Question

Jeri finds a pile of money with at least $\$200$. If she puts $\$50$ of the pile in her left pocket, gives away $\frac23$ 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.)

Answer 1

(200,350)

Explanation:

Let d be the number of dollars in the pile of money. Suppose she puts $\$50$ in her left pocket, gives away $\frac23$ of the rest of the pile, and then puts the rest in her right pocket. She therefore keeps $50$ plus one-third of the remaining $d - 50$ dollars in the pile. Therefore, she has $50 +\frac13(d-50)$ dollars.

If she instead gives away $\$200$ and keeps the rest, she has $d - 200$ dollars. In order for her to have more money in the first case than in the second, we must have

\[50 +\frac13(d - 50) > d - 200.\]Multiplying both sides by 3, we get

\[150 + (d - 50) > 3d - 600.\]This simplifies to $2d < 700,$ so $d < 350.$ Combining this with the given fact that there is at least 200 dollars in the original pile, we see that the number of dollars in the original pile must be in the interval $\boxed{[200,350)}$.

Answer 2

Let say Jeri finds a pile of money = 200 + 3M $

puts $50 of the pile in her left pocket

Rest = 200 + 3M - 50 = 150 + 3M $

Give away = (2/3) (150 + 3M) = 100 + 2M $

Money in right Pocket = 150 + 3M - (100 + 2M)
= 50 + M $

200 + 3M - 200 = 3M $

50 + M > 3M
=> 2M < 50
=> M < 25
=> 3M < 75

Money 200 + 3M
< 275

200 $ ≤ Money in piles < 275 $