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There were three cockfit arena that you want to enter. You will enter the first cockfit arena and you need to pay $10.00 as entrance fee, when you arrive at the arena your money will be doubled, when you go out you need to pay $10.00. When you enter the second arena, you need to pay $10.00 again as entrance fee, then inside it your money will become double, when you leave you need to pay $10.00 again.The third is the same you need to pay $10.00 as entrance fee, when inside your money will be doubled and when you out you will pay your last money ( $10.00 ) this time you dont have money left. How much dollar you need to bring?

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You need to bring $26.25 with you.

[([([(X-10)*2]-10-10)*2]-10-10)*2]-10=0

([([(2X-20)-20]*2)-20]*2)-10=0

[([(2X-40)*2]-20)*2]-10=0

([(4X-80)-20]*2)-10=0

[(4X-100)*2]-10

(8X-200)-10=0

8X-210=0

8X=210

X=26.25

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that doesn't work

30-10 to get into the first one =20

then you double your money =40

40-10 to get out of the first one =30

30-10 to get into the second one=20

then it doubles and it's back to 40

and you can't end with $0

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savagegamer90 answer is correct.

savagegamer's answer is correct indeed...

i simply walked back from the third arena. since $10 had to be paid at the gates while getting out of the arena; and he is left without any money, the amount of money he had after it got doubled inside the 3rd arena is $10..ie, he had $5 at the entrance gates after paying the $10 ticket at the third arena, or $15 after paying and leaving the gates of the 2nd arena.

so, he had $25 inside the 2nd arena after the doubling of the money...

=$12.5 after the entrance into 2nd arena

=$22.5 before paying the tickets at the 2nd arena

=$32.5 inside the 1st arena

=$16.25 after entering the 1st arena

=$26.25 before entering the 1st arena, or at the beginning...

ofcourse, this happens to be the crude way of finding the answer compared to eloquent Xs and Ys...but is very handy when one has to do the whole calculation mentally, without the company of pen and paper...

and moreover from the school algebraic equations used to give jellybeans in the stomach of mathematically challenged people like me!!! :)

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ofcourse, this happens to be the crude way of finding the answer compared to eloquent Xs and Ys...

I think most mathematicians would say that your way is both more elegant and more eloquent, and not the least crude. Simplicity is a virtue in math.

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I think most mathematicians would say that your way is both more elegant and more eloquent, and not the least crude. Simplicity is a virtue in math.

I agree that neither is crude as long as the problem is solved. Of course, I often favor the simplest solution. In this case, it seems they are analogous solutions. savagegamer90 solves the equation from left to right, whereas ashyashwin solves from right to left. When put in savage's terms, ashyashwin's solution looks like this:

[([([(X-10)*2]-10-10)*2]-10-10)*2]-10=0

([([(X-10)*2]-10-10)*2]-10-10)*2=10

[([(X-10)*2]-10-10)*2]-10-10=5

([(X-10)*2]-10-10)*2=25

[(X-10)*2]-10-10=12.5

(X-10)*2=32.5

X-10=16.25

x=26.25

I must admit that, though I solved it the same way savagegamer90 did, it is easier to take ashyashwin's approach when solving it without paper.

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If you win every bet and still lose all your money, I'd say you were a pretty terrible money manager. Try making your wagers higher relative to the entrance fee. And who the hell charges to leave. If you would have lost one of your bets then you'd be stuck there with all those cocks!

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There were three cockfit arena that you want to enter. You will enter the first cockfit arena and you need to pay $10.00 as entrance fee, when you arrive at the arena your money will be doubled, when you go out you need to pay $10.00. When you enter the second arena, you need to pay $10.00 again as entrance fee, then inside it your money will become double, when you leave you need to pay $10.00 again.The third is the same you need to pay $10.00 as entrance fee, when inside your money will be doubled and when you out you will pay your last money ( $10.00 ) this time you dont have money left. How much dollar you need to bring?

I think the previous answers are all wrong...

The wording specifically states that your money is doubled INSIDE arena 2 & 3, but it says your money is doubled when you ARRIVE AT arena 1... meaning your money would be doubled before you enter or pay the entrance fee...

You would only need $21.25 when you arrived at #1.....

then doubled upon your arrival to $$42.50, less $10 to enter, less $10 to exit = $22.50

at #2, less $10 to enter = $12.50, doubled INSIDE to $25, less $10 to exit = $15

at #3, less $10 to enter = $5, doubled Inside to $10, less $10 to exit = $0...

The answer, therefore, is $21.25

The End

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The guy before me, your math is flawed because at the beginning you have to pay $10 to enter, then your money doubles... not your money doubles then you pay for both entering and leaving.

By working backwards it goes like this:

You end with 0 after paying $10 out the last door. 10/2 = $5 (for it to double) + $10 (the entrance fee to #3)= $15

$15 + $10 (the amount you paid to leave # 2) = $25/2 (for it to double) = $12.5 + $10 (entrance fee to #2) = $22.5

$22.5 + $10 (amount you paid to leave # 1) = $32.5/2 (for it to double) = 16.25 + $10 (entrance fee to #1) = $26.25

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Calculating from back:
Third arena:
You come out paying $10 and balance with you becomes Zero.
So you had $5 before the amount was doubled.
So before paying entrance fee you had $15.
Second arena:
You had $15 when you came out of second arena, so you before coming out you had $25.
So you had $12.5 before the amount was doubled.
So before paying entrance fee you had $22.5.
First arena:
When you came out of first arena you had $22.5.
So before leaving the first arena you had $32.5.
So when you came at the first arena (before doubling the amount) you had $16.25.
So before paying the entrance fee to enter the first arena you had $26.25.

Above remains true whether we go for wording of OP in any way.

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Lets do it the other way around
0 at the end
So 10 to leave fight 3 = you had 5$ when entering fight 3 cuz it dubbled.
You had to pay 10 to get in and 5 left to dubble inside so you had 15 before entering fight 3

15+10 you needed to pay to leave fight 2 = 25 when inside fight 2
25/2 as you dubbled the money in fight 2 = 12.5 +10 to enter fight 2 = 22.5 before entering fight 2

10 to leave fight 1 = 32.5. But you need to half that once inside 32.5 = 16.25
16.25+10 because you need to pay to enter fight one

= 26.25€ at the start

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