Difference between revisions of "2020 AMC 10A Problems/Problem 3"
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| − | ==Problem== | + | == Problem == |
Assuming <math>a\neq3</math>, <math>b\neq4</math>, and <math>c\neq5</math>, what is the value in simplest form of the following expression? | Assuming <math>a\neq3</math>, <math>b\neq4</math>, and <math>c\neq5</math>, what is the value in simplest form of the following expression? | ||
<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> | ||
| + | |||
<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> | <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> | ||
| − | == Solution == | + | == Solutions == |
| − | + | === 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>. | ||
| − | ==Solution 2== | + | === Solution 2 === |
Substituting values for <cmath>a, b,\text{and} c</cmath>, we see that if each of them satify the inequalities above, the value goes to be <cmath>-1</cmath>. | Substituting values for <cmath>a, b,\text{and} c</cmath>, we see that if each of them satify the inequalities above, the value goes to be <cmath>-1</cmath>. | ||
Therefore, the product of the given fraction equals <math>(-1)(-1)(-1)=\boxed{\textbf{(A)}-1}</math>. | Therefore, the product of the given fraction equals <math>(-1)(-1)(-1)=\boxed{\textbf{(A)}-1}</math>. | ||
| − | ==Solution 3== | + | === Solution 3 === |
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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~CoolJupiter | ~CoolJupiter | ||
| − | ==Video Solution== | + | === Video Solution 1 === |
https://youtu.be/WUcbVNy2uv0 | https://youtu.be/WUcbVNy2uv0 | ||
~IceMatrix | ~IceMatrix | ||
| + | === Video Solution 2 === | ||
https://www.youtube.com/watch?v=7-3sl1pSojc | https://www.youtube.com/watch?v=7-3sl1pSojc | ||
~bobthefam | ~bobthefam | ||
| + | === Video Solution 3 === | ||
https://youtu.be/ZccL6yKrTiU | https://youtu.be/ZccL6yKrTiU | ||
~savannahsolver | ~savannahsolver | ||
| − | ==See Also== | + | == See Also == |
| − | |||
{{AMC10 box|year=2020|ab=A|num-b=2|num-a=4}} | {{AMC10 box|year=2020|ab=A|num-b=2|num-a=4}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 00:34, 19 October 2020
Contents
Problem
Assuming 
, 
, and 
, what is the value in simplest form of the following expression?
![\[\frac{a-3}{5-c} \cdot \frac{b-4}{3-a} \cdot \frac{c-5}{4-b}\]](http://latex.artofproblemsolving.com/9/2/6/926ff7e4cb0ce3392334f8215dd3aa4c51438f15.png)
Solutions
Solution 1
Note that 
is 
times 
. Likewise, 
is times 
and 
is times 
. Therefore, the product of the given fraction equals 
.
Solution 2
Substituting values for ![\[a, b,\text{and} c\]](http://latex.artofproblemsolving.com/9/d/e/9de61ae8b379a948a6b4159b00dc61ada7aa2b52.png)
, we see that if each of them satify the inequalities above, the value goes to be ![\[-1\]](http://latex.artofproblemsolving.com/f/d/7/fd7927352363a46a9414b895423eb31db1d10ae6.png)
.
Therefore, the product of the given fraction equals .
Solution 3
It is known that 
for 
. We use this fact to cancel out the terms.
~CoolJupiter
Video Solution 1
~IceMatrix
Video Solution 2
https://www.youtube.com/watch?v=7-3sl1pSojc
~bobthefam
Video Solution 3
~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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
