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Ivahew [28]
3 years ago
12

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.)
Mathematics
1 answer:
Marizza181 [45]3 years ago
6 0

Answer:

The possible values of the number of dollars in the original pile of money is ≥ $200 but < $350

Step-by-step explanation:

Here we have, pile of money ≥ $200

Amount in put the left pocket = $50

Fraction given away = 2/3 of rest of pile ≥ 2/3×150 ≥ $100

Amount put in right pocket = ≥ $150 - $100 ≥ $50

Total amount remaining with Jeri = $50 +≥ $50 ≥ $100

Also original pile - $200 < $100

Therefore where maximum amount given away to have more money = $200 we have

2/3× (original pile - 50) = $200

Maximum amount for original pile = $350

Therefore the possible values of the number of dollars in the original pile of money is ≥ $200 but < $350.

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Wewaii [24]
After careful consideration i think I have an answer... 
So if you notice, These two triangles are a reflection of each other. 
Now even though these two triangles equal one another, we still have to consider the fact that they are different. (one side is the opposite of another)
So lets just name all the sides that equal each other. 
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(by the way i'm going to be saying {the reflective side} every time i want to refer to the side with the two lines :P) - so bare with me. 

So therefore this would mean that ∠P = ∠S would be false, since one angle is from the side that is reflected and the other isn't. Makes sense?

So lets look over at the answers...

A.) PN = SQ 
For this answer choice, PN is on the non reflective side, while SQ is on the reflective side. Soooo... This answer choice is wrong. 

B.) NO = QR
NO  is part of the  unreflective side, while QR is part of the reflective side. 
Therefore this answer is wrong. They must be part of the same side. 

C.) ∠P = ∠S 
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There fore this answer is also wrong. 

D.) ∠O = ∠S 
Now ∠O is on the reflective side while ∠S is on the reflective as well. 
There fore this answer is correct . 

YOUR ANSWER IS. 
D.) ∠O = ∠S 

Good Luck! :)


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