Art of Problem Solving

2010 AMC 8 Problems/Problem 1: Difference between revisions

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==Solution==
==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. <math>11+8+9=\boxed{\textbf{(C)}\ 28}</math>
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. <math>11+8+9=\boxed{\textbf{(C)}\ 28}</math>
==Video by MathTalks==
https://youtu.be/EEbksvfujhk


==See Also==
==See Also==
{{AMC8 box|year=2010|before=First Problem|num-a=2}}
{{AMC8 box|year=2010|before=First Problem|num-a=2}}
{{MAA Notice}}
{{MAA Notice}}

Revision as of 18:58, 5 February 2023

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. $11+8+9=\boxed{\textbf{(C)}\ 28}$


Video by MathTalks

https://youtu.be/EEbksvfujhk

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing