2021 AMC Fall 10B Problem 9

The problem never says how many knights there are, so choose a number that makes every fraction come out whole. We need to take $\frac27$ and $\frac16$ of it, so use $42$ (it is $6 \times 7$).

1. Count the knights.
- Red: $\frac27 \times 42 = 12$.
- Blue: $42 - 12 = 30$.
- Magical: $\frac16 \times 42 = 7$.
2. Name the unknown. Say a fraction $x$ of the blue knights is magical. Then a fraction $2x$ of the red knights is magical, because the problem says the red fraction is twice the blue one.
3. Count the magical knights by color.
- Magical red knights: $12 \times 2x = 24x$.
- Magical blue knights: $30 \times x = 30x$.

Together they make all $7$ magical knights:
$$24x + 30x = 7, \qquad 54x = 7, \qquad x = \frac{7}{54}.$$
4. Answer the right question. The problem asks about the red knights, so the answer is $2x$, not $x$:
$$2x = \frac{14}{54} = \frac{7}{27}.$$

Check: $12 \times \frac{7}{27} = \frac{28}{9}$ red magical knights and $30 \times \frac{7}{54} = \frac{35}{9}$ blue ones, total $\frac{63}{9} = 7$. (The counts are not whole numbers with $42$ knights, but the fractions are what matter. With $378$ knights every count is whole.)

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

"Fraction of a group" problems get easier when you pick a total that clears the denominators, because then you can count real knights instead of juggling fractions of fractions. Counting the same group (the magical knights) in two ways, once by color and once all together, gives an equation with just one unknown.

Keep the total as $1$ whole kingdom instead. Red knights are $\frac27$ of it and blue knights are $\frac57$. Magical knights are $\frac16$:
$$\frac27 \times 2x + \frac57 \times x = \frac16, \qquad \frac{9x}{7} = \frac16, \qquad x = \frac{7}{54}.$$
The red fraction is $2x = \frac{7}{27}$.