Teri finds a pile of money with at least $600. If she puts $200 of the pile in her left pocket, gives away 4/5 of the rest of the pile, and then puts the rest in her right pocket, she'll have more money than if she instead gave away $600 of the original pile and kept the rest. What are the possible values of the number of dollars in the original pile of money? (Give your answer in interval notation.)
Jason, what's your AoPS username? You'll be receiving a 3-day ban for attempting to cheat on your Algebra Homework.
You don't get 3-day bans on AoPS. I think you get a warning or your parents get an email. Plus, his name is probably not Jason.
Let's assume the original pile of money is represented by the variable "x".
According to the given information, Teri puts $200 in her left pocket. Therefore, the remaining amount she has is "x - 200".
Teri gives away 4/5 of the remaining amount, which means she gives away (4/5)(x - 200) = (4x - 800)/5.
After giving away this amount, Teri puts the remaining amount in her right pocket, so the amount in her right pocket is (x - 200) - (4x - 800)/5.
According to the next scenario, Teri gives away $600 from the original pile. Therefore, the remaining amount she has is x - 600.
We need to find the values of "x" for which the amount in Teri's right pocket in the first scenario is greater than the amount she has in the second scenario.
((x - 200) - (4x - 800)/5) > (x - 600)
To solve this inequality, we can multiply both sides by 5 to eliminate the fraction:
5(x - 200) - (4x - 800) > 5(x - 600)
5x - 1000 - 4x + 800 > 5x - 3000
x - 200 > 5x - 3000
-4x > -2800
Dividing both sides by -4 and reversing the inequality:
x < 700
Therefore, the possible values of the original pile of money "x" are less than 700.
In interval notation, the solution would be (−∞, 700).
To find the possible values of the number of dollars in the original pile of money, let's break down the problem.
Let's assume that the original pile of money has x dollars.
If Teri puts $200 in her left pocket, she will have (x - $200) dollars remaining.
Then, she gives away 4/5 of the remaining pile, which means she gives away (4/5)(x - $200) dollars. So, she will have (x - $200 - (4/5)(x - $200)) dollars left.
Finally, she puts the remaining money in her right pocket, which means she puts (1/5)(x - $200) dollars in her right pocket.
According to the problem, Teri would have more money if she took this approach, rather than giving away $600 and keeping the rest.
So, let's set up the inequality:
(x - $200 - (4/5)(x - $200)) > ($x - $600)
To simplify, we can distribute the (4/5):
(x - $200 - 4/5x + $160) > ($x - $600)
Now, let's combine like terms:
(1/5)x + $160 > ($x - $600)
To make the inequality easier to work with, let's get rid of the decimals and cents by multiplying both sides of the equation by 5:
x + 800 > 5x - 3000
Now, subtract x from both sides:
800 > 4x - 3000
Add 3000 to both sides:
3800 > 4x
Finally, divide both sides by 4 to solve for x:
x < 950
So, the possible values for the number of dollars in the original pile of money are less than 950.
In interval notation, this can be written as (-∞, 950).