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