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Difference between revisions of "2000 AMC 12 Problems/Problem 11"

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==Video Solution==
 
==Video Solution==

Revision as of 16:28, 1 February 2021

The following problem is from both the 2000 AMC 12 #11 and 2000 AMC 10 #15, so both problems redirect to this page.

Problem

Two non-zero real numbers, $a$ and $b,$ satisfy $ab = a - b$. Which of the following is a possible value of $\frac {a}{b} + \frac {b}{a} - ab$?

$\text{(A)} \ - 2 \qquad \text{(B)} \ \frac { - 1}{2} \qquad \text{(C)} \ \frac {1}{3} \qquad \text{(D)} \ \frac {1}{2} \qquad \text{(E)} \ 2$

Solution 1

$\frac {a}{b} + \frac {b}{a} - ab = \frac{a^2 + b^2}{ab} - (a - b) = \frac{a^2 + b^2}{a-b} - \frac{(a-b)^2}{(a-b)} = \frac{2ab}{a-b} = \frac{2(a-b)}{a-b} =2 \Rightarrow \boxed{\text{E}}$.

Another way is to solve the equation for $b,$ giving $b = \frac{a}{a+1};$ then substituting this into the expression and simplifying gives the answer of $2.$

Solution 2

This simplifies to $ab+b-a=0 \Rightarrow (a+1)(b-1) = -1$. The two integer solutions to this are $(-2,2)$ and $(0,0)$. The problem states than $a$ and $b$ are non-zero, so we consider the case of $(-2,2)$. So, we end up with $\frac{-2}{2} + \frac{2}{-2} - 2 \cdot -2 = 4 - 1 - 1 = 2 \Rightarrow \boxed{\text{E}}$

Video Solution

https://youtu.be/ZWqHxc0i7ro?t=6

~ pi_is_3.1415

Video Solution

https://www.youtube.com/watch?v=8nxvuv5oZ7A&t=3s

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. AMC logo.png

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