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What would be the monthly payment for $20,000 at 6.9% financing for 60 months?

Guest Jul 2, 2014

Best Answer 

 #3
avatar+92806 
+8

that is interesting.  I got the same answer but on the surface our forumulas look different.

$$\begin{array}{rll}
A&=&M \left(\dfrac{1-(1+i)^{-n}}{i}\right)\\\\
A\times \dfrac {i}{1-(1+i)^{-n}}&=&M\\\\
M&=&A\times \dfrac {i}{1-(1+i)^{-n}}\\\\
M&=&20000\times \dfrac {0.00575}{1-(1.00575)^{-60}}\\\\
M&\approx &\$ $395.08
\end{array}$$

Melody  Jul 3, 2014
 #1
avatar+576 
+5

$395.08

Answer found using a calculator at http://www.mortgagecalculator.org/

jboy314  Jul 2, 2014
 #2
avatar+87323 
+8

 

Let's see how jboy314's answer was derived.

The "formula" for the monthly payment is given by:

M = P(1+r)n r / [(1+r)n-1]

Where

M = the Monthly Payment

P  = the amount financed ($20000)

r = the montly interest rate expressed as a decimal = .069/12 = .00575

So we have

M = (20000*(1+.00575)^60*.00575)/ ((1+.00575)^60 -1)) = $395.08

 

CPhill  Jul 2, 2014
 #3
avatar+92806 
+8
Best Answer

that is interesting.  I got the same answer but on the surface our forumulas look different.

$$\begin{array}{rll}
A&=&M \left(\dfrac{1-(1+i)^{-n}}{i}\right)\\\\
A\times \dfrac {i}{1-(1+i)^{-n}}&=&M\\\\
M&=&A\times \dfrac {i}{1-(1+i)^{-n}}\\\\
M&=&20000\times \dfrac {0.00575}{1-(1.00575)^{-60}}\\\\
M&\approx &\$ $395.08
\end{array}$$

Melody  Jul 3, 2014
 #4
avatar+92806 
0

There you go - I challange someone to do the mathas and show why the formulas are indeed the same. 

Melody  Jul 3, 2014
 #5
avatar+26753 
+5

Just take the right-hand side of Chris's formula and divide top and bottom by (1+r)n.  

His equation becomes  M = Pr/(1-1/(1+r)n) or M = Pr/(1-(1+r)-n)

Replace his P by your A and his r by your i.  

Alan  Jul 3, 2014
 #6
avatar+92806 
0

Thanks Alan.

Melody  Jul 3, 2014

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