Difference between revisions of "2020 AMC 10B Problems/Problem 3"
Binderclips1 (talk | contribs) m (fixed typo) |
Binderclips1 (talk | contribs) m (fixed typo) |
||
Line 29: | Line 29: | ||
We have the equations <math>\frac{w}{x}=\frac{4}{3}</math>, <math>\frac{y}{z}=\frac{3}{2}</math>, and <math>\frac{z}{x}=\frac{1}{6}</math>. | We have the equations <math>\frac{w}{x}=\frac{4}{3}</math>, <math>\frac{y}{z}=\frac{3}{2}</math>, and <math>\frac{z}{x}=\frac{1}{6}</math>. | ||
Clearing denominators, we have <math>3w = 4x</math>, <math>2y = 3z</math>, and <math>6z = x</math>. | Clearing denominators, we have <math>3w = 4x</math>, <math>2y = 3z</math>, and <math>6z = x</math>. | ||
− | Since we want <math>\frac{w}{y}</math>, we look to find <math>y</math> in terms of <math>x</math> since we know the relationship between <math>x</math> and <math> | + | Since we want <math>\frac{w}{y}</math>, we look to find <math>y</math> in terms of <math>x</math> since we know the relationship between <math>x</math> and <math>w</math>. |
We begin by multiplying both sides of <math>2y = 3z</math> by two, obtaining <math>4y = 6z</math>. We then substitute that into <math>6z = x</math> | We begin by multiplying both sides of <math>2y = 3z</math> by two, obtaining <math>4y = 6z</math>. We then substitute that into <math>6z = x</math> | ||
to get <math>4y = x</math> . Now, to be able to substitute this into out first equation, we need to have <math>4x</math> on the RHS. | to get <math>4y = x</math> . Now, to be able to substitute this into out first equation, we need to have <math>4x</math> on the RHS. |
Revision as of 19:54, 28 May 2020
- 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 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
Solution 3
We have the equations ,
, and
.
Clearing denominators, we have
,
, and
.
Since we want
, we look to find
in terms of
since we know the relationship between
and
.
We begin by multiplying both sides of
by two, obtaining
. We then substitute that into
to get
. Now, to be able to substitute this into out first equation, we need to have
on the RHS.
Multiplying both sides by
, we have
.
Substituting this into our first equation, we have
, or
, so our answer is
~Binderclips1
Video Solution
https://youtu.be/Gkm5rU5MlOU (for AMC 10) https://youtu.be/WfTty8Fe5Fo (for AMC 12)
~IceMatrix
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 |
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.