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# Cockfit

16 replies to this topic

### #1 agihcam

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Posted 09 July 2007 - 11:47 PM

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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### #2 savagegamer90

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Posted 10 July 2007 - 04:12 PM

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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### #3 TEX

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Posted 10 July 2007 - 09:05 PM

\$30.00
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### #4 savagegamer90

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Posted 10 July 2007 - 09:32 PM

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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### #5 agihcam

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Posted 11 July 2007 - 12:09 AM

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### #6 ashyashwin

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Posted 22 February 2008 - 12:30 PM

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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### #7 Topher

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Posted 10 March 2008 - 08:46 PM

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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### #8 8dividedby12

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Posted 10 March 2008 - 11:53 PM

I don't think I want to go to any "Cockfits" if I have to figure out "How much dollar I have to bring"..
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### #9 footkick

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Posted 12 March 2008 - 04:42 AM

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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### #10 TORFICT

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Posted 14 March 2008 - 01:23 PM

I don't think I want to go to any "Cockfits" if I have to figure out "How much dollar I have to bring"..

Let alone a cockfit where I pay to get in and to go out.
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