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

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<math>\implies \frac{9}{7}b = \frac{1}{6} \implies b=\frac{7}{54}</math>
 
<math>\implies \frac{9}{7}b = \frac{1}{6} \implies b=\frac{7}{54}</math>
  
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We want to find the fraction of red knights that are magical, which is <math>2b = \frac{7}{27} = \boxed{C}</math>
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We want to find the fraction of red knights that are magical, which is <math>2b = \frac{7}{27} = \boxed{\textbf{(C) }\frac{7}{27}}</math>
  
 
~KingRavi
 
~KingRavi
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~Steven Chen (www.professorchenedu.com)
 
~Steven Chen (www.professorchenedu.com)
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== Solution 3 ==
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Suppose there are <math>42</math> knights. Then there are <math>\frac{2}{7}\cdot 42=12</math> red knights and <math>42-12=30</math> blue knights.
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There are <math>\frac{1}{6}\cdot 42=7</math> magical knights. Let <math>x</math> be the number of red magical knights.
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Then we have the equation <math>\frac{x}{12}=2\cdot \frac{7-x}{30}=\frac{7-x}{15}</math>. Solving, we get <math>x=84/27</math>.
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Plugging <math>x</math> back gives us the fraction of red magical knights, which is <math>\frac{\frac{84}{27}}{12}=\frac{84}{27}\cdot \frac{1}{12}=\boxed{\textbf{(C) }\frac{7}{27}}</math>.
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 +
~DairyQueenXD
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 +
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== Solution 4 ==
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Similar to Solution 3, let the total number of knights be <math>42</math> (We can say this because the fractions we know have denominators of <math>7</math> and <math>6</math> which have a LCM of <math>42</math>)
 +
 +
This means that there are <math>\dfrac{2}{7} \times 42 = 12</math> red knights meaning there are <math>30</math> blue knights
 +
 +
This also means that there are <math>\dfrac{1}{6} \times 42 = 7</math> magical knights. This is important information for our next step.
 +
 +
We are given that the fraction of magical red knights is twice the fraction of magical blue knights. As this problem is dealing with the fraction being twice are large and is asking for the fraction, let's just deal with the variable being a fraction. Let <math>\dfrac{p}{q}</math> be the fraction of red magical knights. This means that <math>\dfrac{p}{2q}</math> is the fraction of blue magical knights.
 +
 +
We know that total number of magical knights from what we figured out before is <math>7</math>, so let's use the fraction we figured and make the equation:
 +
 +
<math>(12 \times \dfrac{p}{q}) + (30 \times \dfrac{p}{2q})= 7</math>
 +
 +
Solving this gives us <math>\dfrac{p}{q}=\dfrac{7}{27}</math> or the fraction of red magical knights.
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Hence, the answer is <math>\boxed{\dfrac{7}{27}}</math>
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~ Batmanstark
  
 
==Video Solution by Interstigation==
 
==Video Solution by Interstigation==
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~Education, the Study of Everything
 
~Education, the Study of Everything
  
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==Video Solution by WhyMath==
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https://youtu.be/Ia0Jbe1h_qY
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 +
~savannahsolver
 +
==Video Solution by TheBeautyofMath==
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https://youtu.be/RyN-fKNtd3A?t=1190
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 +
~IceMatrix
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2021 Fall|ab=B|num-a=10|num-b=8}}
 
{{AMC10 box|year=2021 Fall|ab=B|num-a=10|num-b=8}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 00:00, 30 December 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 $2b = \frac{7}{27} = \boxed{\textbf{(C) }\frac{7}{27}}$

~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)

Solution 3

Suppose there are $42$ knights. Then there are $\frac{2}{7}\cdot 42=12$ red knights and $42-12=30$ blue knights.

There are $\frac{1}{6}\cdot 42=7$ magical knights. Let $x$ be the number of red magical knights.

Then we have the equation $\frac{x}{12}=2\cdot \frac{7-x}{30}=\frac{7-x}{15}$. Solving, we get $x=84/27$.

Plugging $x$ back gives us the fraction of red magical knights, which is $\frac{\frac{84}{27}}{12}=\frac{84}{27}\cdot \frac{1}{12}=\boxed{\textbf{(C) }\frac{7}{27}}$.

~DairyQueenXD


Solution 4

Similar to Solution 3, let the total number of knights be $42$ (We can say this because the fractions we know have denominators of $7$ and $6$ which have a LCM of $42$)

This means that there are $\dfrac{2}{7} \times 42 = 12$ red knights meaning there are $30$ blue knights

This also means that there are $\dfrac{1}{6} \times 42 = 7$ magical knights. This is important information for our next step.

We are given that the fraction of magical red knights is twice the fraction of magical blue knights. As this problem is dealing with the fraction being twice are large and is asking for the fraction, let's just deal with the variable being a fraction. Let $\dfrac{p}{q}$ be the fraction of red magical knights. This means that $\dfrac{p}{2q}$ is the fraction of blue magical knights.

We know that total number of magical knights from what we figured out before is $7$, so let's use the fraction we figured and make the equation:

$(12 \times \dfrac{p}{q}) + (30 \times \dfrac{p}{2q})= 7$

Solving this gives us $\dfrac{p}{q}=\dfrac{7}{27}$ or the fraction of red magical knights.

Hence, the answer is $\boxed{\dfrac{7}{27}}$

~ Batmanstark

Video Solution by Interstigation

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

Video Solution

https://youtu.be/LJ9mxTNSRUU

~Education, the Study of Everything

Video Solution by WhyMath

https://youtu.be/Ia0Jbe1h_qY

~savannahsolver

Video Solution by TheBeautyofMath

https://youtu.be/RyN-fKNtd3A?t=1190

~IceMatrix

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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