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Kenton borrows $250,000 on January 1, 2012 to be repaid in 12 annual installments at an effective annual rate of interest of 12%. The first payment is due on January 1, 2013. Instead of annual payment he decides to make monthly payments equal to one-twelfth the annual payment beginning on February 1, 2013. Determine how many months will be needed to pay off the loan. (Ans : 129.381291)

Anyone know how to solve this question?

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Following that would be to use the annuity formula

$239641 = 3363*\dfrac{(1-\frac{1}{(1+r)^t})}{r}$

$\frac{1}{(1+r)^t} = 0.294671$ with r = 0.009489.

$(1+r)^t = 3.393617$

Taking log

$t \times ln(1.009489) = ln(3.393617)$

$t = \frac{ln(3.393617)}{ln(1.009489} = 129.3826$

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  • $\begingroup$ If the interest become monthly, aren't it suppose 12% / 12 ? how u get the r = 0.009489? $\endgroup$ – Liow Wei Sheng Jul 1 '17 at 16:18
  • $\begingroup$ You should take care of compounding and hence it is not simple division but $(1.12)^{\frac{1}{12}} - 1 = 0.009489$ $\endgroup$ – Satish Ramanathan Jul 1 '17 at 16:26
  • $\begingroup$ from line 2, how is the equation come from? I calculated 1 - (239641/3363 x r) but the answer is not 0.294671. Is it the # of significant problem? $\endgroup$ – Liow Wei Sheng Jul 1 '17 at 16:57
  • $\begingroup$ Yes you are correct, if it is assumed that the first payment is paid on the 1st of Jan 2013, then the number of months is 119. If it is not paid on the 1st of Jan 2013, then use the loan value = 250000 and then you would get 129.38. I believe the text book assumes that the first payment is not paid. $\endgroup$ – Satish Ramanathan Jul 2 '17 at 1:53
  • $\begingroup$ Thank you very much for taking time to solve my question. :) $\endgroup$ – Liow Wei Sheng Jul 2 '17 at 4:35
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I misread the question and have revised

He makes the annual payment on 1/1/2013 and then starts paying monthly thereafter.

the annual payment:

$250,000 = P\sum_\limits{n=1}^{12} (1.12)^{-n}\\ 250,000 = P \frac {(1 - 1.12^{-12})}{0.12}\\ P = 250,000 \frac {0.01}{1-1.12^{-12}}\\ P = 40,359$

after making his first months payment he owes

$250,000 \frac {1-1.12^{-11}}{1-1.12^{-12}} = 239,640$

and he will be paying $\frac {40359}{12} = 3363$ per month

$239,640 = 3363\sum_\limits{n=1}^k (1.12)^{\frac {k}{12}}\\ 71.25 = \frac {1 - 1.12^{\frac {-k}{12}}}{(1.12)^{\frac 1{12}} - 1}\\ 1-71.25(1.12^\frac{1}{12} - 1) = (1.12)^{\frac {-k}{12}}\\ \ln 0.324 = \frac {-k}{12}\ln 1.12\\ 119.368 = k$

But $k$ doesn't start counting until one year has passed

$131$ months

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  • $\begingroup$ May i know what is the purpose of line 9? Its 250000 x ( a angle n-1 / a angle n). $\endgroup$ – Liow Wei Sheng Jun 30 '17 at 12:18
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    $\begingroup$ The first payment of 40359 has both principal and interest. The interest is (.12*250000) = 30000. So principal component of the first payment is 40359 - 30000 = 10359. For the second annuity the loan value is 250000 - 10359 = 239641 $\endgroup$ – Satish Ramanathan Jun 30 '17 at 15:42
  • $\begingroup$ Thank you very much for taking time to solve my question. :) $\endgroup$ – Liow Wei Sheng Jul 2 '17 at 4:35

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