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Jerch
PostPosted: Thu May 24, 2007 2:56 am 
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A clever Usul figured out a great way to catch fish in the Underwater Fishing Cavern. (However, in the spirit of fairness, we won't reveal her secret!) The first day she went fishing, she caught exactly 31416 pounds of fish!

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Even though she had more fish than she knew what to do with, she decided to go fishing again the next day. And she caught exactly 40% of what she caught the day before. Which was still a lot of fish.

She continued fishing every day. And every day she caught exactly 40% of what she caught the previous day.

If she continued fishing every day for twenty years, how many pounds of fish, total, would she catch? Please round to the nearest pound.
This *should* be a simple geometric series... [just plug in the number]

If only I can figure out how she managed to catch EXACTLY "12566.4" pounds of fish on the second day, when all fishes are rounded to the nearest 1 pound in Neopia....The fact that the number gets smaller than 1 from the 13th day onwards doesn't help either.


Last edited by Jerch on Thu May 24, 2007 3:45 am, edited 2 times in total.

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Demulesca
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PostPosted: Thu May 24, 2007 3:28 am 
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Oops, I should have paid attention to what you posted.

:S The question doesn't work well with my calculator especially since one of the days she's catching one millionth of a pound. :roflol:


Last edited by Demulesca on Thu May 24, 2007 3:50 am, edited 1 time in total.

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Brainiac
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PostPosted: Thu May 24, 2007 3:49 am 
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are leap years included in the twenty years?

31416 pounds...that's pi to 5 places...
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Jerch
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PostPosted: Thu May 24, 2007 3:49 am 
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That isn't the answer anyway...


Last edited by Jerch on Thu May 24, 2007 3:52 am, edited 2 times in total.

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Demulesca
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PostPosted: Thu May 24, 2007 3:51 am 
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Brainiac wrote:
are leap years included in the twenty years?

31416 pounds...that's pi to 5 places...
Yeah. If you look at Yahoo there are 7304.84 (I think it's 84) days in twenty years. (The .84 comes from the fact that there are actually 365.24 days in a year)
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That isn't the answer anyway...
Yeah sorry, it was a misreading on my part.
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AySz88
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PostPosted: Thu May 24, 2007 3:59 am 
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Hint: Twenty years? She might as well do it forever!

[edit] Oops, well, it's a hint if you don't look up the other info already in the thread... But I'm not sure if the (post-edit) original post is too much of a hint? :P

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Last edited by AySz88 on Thu May 24, 2007 4:03 am, edited 2 times in total.

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Demulesca
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AySz88 wrote:
Hint: Twenty years? She might as well do it forever!

;)

I submitted my answer, I hope I didn't forget high school math. XD
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Anoohilator
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PostPosted: Thu May 24, 2007 6:44 pm 
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Demulesca wrote:
AySz88 wrote:
Hint: Twenty years? She might as well do it forever!

;)

I submitted my answer, I hope I didn't forget high school math. XD
Haha great hint :D

Got quite a neat number thanks to my old pal excel ;)

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Auttypie
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PostPosted: Thu May 24, 2007 7:29 pm 
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Excel seemed the way to go on this one. We will see if I figured it out or if my pregnancy brain got the better of me and I submitted a bogus number, hehe. Good luck!
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Jerch
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PostPosted: Thu May 24, 2007 7:42 pm 
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AySz88 wrote:
Hint: Twenty years? She might as well do it forever!

[edit] Oops, well, it's a hint if you don't look up the other info already in the thread... But I'm not sure if the (post-edit) original post is too much of a hint? :P
Umm...errr...thought it was pretty obvious...:oops:
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Demulesca
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PostPosted: Fri May 25, 2007 1:35 am 
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Auttypie wrote:
Excel seemed the way to go on this one. We will see if I figured it out or if my pregnancy brain got the better of me and I submitted a bogus number, hehe. Good luck!
I used Google :S

Did you know you can google formulas for instant answers? ;)
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AySz88
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PostPosted: Wed May 30, 2007 11:25 pm 
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The formula for the sum of an infinite geometric series is a/(1-r), where

a = first number = 31416
r = ratio between each number = 0.4

The answer is 31416/(1-.4) = 52360, which is if she does it forever.

For the exact answer, you'd have to subtract (31416*(0.4)^7305ish)/0.6, which is what she'd catch from Day 7305ish and going forever. But it's such a small number that it can be safely ignored. :)

The derivation for the formula is very handy - it's kinda how you convert from a repeating decimal back into a fraction. Ex:

x = 0.3333... = 0.3 + 0.03 + 0.003 + ...
10x = 3.3333... = 3 + 0.3 + 0.03 + 0.003 + ...
(10x - x) = (3.3333... - 0.3333....) = 3
9x = 3
x = 1/3

x = a + ar + ar^2 + ar^3 + ...
rx = ar + ar^2 + ar^3 + ...
(x - rx) = (a + (ar - ar) + (ar^2 - ar^2) + (ar^3 - ar^3) + ...) = a
x(1-r) = a
x = a/(1-r)

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Demulesca
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PostPosted: Thu May 31, 2007 5:39 pm 
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AySz88 wrote:
The formula for the sum of an infinite geometric series is a/(1-r), where

a = first number = 31416
r = ratio between each number = 0.4

The answer is 31416/(1-.4) = 52360, which is if she does it forever.

For the exact answer, you'd have to subtract (31416*(0.4)^7305ish)/0.6, which is what she'd catch from Day 7305ish and going forever. But it's such a small number that it can be safely ignored. :)

The derivation for the formula is very handy - it's kinda how you convert from a repeating decimal back into a fraction. Ex:

x = 0.3333... = 0.3 + 0.03 + 0.003 + ...
10x = 3.3333... = 3 + 0.3 + 0.03 + 0.003 + ...
(10x - x) = (3.3333... - 0.3333....) = 3
9x = 3
x = 1/3

x = a + ar + ar^2 + ar^3 + ...
rx = ar + ar^2 + ar^3 + ...
(x - rx) = (a + (ar - ar) + (ar^2 - ar^2) + (ar^3 - ar^3) + ...) = a
x(1-r) = a
x = a/(1-r)
I got that answer but no trophy. :S
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TDG
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PostPosted: Fri Jun 01, 2007 4:30 am 
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Trophies aren'ty awarded to everybody, only the first 250(? - Don't quote me on that number, I'm not sure its right) people.

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Demulesca
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PostPosted: Fri Jun 01, 2007 2:02 pm 
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the_dog_god wrote:
Trophies aren'ty awarded to everybody, only the first 250(? - Don't quote me on that number, I'm not sure its right) people.
Yeah, I know.
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