2020 AMC 10B Problem 3

A ratio only tells you how sizes compare, so you may pick an actual number for one letter and work out the rest.

1. The letter $x$ appears in two of the ratios, so choose a value for $x$ first. The ratios involve thirds ($4:3$) and sixths ($1:6$), so $x = 6$ keeps things simple.
2. "$w$ to $x$ is $4:3$" means $w$ is $\tfrac43$ of $x$. So $w = \tfrac43 \times 6 = 8$.
3. "$z$ to $x$ is $1:6$" means $z$ is $\tfrac16$ of $x$. So $z = 1$.
4. "$y$ to $z$ is $3:2$" means $y$ is $\tfrac32$ of $z$. So $y = \tfrac32$.
5. Now compare $w$ and $y$:
$$\frac{w}{y} = \frac{8}{3/2} = 8 \times \frac23 = \frac{16}{3}.$$
So $w : y = 16 : 3$.

The answer is $\boxed{\textbf{(E)}\ 16:3}$.

A ratio fixes how big things are compared with each other, not how big they actually are. So you are free to choose one number and let the others follow. Pick the letter that appears in the most ratios, and a value that avoids fractions where possible.

Multiply the fractions so that the letters in the middle cancel:
$$\frac{w}{y} = \frac{w}{x} \times \frac{x}{z} \times \frac{z}{y} = \frac43 \times \frac61 \times \frac23 = \frac{48}{9} = \frac{16}{3}.$$
Note that $\frac{x}{z} = 6$ and $\frac{z}{y} = \frac23$ are the given ratios turned upside down.

The variable $x$ appears in two ratios, so choose a value for it that keeps everything integral. Take $x = 6$.

From $w : x = 4 : 3$ we get $w = \tfrac{4}{3} \cdot 6 = 8$.

From $z : x = 1 : 6$ we get $z = 1$.

From $y : z = 3 : 2$ we get $y = \tfrac{3}{2} \cdot 1 = \tfrac{3}{2}$.

Therefore
$$\frac{w}{y} = \frac{8}{3/2} = \frac{16}{3},$$
so $w : y = 16 : 3$.

The answer is $\boxed{\textbf{(E)}\ 16:3}$.