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

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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.
 
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.
  
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<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>
+
<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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~CoolJupiter
  
 
==Video Solution==
 
==Video Solution==

Revision as of 08:07, 1 September 2020

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

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 $(-1)(-1)(-1)=\boxed{\textbf{(A)}-1}$.

Solution 2

Substituting values for \[a, b,\text{and} c\], we see that if each of them satify the inequalities above, the value goes to be \[-1\]. Therefore, the product of the given fraction equals $(-1)(-1)(-1)=\boxed{\textbf{(A)}-1}$.

Solution 3

It is known that $\frac{x-y}{y-x}=-1$ for $x\ne y$. We use this fact to cancel out the terms.

$\frac{\cancel{(a-3)} -1 \cancel{(b-4)} -1 \cancel{(c-5)} -1}{\cancel{(5-c)}\cancel{(3-a)}\cancel{(4-b)}}=(-1)(-1)(-1)=\boxed{\textbf{(A)}~-1}$

~CoolJupiter

Video Solution

https://youtu.be/WUcbVNy2uv0

~IceMatrix

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

~bobthefam

https://youtu.be/ZccL6yKrTiU

~savannahsolver

See Also

2020 AMC 10A (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

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