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

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{{duplicate|[[2014 AMC 12A Problems|2014 AMC 12A #1]] and [[2014 AMC 10A Problems|2014 AMC 10A #1]]}}
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==Problem ==
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What is <math> 10\cdot\left(\tfrac{1}{2}+\tfrac{1}{5}+\tfrac{1}{10}\right)^{-1}? </math>
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<math>\textbf{(A)}\ 3 \qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ \frac{25}{2} \qquad\textbf{(D)}\ \frac{170}{3}\qquad\textbf{(E)}\ 170</math>
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== Solution ==
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We have <cmath>10\cdot\left(\frac{1}{2}+\frac{1}{5}+\frac{1}{10}\right)^{-1}</cmath>
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Making the denominators equal gives
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<cmath>\implies 10\cdot\left(\frac{5}{10}+\frac{2}{10}+\frac{1}{10}\right)^{-1}</cmath>
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<cmath>\implies 10\cdot\left(\frac{5+2+1}{10}\right)^{-1}</cmath>
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<cmath>\implies 10\cdot\left(\frac{8}{10}\right)^{-1}</cmath>
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<cmath>\implies 10\cdot\left(\frac{4}{5}\right)^{-1}</cmath>
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<cmath>\implies 10\cdot\frac{5}{4}</cmath>
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<cmath>\implies \frac{50}{4}</cmath>
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Finally, simplifying gives
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<cmath>\implies \boxed{\textbf{(C)}\ \frac{25}{2}}</cmath>
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== Solution 2 ==
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We have
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<cmath>\left(\frac{1}{10}\right)^{-1}\cdot \left(\frac{1}{2} + \frac{1}{5} + \frac{1}{10}\right)^{-1}</cmath>By Distributive Property,
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<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
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<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>
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<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
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<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
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<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>
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- kante314
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 +
==Video Solution (CREATIVE THINKING)==
 +
https://youtu.be/sbz01QUWY6A
 +
 
 +
~Education, the Study of Everything
  
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==Problem ==
 
  
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What is <math>10 \cdot \left(\tfrac{1}{2} + \tfrac{1}{5} + \tfrac{1}{10}\right)^{-1}?</math>
 
  
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<math> \textbf{(A)}\ 3\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ \frac{25}{2}\qquad\textbf{(D)}}\ \frac{170}{3}\qquad\textbf{(E)}\ 170</math>
 
  
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== Solution ==
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==Video Solution==
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Sum the fractions over the common denominator: <math>\dfrac{1}{2}+\dfrac15+\dfrac1{10}=\dfrac{5+2+1}{10}=\dfrac45</math>
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https://youtu.be/QvkvhIMpXz8
  
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Now the answer is just some arithmetic: <math>10\cdot\left(\dfrac45\right)^{-1}=10\cdot\dfrac{5}{4}=\boxed{\textbf{(C)}\ \dfrac{25}2}</math>
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~savannahsolver
  
 
==See Also==
 
==See Also==
  
 
{{AMC10 box|year=2014|ab=A|before=First Problem|num-a=2}}
 
{{AMC10 box|year=2014|ab=A|before=First Problem|num-a=2}}
 +
{{AMC12 box|year=2014|ab=A|before=First Problem|num-a=2}}
 
{{MAA Notice}}
 
{{MAA Notice}}
 +
 +
[[Category: Prealgebra Problems]]

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

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

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