Difference between revisions of "2000 AMC 12 Problems/Problem 11"
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==Solution 3== | ==Solution 3== | ||
| − | Just realize that two such numbers are <math>a = 1</math> and <math>b = \frac{1}{2}</math>. With this, you can solve and get <math>2 \Rightarrow\boxed{\text{E}}</math> | + | Just realize that two such numbers are <math>a = 1</math> and <math>b = \frac{1}{2}</math>. You can see this by plugging in <math>a = 1</math> and then solving for b. With this, you can solve and get <math>2 \Rightarrow\boxed{\text{E}}</math> |
| − | ==Solution 4 | + | ==Solution 4 == |
Set <math>a</math> to some nonzero number. In this case, I'll set it to <math>4</math>. | Set <math>a</math> to some nonzero number. In this case, I'll set it to <math>4</math>. | ||
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~hastapasta | ~hastapasta | ||
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| + | ==Solution 5== | ||
| + | Notice that <math>a=\frac{a}{b}-1</math> and <math>b=1-\frac{b}{a}</math>. Then, <math>\frac{a}{b}=1+a</math> and <math>\frac{b}{a}=1-b</math>. | ||
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| + | <math>\frac{a}{b}+\frac{b}{a}-ab=(1+a)+(1-b)-(a-b)=2</math>. The answer is <math>\boxed{E}</math>. | ||
== Video Solution == | == Video Solution == | ||
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https://www.youtube.com/watch?v=8nxvuv5oZ7A&t=3s | https://www.youtube.com/watch?v=8nxvuv5oZ7A&t=3s | ||
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| + | ==Video Solution by Daily Dose of Math== | ||
| + | |||
| + | https://youtu.be/Q_th4G-xGLo?si=4VwtJirZjREyyQuO | ||
| + | |||
| + | ~Thesmartgreekmathdude | ||
| + | |||
==See also== | ==See also== | ||
Latest revision as of 23:54, 14 July 2024
- The following problem is from both the 2000 AMC 12 #11 and 2000 AMC 10 #15, so both problems redirect to this page.
Contents
Problem
Two non-zero real numbers, 
and 
satisfy 
. Which of the following is a possible value of 
?
Solution 1
.
Another way is to solve the equation for giving 
then substituting this into the expression and simplifying gives the answer of 
Solution 2
This simplifies to 
. The two integer solutions to this are 
and 
. The problem states than and 
are non-zero, so we consider the case of . So, we end up with 
Solution 3
Just realize that two such numbers are 
and 
. You can see this by plugging in and then solving for b. With this, you can solve and get 
Solution 4
Set to some nonzero number. In this case, I'll set it to 
.
Then solve for . In this case, 
.
Now just simply evaluate. In this case it's 2. So since 2 is a possible value of the original expression, select 
.
~hastapasta
Solution 5
Notice that 
and 
. Then, 
and 
.
. The answer is .
Video Solution
https://www.youtube.com/watch?v=7-RloNHTnXM
Video Solution
https://youtu.be/ZWqHxc0i7ro?t=6
~ pi_is_3.14
Video Solution
https://www.youtube.com/watch?v=8nxvuv5oZ7A&t=3s
Video Solution by Daily Dose of Math
https://youtu.be/Q_th4G-xGLo?si=4VwtJirZjREyyQuO
~Thesmartgreekmathdude
See also
| 2000 AMC 12 (Problems • Answer Key • Resources) | |
| Preceded by Problem 10 |
Followed by Problem 12 |
| 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 12 Problems and Solutions | |
| 2000 AMC 10 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 14 |
Followed by Problem 16 | |
| 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. 
