Difference between revisions of "2020 AMC 10B Problems/Problem 3"
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<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>. | <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> | + | 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 |
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| − | ~quacker88 | ||
==See Also== | ==See Also== | ||
Revision as of 15:54, 7 February 2020
Contents
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
See Also
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
