2021 AMC Fall 10B Problem 9

Since the total is not specified, pick a convenient one: $42$ knights (a common multiple of $7$ and $6$). Then there are $\frac27 \cdot 42 = 12$ red knights, $30$ blue knights, and $\frac16 \cdot 42 = 7$ magical knights.

Let a fraction $x$ of the blue knights be magical; then a fraction $2x$ of the red knights are magical. Counting magical knights,
$$12 \cdot 2x + 30 \cdot x = 7 \quad\Longrightarrow\quad 54x = 7 \quad\Longrightarrow\quad x = \frac{7}{54}.$$
The requested fraction of red knights who are magical is $2x = \frac{7}{27}$.

The answer is $\boxed{\textbf{(C)}\ \frac{7}{27}}$.

"Fraction of a fraction" problems become counting problems once you fix a total that clears every denominator. The magical knights are then counted two ways (by color) and the single unknown rate drops out of a linear equation.

Keep the total as $1$: red $\frac27$, blue $\frac57$. Then $\frac27 (2x) + \frac57 x = \frac16$, so $\frac{9x}{7} = \frac16$ and $x = \frac{7}{54}$; the answer is $2x = \frac{7}{27}$.

• Answering the fraction of blue knights that are magical (7/54) or the fraction of magical knights that are red, instead of the fraction of red knights that are magical.
• Setting the number of magical red knights equal to twice the number of magical blue knights, rather than the fractions.