Cigarretes are expensive(compound interest problem)

  • Thread starter Thread starter Gablar16
  • Start date Start date
  • Tags Tags
    Interest
AI Thread Summary
A typical smoker spending $55 monthly on cigarettes could significantly increase their savings by investing that amount in a savings account at 4.8% interest compounded monthly. The compound interest formula is used to calculate the future value of these monthly deposits, but the complexity arises from the need to account for each deposit compounding over different periods. Various formulas were discussed, leading to a calculated total of approximately $79,679.70 after 40 years, but discrepancies were noted when compared to online calculators. Clarification on the calculations and verification of results from different sources is sought to ensure accuracy.
Gablar16
Messages
44
Reaction score
1
I've been circling his problem for a little while and cannot find how to approach it.

A typical smoker spends 55 dollars a month on cigarrettes. Suppose that the smokers invest at the end of the month that same amount in a savings account at 4.8% compounded monthly. How much money will be in the account at the end of 40 years?


I translated that from spanish so it might be a little unclear, polease don't hesitate to ask for clarifycation.

I know that the compound interest formula is A= P(1+(r/n)^nt but the problem is that everytime I make a deposit I have to add last month amount with interest plus this months amount and it keeps mounting every month. I think the problem lies in P. I Tried plugging in all the numbers and if I invest only the first 55 dollars I will have 2305.60 at the end of the 40 years but I know that's wrong, as the real amount should be much higher. Any push in the right direction would me appreciated greatly. Thanks
 
Physics news on Phys.org
I just scratched this out on paper, and I am trying to get a general solution for this: F = [1+(i/n)^nt]*A + [1+(i/n)^nt-1]*A + [1+(i/n)^nt-2]*A + ...
so its like calculating n different accounts, each with the same principle investment, just with one less compounding term for each successive account
 
Shouldn't A change over time too?

btw thanks for the reply
 
Thanks for the link Ronnin it really helped a lot.

I found a different formula in here
http://mathforum.org/dr.math/faq/faq.interest.html"

but they both gave me the same result.
The website you gave me used (a - ar^(n+1))/(1-r) where;

a=deposit (55)
r=the rate (1+(.048\12))
n=479 (from the geometric sequence?)

I then substituted and

(55-55(1.004)^480)\(-.004)=79679.7


The other formula P = M([1+(i/q)]^nq-1)(q/i) where;

M= 55
i= .048
q=12
n=40

gave me the same result.

My only confusion is that I tried to verify with a web calculator http://www.dinkytown.net/java/CompoundSavings.html" , but it gave a number that it is a bit higher, can someone verify for me? Thanks again
 
Last edited by a moderator:
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks

Similar threads

Replies
8
Views
3K
Replies
3
Views
1K
Replies
13
Views
2K
Replies
16
Views
5K
Replies
29
Views
4K
Replies
9
Views
2K
Back
Top