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

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== Solutions ==  
 
== Solutions ==  
=== Solution 1 ===
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== Solution 1 ==
 
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>.
 
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>.
  

Revision as of 14:25, 22 December 2020

Problem

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

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

Solutions

Solution 1

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 1

https://youtu.be/WUcbVNy2uv0

~IceMatrix

Video Solution

https://youtu.be/Nrdxe4UAqkA

Education, The Study of Everything

Video Solution 2

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

~bobthefam

Video Solution 3

https://youtu.be/ZccL6yKrTiU

~savannahsolver

See Also

2020 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
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All AMC 10 Problems and Solutions

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