Jeri finds a pile of money with at least 200. If she puts $50$ of the pile in her left pocket, gives away $\frac{1}{8}$ 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 $200$ 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 as an interval.)

Question

Jeri finds a pile of money with at least 200.  If she puts $50$ of the pile in her left pocket, gives away $\frac{1}{8}$ 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 $200$ 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 as an interval.)

GPT-4o mini · Answer

Let $ x $ be the amount of money in the original pile. First, we examine the case where Jeri puts $50 in her left pocket and gives away $ \frac{1}{8} $ of the rest of the pile. 1. After putting $50 in her left pocket, she has $ x - 50 $ left. 2. She gives away $ \frac{1}{8} $ of this remaining amount, which is $$ \frac{1}{8}(x - 50). $$ 3. The amount she gives away results in her keeping $$ x - 50 - \frac{1}{8}(x - 50) = x - 50 - \frac{x - 50}{8} = x - 50 - \frac{x}{8} + \frac{50}{8}. $$ Simplifying this, we have: $$ x - 50 - \frac{x}{8} + \frac{50}{8} = \frac{8x - 400 - x + 50}{8} = \frac{7x - 350}{8}. $$ Thus, after putting $50 in her left pocket and giving away $ \frac{1}{8} $ of the remaining money, she has: $$ 50 + \frac{7x - 350}{8}. $$ Next, we consider the second case where she gives away $200 instead of $ \frac{1}{8} $ of the remaining money. 1. She keeps $ x - 200 $. Now, we set up the inequality since she wants the first scenario to yield more money than the second scenario: $$ 50 + \frac{7x - 350}{8} > x - 200. $$ To simplify this inequality, we start by eliminating the fraction. Multiply everything by 8: $$ 8 \cdot 50 + 7x - 350 > 8(x - 200). $$ This simplifies to: $$ 400 + 7x - 350 > 8x - 1600. $$ Reorganizing gives: $$ 400 - 350 + 1600 > 8x - 7x, $$ $$ 1650 > x, $$ or more conveniently, $$ x < 1650. $$ Now we need to consider the lower limit on $ x $, which is given in the problem as $ x \geq 200 $. Combining these inequalities, we find: $$ 200 \leq x < 1650. $$ Thus, the possible values of $ x $ form the interval: $$ \boxed{[200, 1650)}. $$