Question

Matthew opens a store credit card when he purchases a new sound system. The interest rate is 22.45%. He charges $4,120 to the card, and can pay $1,450 per month. What will the total cost of his purchase be?

Answer 1

First off, that interest rate is outrageous! Second, we are assuming a simple interest formula so the approach you want to take is to add 22.45% of each month's total and subtract your monthly payment. When you get to 0 you will have the cost of each month plus the amount of interest paid each month, for your total purchase price.

Answer 2

I got 5,044.94 is that right?

Answer 3

?

Answer 4

I got 5302 but am going back over to see if I made an error.

Answer 5

Again I got 5302.51, but I assumed he made his payments on time so only the residual would be charged interest.
4120-1450 = 2670
2670 + 599.415 = 3269.415
3269.415 - 1450 = 1819.415
1819.415 + 408.459 = 2227.877
2227.877-1450= 777.874
777.874 + 174.633 = 952.506

So three payments of 1450 + a residual of 952.506 = 5302.506
rounded to 5302.51

Answer 6

Something about this just seemed wrong, so I did some searching and concluded that this is the appropriate answer, maybe @myininaya or @amistre64 or @hero could check it.
To start, 22.45% was probably an annual rate, meaning that it was 0.018708333/mo
To find the number of payments necessary I found this formula: N = −log(1−iA/P) / log(1+i)
Where N is number of payments,
i is the interest rate per month,
A is the payment amount
P is the amount borrowed.

\[\large \frac{-\log_{10}(1-\frac{4120*0.018708333}{1450})}{\log_{10}(1+0.01870833) }\] equating to 2.9469 payments. \[2.9469 * 1450 = 4273.03\]

Answer 7

Flint purchased a boat for $10,815. He made a down payment of $1,175. He applied for a six-year installment loan with an interest rate of 9.4% in the amount of $9,640. What is the total cost of the boat after six years? could you help me with this one its given me much trouble

Answer 8

Matthew opens a store credit card when he purchases a new sound system.

The interest rate is 22.45%.

He charges $4,120 to the card, and
can pay $1,450 per month.

What will the total cost of his purchase be?

\[B_n = B_{n-1}(.2245)-1450\]

when Bn = 0 or less, we would have the nth payment and can calculate the overall monies paid is what im thinking

Answer 9

B0 = 4120

B1 = 4120(.2245) - 1450

B2 = (4120(.2245) - 1450)(.2245) - 1450
= 4120(.2245)^2 - 1450(.2245) - 1450

B3 = 4120(.2245)^3 - 1450(.2245)^2 - 1450(.2245) - 1450
= 4120(.2245)^3 - 1450 (.2245^2 + .2245 + 1)

B4 = 4120(.2245)^4 - 1450 (.2245^3 + .2245^2 + .2245 + 1)

\[B_n=4120(.2245)^n -1450(.2245^{n-1} + .2245^{n-2}+... + .2245 + 1)\]

\[B_n=4120(.2245)^n -1450\frac{1-.2245^{n}}{1-.2245}\]

right?

Answer 10

You're taking a summation approach..

Answer 11

yes, a recursive approach is how i might surmise it :)

Answer 12

when Bn = 0 or less we would have the nth payment and can deduce the amount from that in my idea

Answer 13

For @weaver125 's second question I was thinking i=prt, where i is the interest paid, p is the principle amount, r is the rate and t is the term of the loan. Add the interest to the loan amount plus the down.

Answer 14

Initially when I looked at this problem I imagined it as one that a simple i=prt or some other simplistic interest formula would handle. I honestly did not expect it to go this far. :)

Answer 15

i = prt is simple interest

Answer 16

Right, you are offered no payments (for the second question) just a simple "total". Loan amount + interest paid + down payment = total is what I was thinking.

Answer 17

without payments, we might be able to assume that total compounded at the end of the time limit is the amount to be paid ...

Answer 18

Flint purchased a boat for $10,815.

He made a down payment of $1,175.

He applied for a six-year installment loan with an interest rate of 9.4% in the amount of $9,640.

What is the total cost of the boat after six years?

1175 + 9640(1.094)^6 is my guess

Answer 19

That is exactly what I was thinking. :)