Difference between revisions of "2020 AMC 10B Problems/Problem 3"
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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 x since we know the relationship between <math>x</math> and <math>y</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>y</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 19:52, 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.