How do i solve a problem, that asks you to find 40% of something repeditivly without doing it one by one, E,g?

Question

Fred figured out a great way to catch fish. The first day he went fishing, he caught exactly 31416 kilo (units) of fish!

He decided (rather greedily)to go fishing again the next day. And he caught exactly 40% of what she caught the day before. Which was still a lot of fish.

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

If He continued fishing every day for twenty years, how many kilo's of fish, total, would He catch? To the nearest kilo.

With sams answer would it be
= (1-(0.4)^7305) / (0.6)
rather then
= (1-(0.4)^7300) / (0.6)
Since 365.25 (days in a year) x 20 is 7305?


Would that make the final answer (following the same working) 52360?
Rather then 57083?

Sorry im confused i have two seeminly sound workings with two different answers.

Answer 1

Just do {(0.4)^ (2 yrs *365)} * 31416

Got the idea?

0.4 raised to power no of days is the factor you need to use.

Answer 2

Let us assume that first day fred caught "x" kgs of fish.

So the next day he catches 40% of it that means it equals =
X * 40% = X * (40/100) = X * (0.4) = (0.4) * X

Again the next day he catches 40% of the second day that means it equals =
(0.4X) * 40% = (0.4X) * (40/100) = (0.4X) * (0.4) = (0.4)^2 * X

Again the third day he catches 40% of the third day that means it equals =
(0.4)^2 * X * 40% = (0.4)^2 * X * (40/100) =
((0.4)^2 * X) * (0.4) = (0.4)^3 * X

so the total goes like this

Day 1 = X
Day 2 = (0.4) * X
Day 3 = (0.4)^2 * X
Day 4 = (0.4)^3 * X

So in 4 days fred would catch fish equal to
X + (0.4) X + (0.4)^2 * X + (0.4)^3 * X
= X * (1 + 0.4 + (0.4)^2 + (0.4)^3)

So the sequence to be multiplied by X becomes a Geometric Progression.
The sum of first "n" numbers of a Geometric progression is =
(a * (1 - r^n) ) / (1 - r)

where "a" is the first number (in this case it is 1),
"r" is the common ratio (in this case it is 0.4)

Now if we calculate number of days in next 20 years assuming there are 365 days in a year then we can get
n = 20 * 365 = 7300

the sum of first 7300 numbers of the G.P. will be =
1 * (1 - (0.4)^7300) / ( 1 - 0.4)
= (1-(0.4)^7300) / (0.6)

Finally the Kgs of fish that Fred catches in next 20 years is
31416 * (1-(0.4)^7300) / (0.6)
= 57083 Kgs

Answer 3

31416 + 0.4 x 31416 + 0.4² x 31416+ ...

= 31416( 1 + 0.4 + 0.4² + .......

This is a GP a = 1, r = 0.4, the problem is n, the number of days in 20 years. For simplicity we can take this as 20 x 365 = 7300

The sum of the GP = 31416[1 - 0.4^7300] / (1 - 0.4)

Total catch is just less than 52360 kg . My calculator gives 0.4^730 as zero, which it isn't.

Answer 4

a = 31,416 kg
r = 0.4
n = 365*20+20/4 = 7,305
T = 31,416(1 - 0.4^7,305)/(1 - 0.4)
T = 52,360 kg
Actually, he may as well quit after 21 days, as his catch after that is less than 1 gram.

Answer 5

5 times 366 gives you the leap years' days.
15 times 365 gives you normal years' days.
Add them together and that should be your power.

31416 times 0.4^number of days.
That should give you your answer.

Answer 6

Heard of Arthimetic Progression (pardon my spelling)

U define the terms. then use the formula to find sun of terms.

1st term = 31416kg
2nd term = 0.4 x T1
3rd term = 0.4 x T2
.
.
.

so n-th term = 0.4 x T(n-1)

n = the number days in the whole 20yrs.

:) good luck !

Answer 7

Set up a spread sheet using Excel

Answer 8

You use this - http://en.wikipedia.org/wiki/Geometric_progression
There is a formula for a sum of the series somewhere

Answer 9

do 31416 x .4^(number of days after)