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 (CREATIVE THINKING)== | ||
+ | https://youtu.be/sbz01QUWY6A | ||
+ | |||
+ | ~Education, the Study of Everything | ||
+ | |||
+ | |||
+ | |||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/QvkvhIMpXz8 | ||
+ | |||
+ | ~savannahsolver | ||
==See Also== | ==See Also== |
Latest revision as of 00:12, 27 June 2023
Contents
Problem
What is
Solution
We have
Making the denominators equal gives
Finally, simplifying gives
Solution 2
We have
By Distributive Property,
Now, we want to find the least common multiple of
and
so
Converting everything to a denominator of
Now, we use Euclidean Algorithm, to find if this fraction is reducible, so
Thus, both the numerator and denominator are divisible by
so
- kante314
Video Solution (CREATIVE THINKING)
https://youtu.be/sbz01QUWY6A
~Education, the Study of Everything
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.