Difference between revisions of "2020 AMC 10B Problems/Problem 3"

Revision as of 01:29, 8 February 2020

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

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.

Page

Toolbox

Search