Difference between revisions of "2020 AMC 10A Problems/Problem 3"

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==Problem==
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== Problem ==
 
Assuming <math>a\neq3</math>, <math>b\neq4</math>, and <math>c\neq5</math>, what is the value in simplest form of the following expression?
 
Assuming <math>a\neq3</math>, <math>b\neq4</math>, and <math>c\neq5</math>, what is the value in simplest form of the following expression?
 
<cmath>\frac{a-3}{5-c} \cdot \frac{b-4}{3-a} \cdot \frac{c-5}{4-b}</cmath>
 
<cmath>\frac{a-3}{5-c} \cdot \frac{b-4}{3-a} \cdot \frac{c-5}{4-b}</cmath>
<math>\textbf{(A) } -1 \qquad \textbf{(B) } 1 \qquad \textbf{(C) } \frac{abc}{60} \qquad \textbf{(D) } \frac{1}{abc} - \frac{1}{60} \qquad \textbf{(E) } \frac{1}{60} - \frac{1}{abc}</math>
 
  
== Solution ==
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<math>\textbf{(A) } {-}1 \qquad \textbf{(B) } 1 \qquad \textbf{(C) } \frac{abc}{60} \qquad \textbf{(D) } \frac{1}{abc} - \frac{1}{60} \qquad \textbf{(E) } \frac{1}{60} - \frac{1}{abc}</math>
  
Note that <math>a-3</math> is <math>-1</math> times <math>3-a</math>. Likewise, <math>b-4</math> is <math>-1</math> times <math>4-b</math> and <math>c-5</math> is <math>-1</math> times <math>5-c</math>. Therefore, the product of the given fraction equals <math>(-1)(-1)(-1)=\boxed{\textbf{(A)}-1}</math>.
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== Solution 1 (Negatives) ==
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If <math>x\neq y,</math> then <math>\frac{x-y}{y-x}=-1.</math> We use this fact to simplify the original expression:
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<cmath>\frac{\color{red}\overset{-1}{\cancel{a-3}}}{\color{blue}\underset{1}{\cancel{5-c}}} \cdot \frac{\color{green}\overset{-1}{\cancel{b-4}}}{\color{red}\underset{1}{\cancel{3-a}}} \cdot \frac{\color{blue}\overset{-1}{\cancel{c-5}}}{\color{green}\underset{1}{\cancel{4-b}}}=(-1)(-1)(-1)=\boxed{\textbf{(A) } {-}1}.</cmath>
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~CoolJupiter ~MRENTHUSIASM
  
==Solution 2==
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== Solution 2 (Answer Choices) ==
Substituting values for <cmath>a, b,\text{and} c</cmath>, we see that if each of them satify the inequalities above, the value goes to be <cmath>-1</cmath>.
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At <math>(a,b,c)=(4,5,6),</math> the answer choices become
Therefore, the product of the given fraction equals <math>(-1)(-1)(-1)=\boxed{\textbf{(A)}-1}</math>.
 
  
==Solution 3==
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<math>\textbf{(A) } {-}1 \qquad \textbf{(B) } 1 \qquad \textbf{(C) } 2 \qquad \textbf{(D) } {-}\frac{1}{120} \qquad \textbf{(E) } \frac{1}{120}</math>
It is known that <math>\frac{x-y}{y-x}=-1</math> for <math>x\ne y</math>. We use this fact to cancel out the terms.
 
  
<math>\frac{\cancel{(a-3)} -1 \cancel{(b-4)} -1 \cancel{(c-5)} -1}{\cancel{(5-c)}\cancel{(3-a)}\cancel{(4-b)}}=(-1)(-1)(-1)=\boxed{\textbf{(A)}~-1}</math>  
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and the original expression becomes <cmath>\frac{-1}{1}\cdot\frac{-1}{1}\cdot\frac{-1}{1}=\boxed{\textbf{(A) } {-}1}.</cmath>
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~MRENTHUSIASM
  
~CoolJupiter
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== Solution 3 (Fastest) ==
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We can simply set <math>x = a - 3, y = b - 4,</math> and <math>z = 5 - c</math>. Now, the problem simplifies to <cmath>\frac{x}{z}\cdot\frac{-y}{x}\cdot\frac{z}{y}=\frac{-xyz}{xyz}=\boxed{\textbf{(A) } {-}1}.</cmath>
  
==Video Solution==
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Explanation: After substituting <math>x</math>, <math>y</math>, and <math>z</math>, the opposites (for example <math>5 - c</math> and <math>c - 5</math>) can just be written as the negative of each. With the same example, this can be shown by: <math>c - 5 = -(5 - c)</math>.
 +
 
 +
~GREATEST
 +
 
 +
== Video Solution 1 ==
 
https://youtu.be/WUcbVNy2uv0
 
https://youtu.be/WUcbVNy2uv0
  
 
~IceMatrix
 
~IceMatrix
  
 +
== Video Solution 2 ==
 +
 +
https://youtu.be/Nrdxe4UAqkA
 +
 +
Education, The Study of Everything
 +
 +
== Video Solution 3 ==
 
https://www.youtube.com/watch?v=7-3sl1pSojc
 
https://www.youtube.com/watch?v=7-3sl1pSojc
  
 
~bobthefam
 
~bobthefam
  
 +
== Video Solution 4 ==
 
https://youtu.be/ZccL6yKrTiU
 
https://youtu.be/ZccL6yKrTiU
  
 
~savannahsolver
 
~savannahsolver
  
==See Also==
+
== Video Solution 5==
 +
https://youtu.be/ba6w1OhXqOQ?t=956
 +
 
 +
~ pi_is_3.14
  
 +
== See Also ==
 
{{AMC10 box|year=2020|ab=A|num-b=2|num-a=4}}
 
{{AMC10 box|year=2020|ab=A|num-b=2|num-a=4}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 01:04, 11 November 2024

Problem

Assuming $a\neq3$, $b\neq4$, and $c\neq5$, what is the value in simplest form of the following expression? \[\frac{a-3}{5-c} \cdot \frac{b-4}{3-a} \cdot \frac{c-5}{4-b}\]

$\textbf{(A) } {-}1 \qquad \textbf{(B) } 1 \qquad \textbf{(C) } \frac{abc}{60} \qquad \textbf{(D) } \frac{1}{abc} - \frac{1}{60} \qquad \textbf{(E) } \frac{1}{60} - \frac{1}{abc}$

Solution 1 (Negatives)

If $x\neq y,$ then $\frac{x-y}{y-x}=-1.$ We use this fact to simplify the original expression: \[\frac{\color{red}\overset{-1}{\cancel{a-3}}}{\color{blue}\underset{1}{\cancel{5-c}}} \cdot \frac{\color{green}\overset{-1}{\cancel{b-4}}}{\color{red}\underset{1}{\cancel{3-a}}} \cdot \frac{\color{blue}\overset{-1}{\cancel{c-5}}}{\color{green}\underset{1}{\cancel{4-b}}}=(-1)(-1)(-1)=\boxed{\textbf{(A) } {-}1}.\] ~CoolJupiter ~MRENTHUSIASM

Solution 2 (Answer Choices)

At $(a,b,c)=(4,5,6),$ the answer choices become

$\textbf{(A) } {-}1 \qquad \textbf{(B) } 1 \qquad \textbf{(C) } 2 \qquad \textbf{(D) } {-}\frac{1}{120} \qquad \textbf{(E) } \frac{1}{120}$

and the original expression becomes \[\frac{-1}{1}\cdot\frac{-1}{1}\cdot\frac{-1}{1}=\boxed{\textbf{(A) } {-}1}.\] ~MRENTHUSIASM

Solution 3 (Fastest)

We can simply set $x = a - 3, y = b - 4,$ and $z = 5 - c$. Now, the problem simplifies to \[\frac{x}{z}\cdot\frac{-y}{x}\cdot\frac{z}{y}=\frac{-xyz}{xyz}=\boxed{\textbf{(A) } {-}1}.\]

Explanation: After substituting $x$, $y$, and $z$, the opposites (for example $5 - c$ and $c - 5$) can just be written as the negative of each. With the same example, this can be shown by: $c - 5 = -(5 - c)$.

~GREATEST

Video Solution 1

https://youtu.be/WUcbVNy2uv0

~IceMatrix

Video Solution 2

https://youtu.be/Nrdxe4UAqkA

Education, The Study of Everything

Video Solution 3

https://www.youtube.com/watch?v=7-3sl1pSojc

~bobthefam

Video Solution 4

https://youtu.be/ZccL6yKrTiU

~savannahsolver

Video Solution 5

https://youtu.be/ba6w1OhXqOQ?t=956

~ pi_is_3.14

See Also

2020 AMC 10A (ProblemsAnswer KeyResources)
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Problem 2
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Problem 4
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