Question 1199132
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Suppose you put $ 525 a month for retirement into an annuity earning 7% compounded monthly. 
If you need $ 350000 to retire, in how many years will you be able to retire?
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<pre>
The formula for an Ordinary Annuity saving account compounded monthly  is


    FV = {{{P*(((1+r/12)^n-1)/((r/12)))}}}


where FV is the future value, P is the annual payment at the end of each month, 
r is the interest rate per year expressed as decimal, 
n is the number of monthly deposits (of months).


So, we need to find " n " from the equation


    {{{((1+0.07/12)^n-1)/((0.07/12))}}} = {{{FV/P}}} = {{{350000/525}}} = 666.667,  

which is the same as

    {{{((1+0.005833)^n-1)/0.005833}}} = 666.667.


Rewrite it in this form

    {{{1.005833^n-1}}} = 0.005833*666.667,

    {{{(1.005833)^n}}} = 1 + 0.005833*666.667 = 4.88867.


Take the logarithm base 10 of both sides

    n*log(1.005833) = log(4.88867)


and calculate  

     n = {{{log((4.88867))/log((1.005833))}}} = 272.75  months = 273 months (rounded to the nearest greater integer value) = 22 years and 9 months.   <U>ANSWER</U>
    

<U>CHECK</U>.  {{{525*((1+0.07/12)^273-1)/((0.07/12))}}} = 350400, which is slightly greater than 350000;

        {{{525*((1+0.07/12)^272-1)/((0.07/12))}}} = 347846, which is slightly lesser than 350000.


<U>ANSWER</U>.  273 months is needed,  or 22.75 years = 22 years and 9 months.

         If to round to closest year, then  23 years is just enough;  22 years is not enough yet.
</pre>

Solved, checked, explained and completed.


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&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Sequences-and-series/Ordinary-Annuity-saving-plans-and-geometric-progressions.lesson>Ordinary Annuity saving plans and geometric progressions</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/Sequences-and-series/Solved-problem-on-Ordinary-Annuity-saving-plans.lesson>Solved problems on Ordinary Annuity saving plans</A> 


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