2020 AMC 10B Problems/Problem 3

The following problem is from both the 2020 AMC 10B #3 and 2020 AMC 12B #3, so both problems redirect to this page.

Problem

The ratio of $w$ to $x$ is $4:3$, the ratio of $y$ to $z$ is $3:2$, and the ratio of $z$ to $x$ is $1:6$. What is the ratio of $w$ to $y?$

$\textbf{(A)}\ 4:3 \qquad\textbf{(B)}\ 3:2 \qquad\textbf{(C)}\  8:3 \qquad\textbf{(D)}\ 4:1 \qquad\textbf{(E)}\ 16:3$

Solution 1 (One Sentence)

We have \[\frac wy = \frac wx \cdot \frac xz \cdot \frac zy = \frac43\cdot\frac61\cdot\frac23=\frac{16}{3},\] from which $w:y=\boxed{\textbf{(E)}\ 16:3}.$

~MRENTHUSIASM

Solution 2

We need to somehow link all three of the ratios together. We can start by connecting the last two ratios together by multiplying the last ratio by two.

$z:x=1:6=2:12$, and since $y:z=3:2$, we can link them together to get $y:z:x=3:2:12$.

Finally, since $x:w=3:4=12:16$, we can link this again to get: $y:z:x:w=3:2:12:16$, so $w:y = \boxed{\textbf{(E)}\ 16:3}$.

~quacker88

Solution 3

We have the equations $\frac{w}{x}=\frac{4}{3}$, $\frac{y}{z}=\frac{3}{2}$, and $\frac{z}{x}=\frac{1}{6}$. Clearing denominators, we have $3w = 4x$, $2y = 3z$, and $6z = x$. Since we want $\frac{w}{y}$, we look to find $y$ in terms of $x$ since we know the relationship between $x$ and $w$. We begin by multiplying both sides of $2y = 3z$ by two, obtaining $4y = 6z$. We then substitute that into $6z = x$ to get $4y = x$ . Now, to be able to substitute this into out first equation, we need to have $4x$ on the RHS. Multiplying both sides by $4$, we have $16y = 4x$. Substituting this into our first equation, we have $3w = 16y$, or $\frac{w}{y}=\frac{16}{3}$, so our answer is $\boxed{\textbf{(E)}\ 16:3}$.

~Binderclips1

Solution 4

WLOG, let $w=4$ and $x=3$.

Since the ratio of $z$ to $x$ is $1:6$, we can substitute in the value of $x$ to get $\frac{z}{3}=\frac{1}{6} \implies z=\frac{1}{2}$.

The ratio of $y$ to $z$ is $3:2$, so $\frac{y}{\frac{1}{2}}=\frac{3}{2} \implies y=\frac{3}{4}$.

The ratio of $w$ to $y$ is then $\frac{4}{\frac{3}{4}}=\frac{16}{3}$ so our answer is $\boxed{\textbf{(E)}\ 16:3}$.

~quacker88

Solution 5

We have $\frac{w}{x}=\frac{4}{3}, \frac{y}{z}=\frac{3}{2}, \frac{z}{x}=\frac{1}{6}.$ Find LCD: $\frac{w}{x}=\frac{8}{6}, \frac{z}{x}=\frac{1}{6}$, we have $\frac{w}{z}=\frac{8}{1}.$ Similarly, $\frac{w}{z}=\frac{16}{2}, \frac{y}{z}=\frac{3}{2},$ so $\frac{w}{y}=\frac{16}{3},$ $\boxed{\textbf{(E)}\ 16:3}$.

~Yelechi

Video Solution (HOW TO THINK CREATIVELY!!!)

https://www.youtube.com/watch?v=QqkNnsNEgiA (for AMC 10)

https://youtu.be/Lz6XmHr8tJE (for AMC 12)

~Education, the Study of Everything


Video Solution

https://youtu.be/Gkm5rU5MlOU (for AMC 10)

https://youtu.be/WfTty8Fe5Fo (for AMC 12)


https://youtu.be/y4BbRapJepY

~savannahsolver

https://youtu.be/wH7xhYxwaFc

~AlexExplains

See Also

2020 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
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
2020 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
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 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png