Difference between revisions of "2014 AMC 10A Problems/Problem 1"
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Finally, simplifying gives | Finally, simplifying gives | ||
<cmath>\implies \boxed{\textbf{(C)}\ \frac{25}{2}}</cmath> | <cmath>\implies \boxed{\textbf{(C)}\ \frac{25}{2}}</cmath> | ||
| + | |||
| + | == Solution 2 == | ||
| + | We have | ||
| + | <cmath>\left(\frac{1}{10}\right)^{-1}\cdot \left(\frac{1}{2} + \frac{1}{5} + \frac{1}{10}\right)^{-1}</cmath>By Distributive Property, | ||
| + | <cmath>\left(\frac{1}{20}+\frac{1}{50}+\frac{1}{100}\right)^{-1}</cmath>Now, we want to find the least common multiple of <math>20, 50,</math> and <math>100,</math> so | ||
| + | <cmath>\text{lcm}(20,50,100)=\text{lcm}(2^2 \cdot 5,2 \cdot 5^2,2^2 \cdot 5^2)=2^2 \cdot 5^2=100</cmath>Converting everything to a denominator of <math>100,</math> | ||
| + | <cmath>\left(\frac{5}{100}+\frac{2}{100}+\frac{1}{100}\right)^{-1}=\left(\frac{8}{100}\right)^{-1}=\frac{100}{8}</cmath>Now, we use Euclidean Algorithm, to find if this fraction is reducible, so | ||
| + | <cmath>\gcd(100,8)=\gcd(12,8)=\gcd(4,8)=\gcd(4,4)</cmath>Thus, both the numerator and denominator are divisible by <math>4,</math> so | ||
| + | <cmath>\frac{100}{8} \cdot \frac{4}{4}=\frac{100}{4} \cdot \frac{4}{8}=25 \cdot \frac{1}{2}=\boxed{\frac{25}{2}}</cmath> | ||
| + | |||
| + | - kante314 | ||
==Video Solution== | ==Video Solution== | ||
Revision as of 20:17, 19 October 2021
Problem
What is 
Solution
We have ![\[10\cdot\left(\frac{1}{2}+\frac{1}{5}+\frac{1}{10}\right)^{-1}\]](http://latex.artofproblemsolving.com/1/2/e/12e21260e737c4ecc2bf7b4116b1f1e7bcde2b37.png)
Making the denominators equal gives
![\[\implies 10\cdot\left(\frac{5}{10}+\frac{2}{10}+\frac{1}{10}\right)^{-1}\]](http://latex.artofproblemsolving.com/8/5/8/858522583ebacde2c4e851b8fd64c093086a668a.png)
![\[\implies 10\cdot\left(\frac{5+2+1}{10}\right)^{-1}\]](http://latex.artofproblemsolving.com/b/9/c/b9c7e18416c20508fb4b4e0822b1c6183699a746.png)
![\[\implies 10\cdot\left(\frac{8}{10}\right)^{-1}\]](http://latex.artofproblemsolving.com/2/8/3/2832b440e25ccb94bfdd37a6c04898ff0c7544ca.png)
![\[\implies 10\cdot\left(\frac{4}{5}\right)^{-1}\]](http://latex.artofproblemsolving.com/6/d/3/6d309b6b45b8f4e35d741ae32ea81647183f0b5e.png)
![\[\implies 10\cdot\frac{5}{4}\]](http://latex.artofproblemsolving.com/8/6/1/8612999285ee03d02effd49381ab5a5a3aea86a6.png)
![\[\implies \frac{50}{4}\]](http://latex.artofproblemsolving.com/0/1/0/0100cd85a3f1692e5c483ab0559633686ad33cdd.png)
Finally, simplifying gives
![\[\implies \boxed{\textbf{(C)}\ \frac{25}{2}}\]](http://latex.artofproblemsolving.com/5/f/b/5fba06f914e8493aeae1d11cbb060a29311f3b7d.png)
Solution 2
We have
![\[\left(\frac{1}{10}\right)^{-1}\cdot \left(\frac{1}{2} + \frac{1}{5} + \frac{1}{10}\right)^{-1}\]](http://latex.artofproblemsolving.com/b/b/f/bbf8ad5578ed582621b5c63d4182c88e211562c8.png)
By Distributive Property,
![\[\left(\frac{1}{20}+\frac{1}{50}+\frac{1}{100}\right)^{-1}\]](http://latex.artofproblemsolving.com/3/6/c/36ced254e97009416e30a918a5873be9d3776858.png)
Now, we want to find the least common multiple of 
and 
so
![\[\text{lcm}(20,50,100)=\text{lcm}(2^2 \cdot 5,2 \cdot 5^2,2^2 \cdot 5^2)=2^2 \cdot 5^2=100\]](http://latex.artofproblemsolving.com/d/2/a/d2ab602940470b5dd03bd09ffa7c057bfd8cfae0.png)
Converting everything to a denominator of
![\[\left(\frac{5}{100}+\frac{2}{100}+\frac{1}{100}\right)^{-1}=\left(\frac{8}{100}\right)^{-1}=\frac{100}{8}\]](http://latex.artofproblemsolving.com/b/b/3/bb310752666bfc04f6311cae6fd9e37b6f4377b5.png)
Now, we use Euclidean Algorithm, to find if this fraction is reducible, so
![\[\gcd(100,8)=\gcd(12,8)=\gcd(4,8)=\gcd(4,4)\]](http://latex.artofproblemsolving.com/3/6/c/36cf18d63b75b8ad0dd7cbeb2b0b9db88b0217f6.png)
Thus, both the numerator and denominator are divisible by 
so
![\[\frac{100}{8} \cdot \frac{4}{4}=\frac{100}{4} \cdot \frac{4}{8}=25 \cdot \frac{1}{2}=\boxed{\frac{25}{2}}\]](http://latex.artofproblemsolving.com/e/6/1/e616bd29d5fcd3e25ba81f89729a22c1bcfa3283.png)
- kante314
Video Solution
~savannahsolver
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. 
