Difference between revisions of "2020 AMC 10B Problems/Problem 3"
(Created page with "==Problem 3== The ratio of <math>w</math> to <math>x</math> is <math>4:3</math>, the ratio of <math>y</math> to <math>z</math> is <math>3:2</math>, and the ratio of <math>z</...") |
(→Solution 1) |
||
Line 14: | Line 14: | ||
The ratio of <math>w</math> to <math>y</math> is then <math>\frac{4}{\frac{3}{4}}=\frac{16}{3}</math> so our answer is <math>\boxed{\textbf{(E)}\ 16:3}</math> ~quacker88 | The ratio of <math>w</math> to <math>y</math> is then <math>\frac{4}{\frac{3}{4}}=\frac{16}{3}</math> so our answer is <math>\boxed{\textbf{(E)}\ 16:3}</math> ~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. | ||
+ | |||
+ | <math>z:x=1:6=2:12</math>, and since <math>y:z=3:2</math>, we can link them together to get <math>y:z:x=3:2:12</math>. | ||
+ | |||
+ | Finally, since <math>x:w=3:4=12:16</math>, we can link this again to get: <math>y:z:x:w=3:2:12:16</math>, so <math>w:y = \boxed{\textbf{(E)}\ 16:3}</math> ~quacker88 |
Revision as of 15:53, 7 February 2020
Problem 3
The ratio of to is , the ratio of to is , and the ratio of to is . What is the ratio of to ?
Solution 1
WLOG, let and .
Since the ratio of to is , we can substitute in the value of to get .
The ratio of to is , so .
The ratio of to is then so our answer is ~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.
, and since , we can link them together to get .
Finally, since , we can link this again to get: , so ~quacker88