Difference between revisions of "2014 AMC 10A Problems/Problem 1"
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We have <cmath>10\cdot\left(\frac{1}{2}+\frac{1}{5}+\frac{1}{10}\right)^{-1}</cmath> | We have <cmath>10\cdot\left(\frac{1}{2}+\frac{1}{5}+\frac{1}{10}\right)^{-1}</cmath> | ||
+ | Making the denominators equal gives | ||
<cmath>\implies 10\cdot\left(\frac{5}{10}+\frac{2}{10}+\frac{1}{10}\right)^{-1}</cmath> | <cmath>\implies 10\cdot\left(\frac{5}{10}+\frac{2}{10}+\frac{1}{10}\right)^{-1}</cmath> | ||
<cmath>\implies 10\cdot\left(\frac{5+2+1}{10}\right)^{-1}</cmath> | <cmath>\implies 10\cdot\left(\frac{5+2+1}{10}\right)^{-1}</cmath> | ||
Line 13: | Line 14: | ||
<cmath>\implies 10\cdot\frac{5}{4}</cmath> | <cmath>\implies 10\cdot\frac{5}{4}</cmath> | ||
<cmath>\implies \frac{50}{4}</cmath> | <cmath>\implies \frac{50}{4}</cmath> | ||
+ | Finally, simplifying gives | ||
<cmath>\implies \boxed{\textbf{(C)}\ \frac{25}{2}}\ \blacksquare</cmath> | <cmath>\implies \boxed{\textbf{(C)}\ \frac{25}{2}}\ \blacksquare</cmath> | ||
Revision as of 10:19, 9 February 2014
Problem
What is
$\textbf{(A)}\ 3\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ \frac{25}{2} \qquad\textbf{(D)}}\ \frac{170}{3}\qquad\textbf{(E)}\ 170$ (Error compiling LaTeX. Unknown error_msg)
Solution
We have Making the denominators equal gives Finally, simplifying gives
See Also
2014 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.