Difference between revisions of "2014 AMC 10A Problems/Problem 1"
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What is <math> 10\cdot\left(\tfrac{1}{2}+\tfrac{1}{5}+\tfrac{1}{10}\right)^{-1}? </math> | What is <math> 10\cdot\left(\tfrac{1}{2}+\tfrac{1}{5}+\tfrac{1}{10}\right)^{-1}? </math> | ||
− | <math>\textbf{(A)}\ 3\qquad\textbf{(B)}\ | + | <math>\textbf{(A)}\ 3\qquad\textbf{(B)}\ \frac{25}{2} \qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ \frac{170}{3}\qquad\textbf{(E)}\ 170</math> |
== Solution == | == Solution == | ||
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<cmath>\implies \frac{50}{4}</cmath> | <cmath>\implies \frac{50}{4}</cmath> | ||
Finally, simplifying gives | Finally, simplifying gives | ||
− | <cmath>\implies \boxed{\textbf{( | + | <cmath>\implies \boxed{\textbf{(B)}\ \frac{25}{2}}</cmath> |
==See Also== | ==See Also== |
Revision as of 22:17, 23 January 2020
Problem
What is
Solution
We have Making the denominators equal gives Finally, simplifying gives
See Also
2014 AMC 10A (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 AMC 10 Problems and Solutions |
2014 AMC 12A (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 AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.