Difference between revisions of "2014 AMC 10A Problems/Problem 1"

Latest revision as of 23:12, 26 June 2023

Problem

What is $10\cdot\left(\tfrac{1}{2}+\tfrac{1}{5}+\tfrac{1}{10}\right)^{-1}?$

$\textbf{(A)}\ 3 \qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ \frac{25}{2} \qquad\textbf{(D)}\ \frac{170}{3}\qquad\textbf{(E)}\ 170$

Solution

We have $10\cdot\left(\frac{1}{2}+\frac{1}{5}+\frac{1}{10}\right)^{-1}$ Making the denominators equal gives $\implies 10\cdot\left(\frac{5}{10}+\frac{2}{10}+\frac{1}{10}\right)^{-1}$ $\implies 10\cdot\left(\frac{5+2+1}{10}\right)^{-1}$ $\implies 10\cdot\left(\frac{8}{10}\right)^{-1}$ $\implies 10\cdot\left(\frac{4}{5}\right)^{-1}$ $\implies 10\cdot\frac{5}{4}$ $\implies \frac{50}{4}$ Finally, simplifying gives $\implies \boxed{\textbf{(C)}\ \frac{25}{2}}$

Solution 2

We have $\left(\frac{1}{10}\right)^{-1}\cdot \left(\frac{1}{2} + \frac{1}{5} + \frac{1}{10}\right)^{-1}$ By Distributive Property, $\left(\frac{1}{20}+\frac{1}{50}+\frac{1}{100}\right)^{-1}$ Now, we want to find the least common multiple of $20, 50,$ and $100,$ so $\text{lcm}(20,50,100)=\text{lcm}(2^2 \cdot 5,2 \cdot 5^2,2^2 \cdot 5^2)=2^2 \cdot 5^2=100$ Converting everything to a denominator of $100,$ $\left(\frac{5}{100}+\frac{2}{100}+\frac{1}{100}\right)^{-1}=\left(\frac{8}{100}\right)^{-1}=\frac{100}{8}$ Now, we use Euclidean Algorithm, to find if this fraction is reducible, so $\gcd(100,8)=\gcd(12,8)=\gcd(4,8)=\gcd(4,4)$ Thus, both the numerator and denominator are divisible by $4,$ so $\frac{100}{8} \cdot \frac{4}{4}=\frac{100}{4} \cdot \frac{4}{8}=25 \cdot \frac{1}{2}=\boxed{\frac{25}{2}}$

- kante314

Video Solution (CREATIVE THINKING)

https://youtu.be/sbz01QUWY6A

~Education, the Study of Everything

Video Solution

https://youtu.be/QvkvhIMpXz8

~savannahsolver

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.

@@ Line 15: / Line 15: @@
 <cmath>\implies \frac{50}{4}</cmath>
 Finally, simplifying gives
-<cmath>\implies \boxed{\textbf{(B)}\ \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
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