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

Answer 1

Answer:

The monthly payment is $316.54.

Step-by-step 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%.

$7680\times \frac{20.45}{100}\times \frac{1}{2}=785.28$

The principle amount after six-month deferred payment is

$7680+785.28=8465.28$

$PV=C\times [\frac{1-(1+r)^{-n}}{r}]$

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

$8465.28=C\times [\frac{1-(1+(\frac{0.2045}{12})^{-36}}{\frac{0.2045}{12}}]$

$C=316.5444\approx \ 316.54$

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