Difference between revisions of "2020 AMC 10B Problems/Problem 3"
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{{duplicate|[[2020 AMC 10B Problems|2020 AMC 10B #3]] and [[2020 AMC 12B Problems|2020 AMC 12B #3]]}} | {{duplicate|[[2020 AMC 10B Problems|2020 AMC 10B #3]] and [[2020 AMC 12B Problems|2020 AMC 12B #3]]}} | ||
− | ==Problem | + | ==Problem== |
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</math> to <math>x</math> is <math>1:6</math>. What is the ratio of <math>w</math> to <math>y?</math> | 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</math> to <math>x</math> is <math>1:6</math>. What is the ratio of <math>w</math> to <math>y?</math> | ||
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<math>\textbf{(A)}\ 4:3 \qquad\textbf{(B)}\ 3:2 \qquad\textbf{(C)}\ 8:3 \qquad\textbf{(D)}\ 4:1 \qquad\textbf{(E)}\ 16:3</math> | <math>\textbf{(A)}\ 4:3 \qquad\textbf{(B)}\ 3:2 \qquad\textbf{(C)}\ 8:3 \qquad\textbf{(D)}\ 4:1 \qquad\textbf{(E)}\ 16:3</math> | ||
− | ==Solution 1== | + | ==Solution 1 (One Sentence)== |
+ | We have <cmath>\frac wy = \frac wx \cdot \frac xz \cdot \frac zy = \frac43\cdot\frac61\cdot\frac23=\frac{16}{3},</cmath> from which <math>w:y=\boxed{\textbf{(E)}\ 16:3}.</math> | ||
− | + | ~MRENTHUSIASM | |
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==Solution 2== | ==Solution 2== | ||
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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> ~quacker88 | + | 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>. |
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+ | ~quacker88 | ||
==Solution 3== | ==Solution 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 <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. | ||
Multiplying both sides by <math>4</math>, we have <math>16y = 4x</math>. | Multiplying both sides by <math>4</math>, we have <math>16y = 4x</math>. | ||
− | Substituting this into our first equation, we have <math>3w = 16y</math>, or <math>\frac{w}{y}=\frac{16}{3}</math>, so our answer is <math>\boxed{\textbf{(E)}\ 16:3}</math> | + | Substituting this into our first equation, we have <math>3w = 16y</math>, or <math>\frac{w}{y}=\frac{16}{3}</math>, so our answer is <math>\boxed{\textbf{(E)}\ 16:3}</math>. |
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~Binderclips1 | ~Binderclips1 | ||
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+ | ==Solution 4== | ||
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+ | WLOG, let <math>w=4</math> and <math>x=3</math>. | ||
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+ | Since the ratio of <math>z</math> to <math>x</math> is <math>1:6</math>, we can substitute in the value of <math>x</math> to get <math>\frac{z}{3}=\frac{1}{6} \implies z=\frac{1}{2}</math>. | ||
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+ | The ratio of <math>y</math> to <math>z</math> is <math>3:2</math>, so <math>\frac{y}{\frac{1}{2}}=\frac{3}{2} \implies y=\frac{3}{4}</math>. | ||
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+ | 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>. | ||
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+ | ~quacker88 | ||
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+ | ==Solution 5== | ||
+ | We have <math>\frac{w}{x}=\frac{4}{3}, \frac{y}{z}=\frac{3}{2}, \frac{z}{x}=\frac{1}{6}.</math> Find LCD: <math>\frac{w}{x}=\frac{8}{6}, \frac{z}{x}=\frac{1}{6}</math>, we have <math>\frac{w}{z}=\frac{8}{1}.</math> Similarly, <math>\frac{w}{z}=\frac{16}{2}, \frac{y}{z}=\frac{3}{2},</math> so <math>\frac{w}{y}=\frac{16}{3},</math> <math>\boxed{\textbf{(E)}\ 16:3}</math>. | ||
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+ | ~Yelechi | ||
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+ | ==Video Solution (HOW TO THINK CREATIVELY!!!)== | ||
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+ | https://www.youtube.com/watch?v=QqkNnsNEgiA (for AMC 10) | ||
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+ | https://youtu.be/Lz6XmHr8tJE (for AMC 12) | ||
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+ | ~Education, the Study of Everything | ||
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==Video Solution== | ==Video Solution== | ||
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https://youtu.be/Gkm5rU5MlOU (for AMC 10) | https://youtu.be/Gkm5rU5MlOU (for AMC 10) | ||
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https://youtu.be/WfTty8Fe5Fo (for AMC 12) | https://youtu.be/WfTty8Fe5Fo (for AMC 12) | ||
− | ~ | + | |
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+ | https://youtu.be/y4BbRapJepY | ||
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+ | ~savannahsolver | ||
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+ | https://youtu.be/wH7xhYxwaFc | ||
+ | |||
+ | ~AlexExplains | ||
==See Also== | ==See Also== |
Latest revision as of 15:38, 3 November 2024
- The following problem is from both the 2020 AMC 10B #3 and 2020 AMC 12B #3, so both problems redirect to this page.
Contents
Problem
The ratio of to is , the ratio of to is , and the ratio of to is . What is the ratio of to
Solution 1 (One Sentence)
We have from which
~MRENTHUSIASM
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
Solution 4
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 5
We have Find LCD: , we have Similarly, so .
~Yelechi
Video Solution (HOW TO THINK CREATIVELY!!!)
https://www.youtube.com/watch?v=QqkNnsNEgiA (for AMC 10)
https://youtu.be/Lz6XmHr8tJE (for AMC 12)
~Education, the Study of Everything
Video Solution
https://youtu.be/Gkm5rU5MlOU (for AMC 10)
https://youtu.be/WfTty8Fe5Fo (for AMC 12)
~savannahsolver
~AlexExplains
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.