# 2014 AMC 10A Problems/Problem 1

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## 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

~savannahsolver

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