can someone help with this equation?

Question

this guy is fishing and on his first day of fishing he catches 31416 pounds of fish...the next day he goes to catch more fish and he catches 40% the amount of fish he caught the previous day...if he fishes for 20 years and everyday catchs 40% of what he caught the previous day...how many pounds of fish will he have caught??

so far ive thought that 365 days times 20 years is 7300 days...so what would 7300 days and 40% less of 31416 each day be?

i think excel can do it correct but i forgot how to use it

and part of it is not how many he caught on the last day of the 20 years its how many in total she caught

and im in 9th grade first year math so i dont really know any of these fancy equations haha so umm does anyone know a formula that i can use in excel?

yes after 13 days he doesnt catch any fish but im guessing that you have to add up the decimals? like .08122 + .0032123 or whatever

Answer 1

it depends if you have to count the leap years.

(either way i don't know the answer, but you can respond that it is impossible to answer due to not knowing if you must calculate the leap years or not, and if so, the number would be different depending on what day of what year he started. i liked to think up dumb reasoning like this to get out of work when i was in school)

(don't really use my response as answer - i flunked school)

Answer 2

He catches 40% of what he caught the day before.
The first day he catches n.
The second day he catches .4n
Each day (d) he catches .4^(d-1)n

I’ll let you in on a little secret. It’s not as much as one would think. After the 22 day if he caught the same each day for the rest of the 20 years he wouldn’t add a pound to the catch. His total is very close to 52360,

Sorry somebody current with integration can figure out the formula to find out exactly.
Σ.4^(d-1)n d=1 to 7305

Um... thanks to all those who did the integration. I was afraid I might have to look it up. It's been a while. :-)

And Alex, apparently there are no decimals. Apparently, it's exact. Hard to believe.
If you want, write a macro that runs through the process 7305 times.
Sub fish()
Dim a, b As Double
Dim i As Long
a = 31416
b = a
For i = 2 To 7305
a = 0.4 * a
b = b + a
Next i
Range("A1").Value2 = b
End Sub

Answer 3

quilasheila is correct, but it takes only 19 days for him to catch 52359.998560 lb. After that it's pretty much a waste of time.

As to how many she caught, I would say none, as you don't even mention a she until 'way down in the additional details.

edit:
OK, to do it in Excel:
Format columns A and B for 10 decimal places.
In cell A2 enter 31416
In cell A3 enter =0.4*A1
In cell B2 enter =B1+A2
Copy B2 to B3 (or select B2 and fill down to B3)
Select A3 and B3
Fill down for at least 28 more rows. Beyond that nothing will change.
Excel can handle column A for 771 rows, but even Excel can't handle the entire 7300 days without resorting to tricks (It returns 0 for numbers smaller than 10^-307). With Column B, the sum column, you would have to resort to tricks after the 28th row to display the decimal places that change.

This kind of sum is a geometric series. I found the derivation of the formula to be shockingly simple when I finally looked at it.

Answer 4

Using Excel enter 31416 into cell A1, in cell A2 enter the formula =A1*.4. Copy this formula and now highlight A3 - A7300, hit enter (this should cause the formula to cascade all the way down). In cell A7301 enter =SUM(A1:A7300).

You will notice that starting in cell A786 the number is 0, because the number is too small to reduce any further. The final answer you should get is 52360, as mentioned in a previous answer.

Happy Fishing!

Answer 5

SUM OF GEOMETRIC SERIES

1st day 31416
2nd day 31416 * 0.4
3rd day (31416 * 0.4)* 0.4
4th day (31416 *0.4*0.4) * 0.4
...
so on the nth day 31416 * (0.4)^(n-1)

This is a geometric series and we will have 7300 days

so we want to sum the series 31416 * .4^(n-1) for n=1 to 7300

The fraaction for the sum of a geometric series is sum =
a(1-r^n)
_______
1 - r

where a = first term 31416 and r is the ratio 0.4

31416 (1-(.4^7300)
_______________
1 - 0.4

52,360 fishies!

________________________________________
If this is above your level of study, then we should be a ble to list out the fish caught daily until he catches none.
Eventually, and not after too long, 40% of 40% of 40% dwindles down to no fish caught.
Here is the list (each term is 40% of the previous)
31416 + 12566 + 5026 + 2011 + 804 + 322 + 129 + 51 + 21 + 8 + 3 + 1 + 1 ... these numbers were rounded to the nearest fish.. add'em up!
52,351 fishies!
After 13 days, he gets no mo' fish!

=]

Answer 6

This is simply a geometric series, and can be modeled by

S(sub n) = [a((r^n)-1)] / r-1

'a' = beginning
'r' = common ratio (in this case 0.4)
'n' would be your 7300

Solve for S(sub 7300).

Answer 7

31416/7300*.40

approx 1.72

Answer 8

oh my gosh.....uhhh, i dont hink i can help, but good luck!