2020 AMC 10A Problems/Problem 3

Revision as of 21:17, 17 April 2021 by MRENTHUSIASM (talk | contribs) (Solution 3)

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

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 satisfy 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: \[\frac{\overset{-1}{\cancel{a-3}}}{\underset{1}{\xcancel{5-c}}} \cdot \frac{\overset{-1}{\bcancel{b-4}}}{\underset{1}{\cancel{3-a}}} \cdot \frac{\overset{-1}{\xcancel{c-5}}}{\underset{1}{\bcancel{4-b}}}=(-1)(-1)(-1)=\boxed{\textbf{(A) } -1}.\]

~CoolJupiter (Solution)

~MRENTHUSIASM ($\LaTeX$ Adjustment)

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)
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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