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**davidtrinh****Member**- Registered: 2016-12-03
- Posts: 10

How long does it take to pay off a $125000 mortgage if the annual interest rate is 5.5% and compounded monthly with a monthly payment of $710?

Below is my attempt.

I use the formula

FV = P* ((1+r/n)^(nt)-1)/(r/n)

where FV is future value, r is the interest rate, n is the number of compounding periods in a year, t is the elapsed time, and P is the principal.

12500 = 710*((1+0.055/12)^(12t)-1)/(0.055/12)

Next, I divide both sides by 710 and then multiply both 0.055/12.

3841.24 = (1.0045833333)^(12t)

Then, I take the log of both sides.

log3841.24 = 12t*log1.004583

Lastly, I solve for t and get t = 150 years.

But, the answer key says 30 years.

Can someone explain how to do the problem? Thanks a lot.

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**Grantingriver****Member**- Registered: 2016-02-01
- Posts: 100

The problem is probably incorrect!! How are you goning to pay $125000 in 30 years with a monthly payment of $710 along with the interest, while the amount paid in this period with the $710 only is 710×12×30=$255600 ??!!!. It would be a band news if you are an accountant...

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**Mathegocart****Member**- Registered: 2012-04-29
- Posts: 1,941

I presume you mean 710 to be the principal, P. Now let's solve for t.

125,000 = 710(1 + 0.055/12)^(12t)

125,000 = 710(1.00458)^(12t)

log(176.056338028) = log((1.00458)^12t )

(divide by the logs)

1131.58 = 12t.

And with a swift but succinct slice, we find..

t = 94.298 yrs or 94.3 yrs.

Somewhat peculiar that both you and the book are wrong, but WolframAlpha and Desmos confirms it.

*Last edited by Mathegocart (2018-05-10 23:35:35)*

The integral of hope is reality.

May bobbym have a wonderful time in the pearly gates of heaven.

He will be sorely missed.

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**Grantingriver****Member**- Registered: 2016-02-01
- Posts: 100

Dear Davidtrinh and Mathegocart, the question simply has something wrong with it. The numbers “30 years” and “94.3 years” are irrational (it is like calculating the length of a person, according to some data, and finding it to be 3 km!!). There is no “bank” that accepts this contract. Actually, in normal circumstances, the bank increases the period of payment if the costumer borrows more money until the period reaches a particular number of years (usually 6-8 years), then they would not increase the period no matter what is the amount that was borrowed. Therefore the problem is unrealistic which is not expected from a textbook. The number “30 years” is probably a typo and the correct number seems to be “3 years”. On the other hand, the formula used by “Mathegocart” is probably not the correct one, since this equation is derived to calculate the saving account balance after a period of time in the case that an amount “P” is deposited in the bank (calculating the monthly payment of a loan in this way is a little bit strange). Finally, even if we assume that the number of years was not “30 years”, there would be a number of possible ways to calculate the period depending on the method applied for calculation. Therefore, without any further information, your question can not be solved “correctly”. This is, in fact, the reason why the brilliant mathematicians in this site did not answer this question.

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**Mathegocart****Member**- Registered: 2012-04-29
- Posts: 1,941

Grantingriver wrote:

Dear Davidtrinh and Mathegocart, the question simply has something wrong with it. The numbers “30 years” and “94.3 years” are irrational (it is like calculating the length of a person, according to some data, and finding it to be 3 km!!). There is no “bank” that accepts this contract. Actually, in normal circumstances, the bank increases the period of payment if the costumer borrows more money until the period reaches a particular number of years (usually 6-8 years), then they would not increase the period no matter what is the amount that was borrowed. Therefore the problem is unrealistic which is not expected from a textbook. The number “30 years” is probably a typo and the correct number seems to be “3 years”. On the other hand, the formula used by “Mathegocart” is probably not the correct one, since this equation is derived to calculate the saving account balance after a period of time in the case that an amount “P” is deposited in the bank (calculating the monthly payment of a loan in this way is a little bit strange). Finally, even if we assume that the number of years was not “30 years”, there would be a number of possible ways to calculate the period depending on the method applied for calculation. Therefore, without any further information, your question can not be solved “correctly”. This is, in fact, the reason why the brilliant mathematicians in this site did not answer this question.

Hi, the question is ambiguously worded, but at a second reasing I realize I misconstrued the problem. Upon utilizing the mortgage formula, I got a time of 359.65 months.. or 30 years, confirming the book.

The compound interest formula for paying off a mortgage is

payment = P [ i(1 + i)^n ] / [ (1 + i)^n - 1]

where..

payment = 710

i = interest divided by 12(monthly)(0.055/12)

n(time in months)

*Last edited by Mathegocart (2018-05-12 14:41:17)*

The integral of hope is reality.

May bobbym have a wonderful time in the pearly gates of heaven.

He will be sorely missed.

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**Grantingriver****Member**- Registered: 2016-02-01
- Posts: 100

OK, brilliant!! The problem should then be restated as follows:

``How long does it take to pay off a $125000 mortgage if the annual interest rate is 5.5% and compounded monthly with a monthly payment of $710?. hint:

where c , P, r and n are the monthly payment, the mortgage, the interest (per month) and the pay off period (in months), respectively?''

and hence the solution goes as follows:

which is the final formula that you have utilized. However, solving for n and plugging the numbers we get:

Note: The students should have been taught that:

Otherwise, the hint should be included. Thank you very much Mathegocart. I hope that will help you Davidtrinh.

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**davidtrinh****Member**- Registered: 2016-12-03
- Posts: 10

Thank you, Grintingriver.

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