2021 AMC 10 B Fall

Complete problem set with solutions and individual problem pages

Problem 9 Easy

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?(2021 AMC Fall 10B, Question #9)

  • A.

    \frac{2}{9}

  • B.

    \frac{3}{13}

  • C.

    \frac{7}{27}

  • D.

    \frac{2}{7}

  • E.

    \frac{1}{3}

Answer:C

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 2 b is the fraction of red knights that are magical. Thus we can write the equation b \cdot \frac{5}{7} k+2 b \cdot \frac{2}{7} k=\frac{k}{6} \Longrightarrow \frac{5}{7} b+\frac{4}{7} b=\frac{1}{6} \Longrightarrow \frac{9}{7} b=\frac{1}{6} \Longrightarrow b=\frac{7}{54} We want to find the fraction of red knights that are magical, which is 2 b=\frac{7}{27}=C

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 (C) \frac{7}{27}.

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