problem 1"

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==Problem==
 
==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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At Euclid Middle School the mathematics teachers are Mrs. 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 <math>8</math> this year. How many mathematics students at Euclid Middle School are taking the contest?  
  
 
<math> \textbf{(A)}\ 26 \qquad\textbf{(B)}\ 27\qquad\textbf{(C)}\ 28\qquad\textbf{(D)}\ 29\qquad\textbf{(E)}\ 30 </math>
 
<math> \textbf{(A)}\ 26 \qquad\textbf{(B)}\ 27\qquad\textbf{(C)}\ 28\qquad\textbf{(D)}\ 29\qquad\textbf{(E)}\ 30 </math>

Revision as of 17:04, 6 January 2018

Problem

At Euclid Middle School the mathematics teachers are Mrs. Germain, Mr. Newton, and Mrs. Young. There are $11$ students in Mrs. Germain's class, $8$ students in Mr. Newton's class, and $9$ students in Mrs. Young's class taking the AMC $8$ this year. How many mathematics students at Euclid Middle School are taking the contest?

$\textbf{(A)}\ 26 \qquad\textbf{(B)}\ 27\qquad\textbf{(C)}\ 28\qquad\textbf{(D)}\ 29\qquad\textbf{(E)}\ 30$

Solution

Given that these are the only math teachers 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. $11+8+9=\boxed{\textbf{(C)}\ 28}$

See Also

2010 AMC 8 (ProblemsAnswer KeyResources)
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

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