My Math Forum Interesting Loan case...

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 March 8th, 2017, 11:34 AM #1 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 9,746 Thanks: 651 Interesting Loan case... A loan is obtained, at monthly payments, over 12 months. The amount that went to principal at 3rd pay't = 804.34 The amount that went to principal at 7th pay't = 837.00 How much was borrowed, at what rate, at what monthly pay't? This'll probably wake up Sir Jonah
 March 12th, 2017, 06:33 AM #2 Senior Member   Joined: May 2008 Posts: 250 Thanks: 45 Beer soaked ramblings follow. Insuficient data to resolve problem. Suppose P3=804.34 & I3=amount that went to interest at 3rd pay't, we then have (P3+I3)(1+i)^[-(12-3+1)]=P3 OR (P3+I3){1-(1+i)^[-(12-3+1)]}=I3. Solving for I3 in both equations leads to an identity of I3=P3[(1+i)^10-1] for both. Methinks I3 should be given to solve amount borrowed and rate. I would be very interested indeed as to how you propose to solve this one Sir D with the given info. Last edited by jonah; March 12th, 2017 at 06:46 AM.
 March 12th, 2017, 10:18 AM #3 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 9,746 Thanks: 651 To eliminate "back and forth"'s, here's the resulting loan: (the givens are earmarked with *) Code: MTH [PAYMENT] PRINCIPAL INTEREST BALANCE 0 10,000.00 1 888.48 788.48 100.00 9,211.52 ... 3* 888.48 804.34* 84.14 7,610.81 ... 7* 888.48 837.00* 51.48 4,312.21 ... 11 888.48 870.98 17.50 879.69 12* 888.48 879.69 8.79 .00 Since PRINCIPAL amounts increase by a factor of 1+i, (i being the monthly loan rate), then: 804.34*(1 + i)^(7-3) = 837.00 : solve to get i = .01 The rest is fairly straightforward. OK Sir Jonah? Thanks from jonah
March 14th, 2017, 06:16 AM   #4
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Brandy soaked revelation follows.
Quote:
 Originally Posted by Denis To eliminate "back and forth"'s, here's the resulting loan: (the givens are earmarked with *) Code: MTH [PAYMENT] PRINCIPAL INTEREST BALANCE 0 10,000.00 1 888.48 788.48 100.00 9,211.52 ... 3* 888.48 804.34* 84.14 7,610.81 ... 7* 888.48 837.00* 51.48 4,312.21 ... 11 888.48 870.98 17.50 879.69 12* 888.48 879.69 8.79 .00 Since PRINCIPAL amounts increase by a factor of 1+i, (i being the monthly loan rate), then: 804.34*(1 + i)^(7-3) = 837.00 : solve to get i = .01 The rest is fairly straightforward. OK Sir Jonah?
It is indeed OK Sir D.
I was about to correct myself but I see that you beat me to it.
Turns out I be wrong.
Took a day to abstain from me anti allergy meds and gave myself me favorite brandy and it all seemed to be clear somehow.
Don't know why I did't notice it but it should have been apparent that with R as the monthly payment, we should have the following system of equations: R(1+i)^[-(12-3+1)]=804.34 & R(1+i)^[-(12-7+1)]=837.
A little algebra and we get something like i=.01000017213 or .01and the rest is history.

March 14th, 2017, 06:35 AM   #5
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Quote:
 Originally Posted by jonah Took a day to abstain from me anti allergy meds and gave myself me favorite brandy and it all seemed to be clear somehow.
Geezzz...in my days, I'd simply drink the brandy steadily,
so no meds required

 March 18th, 2017, 10:33 AM #6 Senior Member   Joined: May 2008 Posts: 250 Thanks: 45 Beer soaked ramblings follow. Somebody just asked me how he or she can solve for the amount borrowed given Sir D's insight. Sorry for the delay, been traveling a lot. Given my system of equations, you solve for either i (or R), then R and then A, the amount borrowed using the the present value for the ordinary annuity. Using Sir D's equation, as you have seen, you simply solve for i, then A (if you're not aware of the formula I used to come up with them system of equations), and then R using the aforementioned present value formula for the ordinary annuity. Accordingly, for A, we should have A=804.34(1.01)^(1-3)+ ...+804.34+...+804.34 (1.01)^(12-3) A=804.34(1.01)^(1-3)*[(1.01)^12-1]/.01 Last edited by jonah; March 18th, 2017 at 10:37 AM.
March 18th, 2017, 11:01 AM   #7
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Quote:
 Originally Posted by jonah A=804.34(1.01)^(1-3)*[(1.01)^12-1]/.01
Youzza right on, Sir J!

 March 18th, 2017, 08:42 PM #8 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 9,746 Thanks: 651 Code: MTH [PAYMENT] PRINCIPAL INTEREST BALANCE 0 10,000.00 1 888.48 788.48 100.00 9,211.52 ... 3(e) 888.48 804.34(u) 84.14 7,610.81 ... 7(f) 888.48 837.00(v) 51.48 4,312.21 ... 11 888.48 870.98 17.50 879.69 12(n) 888.48 879.69 8.79 .00 Using e,f,n,u,v to represent the 5 givens (as shown above) These are my formulas: i = (v/u)^[1 / (f-e)] - 1 : monthly interest rate = .01 p = u(1+i)^(n-e+1) : monthly payment = 888.48 a = p[1 - (1+i)^(-n)] / i : loan amount = 10000.00 Agree? Last edited by Denis; March 18th, 2017 at 08:48 PM.
March 19th, 2017, 06:12 AM   #9
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Quote:
 Originally Posted by Denis Agree?
Absolutely.

 March 19th, 2017, 06:14 AM #10 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 9,746 Thanks: 651 You "absolutely" has no booze this weekend?

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