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  #1  
Old 10-24-2019, 07:15 PM
ManuelG ManuelG is offline
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Hi guys, can you help me with this problem, I've tried to figure it out but I'm stucked.

A loan of 5000 can be repaid by payments of 117.38 at the end of each month for n years (12n payments), starting one month after the loan is made. At the same rate of interest, 12n monthly payments of 113.40 each accumulate to 10,000 one month after the final payment. Find the equivalent effective annual rate of interest.

Thank you in advance!
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Old 10-24-2019, 10:43 PM
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Originally Posted by ManuelG View Post
A loan of 5000 can be repaid by payments of 117.38 at the end of each month for n years (12n payments), starting one month after the loan is made. At the same rate of interest, 12n monthly payments of 113.40 each accumulate to 10,000 one month after the final payment.
The second statement lets you determine the the dollar amount that 117.38 per month would accumulate to then. The first sentence tells you what a single payment of 5000 would accumulate to then, in terms of (1+i)^n. Thus you can solve for (1+i)^n

Then, going back to the second sentence, you should be able to write an equation involving (1+i)^n, i, 113.40 and and 10,000. Plug in the value of (1+i)^n; solve for i.

Pretty much that's what to do. Maybe slightly more complicated.
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Old 10-24-2019, 11:57 PM
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The first statement will give a angle 12n. The second will give s double dot angle 12n. Take the ratio to give (1+i)^(12n+1) where i is the monthly rate. Add 1 to the s to convert it to 12n+1 payments. Use the formula to solve for the monthly rate. Convert to an annual rate.
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Old 10-25-2019, 10:18 PM
ManuelG ManuelG is offline
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Thank you guys, I've got an annual interest rate of 14.74%
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Old 10-26-2019, 12:59 AM
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Close but not correct. Try the monthly rate you found and use the present value to compute N on the calculator. Then use that N to see it gives the correct Future Value.
With the correct interest rate you should find 60 payments.
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Old 10-26-2019, 02:24 PM
Sullinator Sullinator is offline
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Originally Posted by ManuelG View Post
Hi guys, can you help me with this problem, I've tried to figure it out but I'm stucked.

A loan of 5000 can be repaid by payments of 117.38 at the end of each month for n years (12n payments), starting one month after the loan is made. At the same rate of interest, 12n monthly payments of 113.40 each accumulate to 10,000 one month after the final payment. Find the equivalent effective annual rate of interest.

Thank you in advance!
I’m not an actuary or anything but I think it’s 5000=117.38a-angle12n, 10000=113.4s-angle12n, 1000=113.4s-angle12n is the same thing as 1000V^(12n)=113.4a-angle12n, then solve for a-angle12n and once you do that plug it into 5000=117.38a-angle12n to solve for monthly rate, once you do that you convert it to annual rate. Monthly rate comes out to be .012136, annual rate is .155757, I could be wrong but I think that’s it.
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Old 10-26-2019, 05:36 PM
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Iím not an actuary or anything but I think itís 5000=117.38a-angle12n, 10000=113.4s-angle12n, 1000=113.4s-angle12n is the same thing as 1000V^(12n)=113.4a-angle12n, then solve for a-angle12n and once you do that plug it into 5000=117.38a-angle12n to solve for monthly rate, once you do that you convert it to annual rate. Monthly rate comes out to be .012136, annual rate is .155757, I could be wrong but I think thatís it.
Closer but still off by a little.
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Old 10-26-2019, 05:46 PM
Sullinator Sullinator is offline
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Closer but still off by a little.
Oh it says accumulate to 10000 one month after the final payment so it’s 113.4s-Doubledot angel-12n since it’s a accumulation due and not immediate, I read it the wrong way thinking it’s immediate.
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