1. ## Money Problem

Jack's parents have been paying $450 per month into a retirement fund for the last 30 years. The fund is now worth$450 000. What annual interest rate, compounded monthly are Jack's parents earning? Round to two decimal places.

Please any one, how do you solve this?

2. ## Re: Money Problem

Hello, sakonpure6!

Where did this question come from?
It cannot be solved by elementary means.

Jack's parents have been paying $450 per month into a retirement fund for the last 30 years. The fund is now worth$450,000. .What annual interest rate, compounded monthly,
are Jack's parents earning? .Round to two decimal places.

This is an annuity.
The formula is: . $A \;=\;D\,\frac{(1+i)^n-1}{i}$

. . where: . $\begin{Bmatrix} A &=& \text{final balance} \\ D &=& \text{periodic deposit} \\ i &=& \text{periodic interest rate} \\ n &=& \text{number of periods} \end{Bmatrix}$

We have: . $A \,=\,450,\!000;\;D \,=\,450;\;n \,=\,360$

We wish to solve for $i$, the monthly interest rate.

Substitute: . $45,\!000 \;=\;450\,\frac{(1+i)^{360} - 1}{i}$

. . $\frac{(1+i)^{360}-1}{i} \:=\:100 \quad\Rightarrow\quad (1+i)^{360}-1 \:=\:100i$

. . $(1+i)^{360} - 100i - 1 \:=\:0$

Now you must solve a 360-degree polynomial equation.

Good luck!

3. ## Re: Money Problem

When I was younger I had a sweet formula for these problem, however I cant remeber it...
I'll guess that 5% interest calculated monthly mean that after 1 month you have 105% of what you first had. If not.... well... =)

Now to find back the formula let's try this:

$A_n = \frac{100+x}{100}*A_{n-1}+450$

we know that $A_{0}=450$, we'll use n as the nomber of month that have passed. in 30 years there is 360 month. So we know that $A_{360}=450000$

we need to find the recurence equation solution that goes with $A_n$

so.. there I cheated I asked wolfram (I'm not that good with recurence equation solution... You can give it a try on your side maybe you're better than me at it!)

anyway here it is..

$A_n=\frac{100(((c_{1}+450)x+45000)(\frac{x}{100}+1 )^{n}-450(x+100))}{x(x+100)}$

You know the value for n=0 and n=360. so you could put them in there and find the value of c and x.

I think this MAY work

If you've not seen recurrence equation yet, I don't know how to solve it without them. =(

edit: Soroban solution seem much more sweet!

4. ## Re: Money Problem

alright I will try it both ways , thank you very much!

5. ## Re: Money Problem

I have no idea how you could solve a 360 degre polynomial equation. I know that wolfram doesn't have enough computer time for free users and I dont have maple/mathlab/etc.. on this computer..

Altough I guess you could try brute forcing it. since they want the answer rounded to 2 decimal. it wouldn't be too long to find an answer that fit there.

so I did take 10 min to write a code in C# and took a minute to find a solution that worked.

Spoiler:
The interest should be 0.49804

I added the code below its very basic in C#.
Code:
namespace MoneyProblem{
class Program
{
const int NUMBER_MONTH = 360
const float MONTLY_PAYMENT = 450f;
const float THE_NUMBER_WE_LOOKING_FOR= 0.49804f;

static void Main(string[] args)
{
float[] Tableau = new float[NUMBER_MONTH];

Tableau[0] = 450f;

for (int i = 0; i < NUMBER_MONTH-1; i++)
{
Tableau[i + 1] = FonctionRecursive(Tableau[i]);
}
Console.WriteLine(Tableau[NUMBER_MONTH]);
}

static float FonctionRecursive(float val)
{
float Reponse;
Reponse =((100f+THE_NUMBER_WE_LOOKING_FOR)/100f)*val+MONTLY_PAYMENT;
return Reponse;
}
}
}
edit: I substitued my value in soroban equation and it does work!

6. ## Re: Money Problem

An easy way to find i in this problem is to change monthly payments to yearly.Then you look at regular 100 year interest tables for a sinking fund factor R/S=5400/45000=0.012 and find that 6% table has a SFF of 0.01265 at 30 years.Use of the standard formula given by Soroban can then be used to find monthly interest to three significant figures (0.498) Annual (5.98)using successive approximations