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bearhunter [10]
3 years ago
6

Can someone please help me solve this problem????

Mathematics
2 answers:
marshall27 [118]3 years ago
7 0
For this problem, we will be using the formula for loan:

PMT=P[(r/n)/1-(1+r/n)^-ny]
where: 
P=Principal Value
r=rate
n=number of compoundings/year
y=year

To solve:
*Weston is only financing $185,000.

P=$185,000
r=6.525% or 0.06525
n=12 (monthly)
year=30

PMT=185000[(0.06525/12)/1-(1+0.6525/12)^-12*30
        = 185000[(.0054375)/1-(1+.0054375)^-360
        = 185000[(.0054375)/1-(1.0054375)^-360
        = 185000[(.0054375)/1-(0.142)
        = 185000[(.0054375)/(.858)
        = 185000(0.00634)
        = 1,172.37

Answer: $1,172.37

vodka [1.7K]3 years ago
3 0
<span>$1,172.37 The formula for calculating the payment on a loan is P = r(PV)/(1 - (1 + r)^(-n)) where P = Payment PV = Present value r = interest per period n = number of periods Since there's 12 months per year and the loan is for 30 years, there will be 12 * 30 = 360 periods. The interest rate per month will be 0.06525 / 12 = 0.0054375, and finally, the present value is the size of the loan at 185000. Now let's substitute and solve. P = r(PV)/(1 - (1 + r)^(-n)) P = 0.0054375(185000)/(1 - (1 + 0.0054375)^(-360)) P = 1005.9375/(1 - (1.0054375)^(-360)) P = 1005.9375/(1 - 0.141961802) P = 1005.9375/0.858038198 P = 1172.369134 P = 1172.37 So the monthly interest and principle payments are $1,172.37 Note: The actual payments will be higher since the above figure doesn't include insurance and taxes.</span>
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Simple Interest

This is a simple interest problem.

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In this question:

We divide in two intervals, the first year and the second year. We have to know that the amount after the end of the first year is the principal at the start of the second year.

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Amount of 2% interest, during 1 year.

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