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

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(Sol 1 already says something very similar to Sol 3, and Sol 1 used color.)
 
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<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>
  
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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>
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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>
  
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== Solution 1 ==
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== Solution 1 (Negatives) ==
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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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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
  
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== Solution 2 ==
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== Solution 2 (Answer Choices) ==
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Substituting values for <math>a, b,\text{ and } c</math>, we see that if each of them satisfy the inequalities above, the value goes to be <math>-1</math>.
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At <math>(a,b,c)=(4,5,6),</math> the answer choices become
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Therefore, the product of the given fraction equals <math>(-1)(-1)(-1)=\boxed{\textbf{(A) } -1}</math>.
 
  
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== 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>
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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: <cmath>\frac{\overset{-1}{\cancel{a-3}}}{\underset{1}{\xcancel{5-c}}} \cdot \frac{\overset{-1}{\bcancel{b-4}}}{\underset{1}{\cancel{3-a}}} \cdot \frac{\overset{-1}{\xcancel{c-5}}}{\underset{1}{\bcancel{4-b}}}=(-1)(-1)(-1)=\boxed{\textbf{(A) } -1}.</cmath>
 
  
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~CoolJupiter (Solution)
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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
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~MRENTHUSIASM (<math>\LaTeX</math> Adjustment)
 
  
 
== Video Solution 1 ==
 
== Video Solution 1 ==

Latest revision as of 23:56, 30 September 2023

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

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 (Problems • Answer Key • Resources)
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Problem 2 		Followed by
Problem 4
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All AMC 10 Problems and Solutions

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