+725 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.)
+23 To find the possible values of the original pile of money Jeri finds, let's denote the original amount of money as x.
Step 1: Analyze the first scenario
1. Jeri takes $50 from the pile, leaving her with x−50.
2. She then gives away 81 of the remaining money, which is 81(x−50).
3. The amount she gives away is 8x−50.
4. Therefore, the amount left after giving away that portion is: x−50−8x−50=x−50−8x+850
5. To simplify this, we can find a common denominator (which is 8): x−50−8x−50=88x−400−x+50=87x−350
6. Finally, the amount in her right pocket is: 50+87x−350
Step 2: Analyze the second scenario
If Jeri gives away $200 directly from the original pile, the amount she keeps is: x−200
Step 3: Set up the inequality
The condition states that the amount she has in the first scenario is greater than in the second scenario. Therefore, we set up the inequality: 50+87x−350>x−200
Step 4: Solve the inequality
1. First, clear the fraction by multiplying everything by 8: 8⋅50+7x−350>8(x−200)
2. Simplifying gives: 400+7x−350>8x−1600
3. Rearranging terms leads to: 400−350+1600>8x−7x 4. Thus: 1650>x 5. This means x<1650.
Step 5: Combine with the initial condition
Since we know that x must be at least 200, we have: 200≤x<1650 Final Answer: The possible values for the number of dollars in the original pile of money is the interval: [200,1650)
Hope you understood!
