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Difference between revisions of "2020 AMC 10A Problems/Problem 3"
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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{\text{(A) }-1}</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{\text{(A) }-1}</math>.
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 +
==Video Solution==
 +
https://youtu.be/WUcbVNy2uv0
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 +
~IceMatrix
  
 
==See Also==
 
==See Also==

Revision as of 22:08, 31 January 2020

Problem 3

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

Note that $a-3$ is $-1$ times $3-a$. Likewise, $b-4$ is $-1$ times $4-b$ and $c-5$ is $-1$ times $5-c$. Therefore, the product of the given fraction equals $(-1)(-1)(-1)=\boxed{\text{(A) }-1}$.

Video Solution

https://youtu.be/WUcbVNy2uv0

~IceMatrix

See Also

2020 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
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All AMC 10 Problems and Solutions

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