Question

PLEASE SHOW HOW YOU GET THE ANSWER STEP BY STEP I DONT KNOW THE PROCESS I NEED TO KNOW HOW TO DO IT PLEASE I WILL MEDAL

Charlotte purchased a pool for $7680 using a six-month deferred payment plan with an interest rate of 20.45%. She did not make any payments during the deferment period. What will Charlotte's monthly payment be if she must pay off the pool within three years after the deferment period?

Answer 1

The monthly payment is $316.54.

Explanation:

It is given that Charlotte purchased a pool for $7680. The rate of interest is 20.45%.

She use a six-month deferred payment plan with an interest rate of 20.45%.

`7680* (20.45)/(100)* (1)/(2)=785.28`

The principle amount after six-month deferred payment is

`7680+785.28=8465.28`

`PV=C* [(1-(1+r)^(-n))/(r)]`

Where, PV is present value, C is monthly payment, r is rate of interest and n is number of years.

`8465.28=C* [(1-(1+((0.2045)/(12))^(-36))/((0.2045)/(12))]`

`C=316.5444\approx \$316.54`

Therefore the monthly payment is $316.54.

Answer 2

Initially, Charlotte owes $7680. She finishes her payments after a total of 6 + 36 = 42 months. Using a simple compounding formula, the amount she owes is worth P at the end of 42 months, where P is:
P = 7680 * (1 + .2045/12)^42 = 15616.67379
Now, the first installment she pays (at the end of six months) is paid 35 months in advance of the end, so it is worth x * (1 + .2375/12)^35 at the end of her loan period.
Similarly, the second installment is worth x * (1 + .2375/12)^34 at the end of the loan period.
Continuing, this way, the last installment is worth exactly x at the end of the loan period.
So, the total amount she paid equals:
x [(1 + .2375/12)^35 + (1 + .2375/12)^34 + ... + (1 + .2375/12)^0]
To calculate this, assume that 1+.2045/12 = a. Then the amount Charlotte pays is:
x (a^35 + a^34 + ... + a^0) = x (a^36 - 1)/(a - 1)
Clearly, this value must equal P, so we have:
x (a^36 - 1)/(a - 1) = P = 15616.67379
Substituting, a = 1 + .2045/12 and solving, we get
x = 317.82