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Difference between revisions of "2021 Fall AMC 10B Problems/Problem 9"

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~Steven Chen (www.professorchenedu.com)
 
~Steven Chen (www.professorchenedu.com)
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==Video Solution 1==
 
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~Education, the Study of Everything
 
  
 
==Video Solution by Interstigation==
 
==Video Solution by Interstigation==

Revision as of 17:53, 16 August 2022

Problem

The knights in a certain kingdom come in two colors. $\frac{2}{7}$ of them are red, and the rest are blue. Furthermore, $\frac{1}{6}$ of the knights are magical, and the fraction of red knights who are magical is $2$ times the fraction of blue knights who are magical. What fraction of red knights are magical?

$\textbf{(A) }\frac{2}{9}\qquad\textbf{(B) }\frac{3}{13}\qquad\textbf{(C) }\frac{7}{27}\qquad\textbf{(D) }\frac{2}{7}\qquad\textbf{(E) }\frac{1}{3}$

Solution 1

Let $k$ be the number of knights: then the number of red knights is $\frac{2}{7}k$ and the number of blue knights is $\frac{5}{7}k$.

Let $b$ be the fraction of blue knights that are magical - then $2b$ is the fraction of red knights that are magical. Thus we can write the equation $b \cdot \frac{5}{7}k + 2b \cdot \frac{2}{7}k = \frac{k}{6}\implies \frac{5}{7}b + \frac{4}{7}b = \frac{1}{6}$ $\implies \frac{9}{7}b = \frac{1}{6} \implies b=\frac{7}{54}$

We want to find the fraction of red knights that are magical, which is $2b = \frac{7}{27} = \boxed{C}$

~KingRavi

Solution 2

We denote by $p$ the fraction of red knights who are magical.

Hence, \[ \frac{1}{6} = \frac{2}{7} p + \left( 1 - \frac{2}{7} \right) \frac{p}{2} . \]

By solving this equation, we get $p = \frac{7}{27}$.

Therefore, the answer is $\boxed{\textbf{(C) }\frac{7}{27}}$.

~Steven Chen (www.professorchenedu.com)

Video Solution by Interstigation

https://youtu.be/p9_RH4s-kBA?t=1274

See Also

2021 Fall AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
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

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