Mathematics Homework Help
MATH 1030 Walden University Interest Increase and Total Payable Discussion
Respond to 2 classmates and answer all prompts for each required response.
Response 1: Respond to at least one classmate using the following:
- Lower the interest rate provided in your classmate’s post by one percentage point.
- Determine the amount of interest your classmate will pay with this new interest rate and how much money will be saved over the length of the loan.
- Write a 1-paragraph memo to your classmate explaining why the new interest rate is the better choice.
For response 1 you will respond to Sarah:
Sarah,
My husband and I have been throwing around the idea of purchasing a new pontoon boat. So, when this discussion came up, I figured it was perfect timing to do some figuring! The pontoon boat I chose had a sale price of $24,995.00, and a middle of the field interest rate for watercraft right now is 4.25%. Boat loans can range from 3-15 years depending on price, so I chose a term of 5 years or 60 months. I calculated the amount of interest that will be paid throughout the term of the loan using the following:
Interest= Principal*Rate*Time
I=$24,995*0.0425*5
The total amount of interest for the length of the loan is: $5311.44 (this is rounded for dollars and cents)
The final cost of the loan is determined by using this formula: principal + interest. Furthermore, $24,995+$5311.44= $30,306.44
The monthly payment for the pontoon is calculated with the following formula:
m= p*r(1+r) raised to n
(1+r) raised to n-1
m=24,995(.0425/12)(1+.0425/12) raised to 60
(1+.0425/12) raised to 60-1
m= (88.52)(1.23630214538)
.23630212
Monthly payment= $463.13 (rounded to dollars and cents)
If the loan term were lowered by one year, meaning the loan was taken out for 4 years instead of 5, the interest would be the following:
I= 24,995*.0425*4
I= $4,249.15
The total amount when paid in full after 4 years is: $29,244.15. This was calculated by adding $24,995+$4,249.15.
By using the above formula for calculating monthly payments, I simply changed the 60th power to 48th for 4 years in months is 48. The payment for 4 years would be: $567.14.
I am shocked that the payment amount was higher than I anticipated. However, I am not shocked that the four-year payment was $100 a month higher than the 5 year. Decreased time for the same amount of money equals a higher payment. My car is now paid off, but that payment was only $215 a month. Yes, that would contribute toward half of the pontoon payment, but I do not think I will be getting a pontoon anytime soon. Or at least a new one!!
Response 2: Respond to at least one other classmate using the following:
- Imagine that your classmate has signed up for a variable rate loan, which means that the interest rate can change.
- Assume that your classmate has paid back half of the loan when the interest rate increases by 5%.
- How much time is left on the loan?
- How much will need to be paid in interest during the second half of the loan?
- Explain to your classmate how much this interest rate change will affect the overall balance by comparing the original total cost to the new total cost. (Hint: Be sure to consider both the first half of the loan and second half when finding the new total payment.)
For the second response you will respond to Valerie:
Valerie,
The area that we live in you can find a really nice decent house for a good price, leave it to me to fall in love with a 550,000 house. This is out of our price range, by the way. Lol! Below is the cost after it is all figured out and the monthly payments as well.
Cost: $550,000
Interest Rate: 8%
Years: 10
Interest= Principal*Rate*Time
I=$550,000*.08*10=$440,000
$550,000+$440,000=$990,000
$990,000/120 months=$8,250 a month
Part 2;
Cost: $550,000
Interest Rate: 8%
Years: 9
I=$550,000*.08*9=$396,000
$550,000+$396,000=$946,000
$946,000/108 months=$8,759.26
Difference in monthly mortgage: $509.26
One thing I would do different is not purchase a house at this price. The difference in cost I thought would be more when you took a year off of the original equation.