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 3

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

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 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 $y$ . 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

Video Solution

https://youtu.be/Gkm5rU5MlOU (for AMC 10) https://youtu.be/WfTty8Fe5Fo (for AMC 12)

~IceMatrix

2020 AMC 10B (Problems • Answer Key • Resources)
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 (Problems • Answer Key • Resources)
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.

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