Answer to Question #120139 in Financial Math for royal sam

Question

: LOAN AMORTIZATION

I. A family buys a house worth $326,000. They pay $75,000 deposit and take a mortgage for the balance at J12=9% p.a. to be amortized over 30 years with monthly payments.

A. Find the value of the mortgage on their house?
B. Find the value of the monthly payment?
C. Find the loan outstanding after making 20 payments?
D. Find the principal repaid in the 21st payment?
II. Fill out the loan amortization schedule provided in the solution template for the first 5 loan payments. What do you notice about the composition of the payment amount?

Answer 1

cost of the house : $ 326,000

Initial deposit : $ 75,000

mortgage value : $(326,000 - 75,000) = $ 251,000

Answer:A $ 251,000

Now Amortizing period = 360 months (30 years)

rate of interest is 9%.

and per month rate of interest is (9/12)= 0.75%.

B.

To calculate per month payment we will use the following formula :

$$P_n=x*\frac{(1-(1+r)\raisebox{0.25em}{-n})}{r}$$

x: monthly payment

P: principal amount.

r: rate of interest converted to per month equivalent.

n: number of periods.

Now here

P = $251,000

r = 9/12=0.75% = 0.0075

n= 30 years = 360 months

​

or $x= \frac{251000*0.0075}{1-(1.0075)\raisebox{0.15em}{-360}}$

$= \frac{1882.5}{0.932113}$

$=2019.6$

​

Answer:B monthly payment is $ 2019.6

C.

After making 20 payments ,number of months left to be paid is (360-20)=340

so we need to calculate the outstanding balance of left 340 months when we know family is paying $2019.6 every month @ 9% annually.

again,

$$P_n=x*\frac{(1-(1+r)\raisebox{0.25em}{-n})}{r}$$

where x is the monthly payemnt ,here x = $2019.6

r = rate of interest ,here r= 0.0075

and n = number of months left ,here n = 340

so,

$P_n=2019.6\frac{(1-(1+0.0075)\raisebox{0.25em}{-340})}{0.0075}$

$= 2019.6*122.8228$

$= 248052.92$

Answer : C the balance outstanding after making 20 payments is $ 248052.92

D.

Principal repaid in the 21st payment:

This can be calculated simply by applying P₃₄₀ − P₃₃₉

Now

P₃₃₉ = $= 2019.6 * \frac{1-(1+0.0075)\raisebox{0.15em}{-339}}{0.0075}$

$= 2019.6∗ \frac{1-0.0794}{0.0075}$

$=2019.6*\frac{0.9206}{0.0075}$

$=2019.6∗122.747$

$= 247899.84$

so principal paid in 21st payment is

P₃₄₀ − P₃₃₉ $= 248052.92-247899.84$

$= 153.08$

Answer : D: $ 153.08

Part II

here the Principal is $ 251000,rate of interest is 9% p.a and amortization period is 30 years

monthly payment (x) can be calculated using the formula

$$P_n=x* \frac{(1-(1+r)\raisebox{0.5em}{-n})}{r}$$

​

P =251000

r = 9% = 0.0075

n = 30*12 = 360

after putting all values we will get x= 2019.6

x=2019.6

Now the amortization schedule of 1st 5 payments is attached as an image

In table, interest payment of every month is calculated as (0.0075*outstanding principal balance)