Question 1200385: Consider a student loan of $15,000 at a fixed APR of 9% for 15 years.
A. calculate the monthly payment.
B. determine the total amount paid over the term of the loan.
C. Of total amount paid, what percentage is paid toward the principal and what percentage is paid for interest.
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Answers:- $152.14
- $27,385.20
- Principal = 54.77%, Interest = 45.23%, these percentages are approximate.
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Work Shown:
Part A
Monthly payment formula
P = (L*i)/( 1 - (1+i)^(-n) )
where,
L = loan amount
i = monthly interest rate in decimal form
n = number of months
In this problem:
L = 15000
i = 0.09/12 = 0.0075
n = 15*12 = 180 months (equivalent to 15 years)
Let's find the monthly payment
P = (L*i)/( 1 - (1+i)^(-n) )
P = (15000*0.0075)/( 1 - (1+0.0075)^(-180) )
P = 152.139987624268
P = 152.14
The monthly payment is $152.14
To check your work, you can use an online calculator such as this one:
https://www.calculator.net/loan-calculator.html
Or you can use this TVM (time value of money) calculator
https://www.geogebra.org/m/mvv2nus2
The inputs will be:- N = 180
- I% = 9
- PV = 15000
- PMT = leave blank
- FV = 0
- P/Y = 12
- C/Y = 12
After the inputs are set, press the "solve for PMT" button to have -152.14 show up in the PMT box. The value is negative to represent a cash outflow.
See this question
https://www.algebra.com/algebra/homework/Finance/Finance.faq.question.1199129.html
for another example of using the TVM solver.
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Part B
monthly payment = $152.14 found in part A
number of months = 180 which is equivalent to 15 years
152.14*180 = $27,385.20 is the total amount paid back.
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Part C
A total of $27,385.20 is paid back.
$15,000 is the principal aka amount loaned.
(15000)/(27385.20) = 0.5477411156391
= 0.5477
= 54.77% is paid toward principal. This percentage is approximate.
The remaining percentage is toward interest.
100% - 54.77% = 45.23%
Similar question
https://www.algebra.com/algebra/homework/word/misc/Miscellaneous_Word_Problems.faq.question.1199180.html
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