Difference between revisions of "2010 AMC 8 Problems/Problem 1"
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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. 11+8+9=\boxed{\textbf{(C)}\ 28} | + | 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> |
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| + | ==Video by MathTalks== | ||
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| + | https://youtu.be/EEbksvfujhk | ||
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| + | ==Video Solution by WhyMath== | ||
| + | https://youtu.be/XMuirbJA4ZU | ||
==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}} | ||
Latest revision as of 10:22, 18 November 2024
Problem
At Euclid Middle School the mathematics teachers are Mrs. 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 this year. How many mathematics students at Euclid Middle School are taking the contest?
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. 
Video by MathTalks
Video Solution by WhyMath
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
