Difference between revisions of "2010 AMC 8 Problems/Problem 1"
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+ | ==Problem== | ||
+ | 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? | ||
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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 | + | 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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+ | ==See Also== | ||
+ | {{AMC8 box|year=2010|before=First Problem|num-a=2}} | ||
+ | {{MAA Notice}} |
Latest 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 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
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.