Question 1209106
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x = value, in dollars, of the original pile of money


x >= 200 since there's at least $200 in the pile.
>= means "greater than or equal to"


Scenario A) She pockets $50, gives 9/10 of the remaining pile away and keeps 1/10 of it.


Scenario B) Gives away $200 and keeps the rest




Let's trace out the path of scenario A.
x = starting amount of money
50 = amount she puts in her left pocket
x-50 = remaining amount after putting the $50 away
0.9*(x-50) = amount given away
0.1*(x-50) = amount she puts in her right pocket
0.1*(x-50)+50 = total amount in both pockets
We'll use this expression later.



Now consider scenario B
x = starting amount of money
x-200 = total amount in Jeri's pocket after giving away $200


Let's see when the outcome of scenario A is larger than the outcome of B.
A > B
0.1*(x-50)+50 > x-200 
0.1*(x-50) > x-200-50
0.1*(x-50) > x-250
0.1x-5 > x-250
-5+250 > x-0.1x
245 > 0.9x
0.9x < 245
x < 245/0.9
x < 272.22222... where the 2's go on forever
x < 272.22 approximately when rounding to the nearest penny


Let's see what happens when x = 272.22
At the end of scenario A, Jeri will have 0.1*(x-50)+50 = 0.1*(272.22-50)+50 = 72.22 dollars
At the end of scenario B, she will have x-200 = 272.22-200 = 72.22 dollars.
She ends up with the same amount of money in either in scenario.
So this will show that we do not include 272.22 as part of the interval.


The next highest amount would be 272.21
which means x < 272.22 is equivalent to x <= 272.21 when considering the penny as the smallest unit of currency.


We start with the condition that x >= 200 aka 200 <= x, and then determined x <= 272.21
Those glue together to get the compound inequality
<font color=red>200 <= x <= 272.21</font>


The starting amount of money in the pile is between $200 and $272.21 where both endpoints are included.
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