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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]]}}
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==Problem 3==
 
==Problem 3==
  
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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> ~quacker88
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==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>.
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Clearing denominators, we have <math>3w = 4x</math>, <math>2y = 3z</math>, and <math>6z = x</math>.
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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>.
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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>
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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.
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Multiplying both sides by <math>4</math>, we have <math>16y = 4x</math>.
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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
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==Video Solution==
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https://youtu.be/Gkm5rU5MlOU (for AMC 10)
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https://youtu.be/WfTty8Fe5Fo (for AMC 12)
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~IceMatrix
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https://youtu.be/y4BbRapJepY
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~savannahsolver
  
 
==See Also==
 
==See Also==
  
 
{{AMC10 box|year=2020|ab=B|num-b=2|num-a=4}}
 
{{AMC10 box|year=2020|ab=B|num-b=2|num-a=4}}
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{{AMC12 box|year=2020|ab=B|num-b=2|num-a=4}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 17:01, 16 June 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 $w$ to $x$ is $4:3$, the ratio of $y$ to $z$ is $3:2$, and the ratio of $z$ to $x$ is $1:6$. What is the ratio of $w$ to $y?$

$\textbf{(A)}\ 4:3 \qquad\textbf{(B)}\ 3:2 \qquad\textbf{(C)}\ 8:3 \qquad\textbf{(D)}\ 4:1 \qquad\textbf{(E)}\ 16:3$

Solution 1

WLOG, let $w=4$ and $x=3$.

Since the ratio of $z$ to $x$ is $1:6$, we can substitute in the value of $x$ to get $\frac{z}{3}=\frac{1}{6} \implies z=\frac{1}{2}$.

The ratio of $y$ to $z$ is $3:2$, so $\frac{y}{\frac{1}{2}}=\frac{3}{2} \implies y=\frac{3}{4}$.

The ratio of $w$ to $y$ is then $\frac{4}{\frac{3}{4}}=\frac{16}{3}$ so our answer is $\boxed{\textbf{(E)}\ 16:3}$ ~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.

$z:x=1:6=2:12$, and since $y:z=3:2$, we can link them together to get $y:z:x=3:2:12$.

Finally, since $x:w=3:4=12:16$, we can link this again to get: $y:z:x:w=3:2:12:16$, so $w:y = \boxed{\textbf{(E)}\ 16:3}$ ~quacker88

Solution 3

We have the equations $\frac{w}{x}=\frac{4}{3}$, $\frac{y}{z}=\frac{3}{2}$, and $\frac{z}{x}=\frac{1}{6}$. Clearing denominators, we have $3w = 4x$, $2y = 3z$, and $6z = x$. Since we want $\frac{w}{y}$, we look to find $y$ in terms of $x$ since we know the relationship between $x$ and $w$. We begin by multiplying both sides of $2y = 3z$ by two, obtaining $4y = 6z$. We then substitute that into $6z = x$ to get $4y = x$ . Now, to be able to substitute this into out first equation, we need to have $4x$ on the RHS. Multiplying both sides by $4$, we have $16y = 4x$. Substituting this into our first equation, we have $3w = 16y$, or $\frac{w}{y}=\frac{16}{3}$, so our answer is $\boxed{\textbf{(E)}\ 16:3}$ ~Binderclips1

Video Solution

https://youtu.be/Gkm5rU5MlOU (for AMC 10) https://youtu.be/WfTty8Fe5Fo (for AMC 12)

~IceMatrix

https://youtu.be/y4BbRapJepY

~savannahsolver

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. AMC logo.png

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