Application of Series Help! (1 Viewer)

[SGSII]

Hi guys
Could someone please give me hand in answering these two questions with working out:

For this one I was able to find the percentage of water in the first week, but cant seem to get the right answer for the second and third week :S

And for this one, I just cant seem to get it right either...
Please help!
Thanks

 

[BabzBrah]

First question
a)
i) 100% x (0.93) = 93%

ii) 93 x (0.93) = 86.49%

iii) 86.49 x (0.93) = 80.4357

b) the series is 93 + 86.49 + 80.4357 ......
Using Formula Tn = ar^(n-1)
a = 93 r = (0.93) n = 15
T15 = 93 x (0.93) ^14
T15 = 33.67%

c) find n when 93 (0.93) ^(n-1) < 25
(0.93)^(n-1) < 0.27
From that point personally I just test n values in my calculator
I think there are other methods to do this but Im yet to learn them.
But if you wished you since (0.93) ^(n-1) is equivalent to 0.93^n / 0.93^1 (when you multiply powers with the same bases the bases are added and when you divide its the first - the second power) you can multiply 0.27 to get

0.93^n < 0.29

0.93^19 = 0.27 Therefore n = 19


Edit: Question 9

a) a = 98 r = 0.98 n = 2
T2 = 98 x (0.98)^1
T2 = 96.04%

b) Tn = 98(0.98)^n-1) < 50 this is similar to other question
0.98^(n-1) < 0.51
from testing point in calc n=34 days

c) 98x(0.98)^(n-1) < 25
0.98^n-1 < 0.26
test n values... n= 68days

 

[TheMathStudent]

just a little note to above; think it would be neater to do a=0.98 as opposed to 98 and deal with decimals so that




just remember to round UP whatever n you get, so if you get 51.1 it needs to be 52 days as on the 51st day the price is still above 50%

 

[SGSII]

Alrightie, thank you so much!