# Math Help - TV or Not TV finding the interest

1. ## TV or Not TV finding the interest

Two banks are offering car loans. You wish to barrow $5000. The fixed payments for each loan are$100 per months. Bank A changes 1 percent interest per month on the unpaid balance, the money you still owe the bank. Bank B charges 1.5 percent per month on the unpaid balance and will throw in a television set valued at $1000. If you could use the TV set, which loan would you pick? 2. Hello, aznmartinjai! A strange problem ... I hope I interpreted it correctly. Two banks are offering car loans. .You wish to borrow$5000.
The fixed payments for each loan are $100 per month. Bank A changes 1 percent interest per month on the unpaid balance. Bank B charges 1.5 percent per month on the unpaid balance and will throw in a television set valued at$1000.
If you could use the TV set, which loan would you pick?
. . . . . . . . . . . . . . . . . . . . . . . i(1 + i)^n
Amortization Formula: . A .= .P ---------------
. . . . . . . . . . . . . . . . . . . . . . (1 + i)^n - 1

where: P = amount of loan
. . . . . .i = perioidic interest rate
. . . . . n = number of periods
. . . . . A = periodic payment

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Bank A: .P = 5000, i = 0.01, A = 100

. . . . . . . . . . . . . . . . . .0.01(1.01)^n
We have: . 100 .= .5000 ----------------
. . . . . . . . . . . . . . . . . . (1.01)^n - 1

Then we have: . (1.01)^n - 1 .= .½(1.01)^n

. . . . . . . . . . . . . ½(1.01)^n .= .1

. . . . . . . . . . . . . . (1.01)^n .= .2

Take logs: . . . . . .n·ln(1.01) .= .ln(2)

. . . . . . . . . . . . . . . . . . . . . . . . ln(2)
. . . . . . . . . . . . . . . . . . . n .= .--------- .= .69.6607...
. . . . . . . . . . . . . . . . . . . . . . . ln(1.01)

You will be paying $100 per month for 70 months . . a total of$7,000.
You will pay $2,000 in interest charges. ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ Bank B: .P = 5000, i = 0.015, A = 100 . . . . . . . . . . . . . . . . . .0.015(1.015)^n We have: . 100 .= .5000 ------------------- . . . . . . . . . . . . . . . . . . (1.015)^n - 1 Then we have: . (1.015)^n - 1 .= .¾(1.015)^n . . . . . . . . . . . . . ¼(1.015)^n .= .1 . . . . . . . . . . . . . . (1.015)^n .= .4 Take logs: . . . . . .n·ln(1.015) .= .ln(4) . . . . . . . . . . . . . . . . . . . . . . . . . ln(4) . . . . . . . . . . . . . . . . . . . . n .= .----------- .= .93.11105... . . . . . . . . . . . . . . . . . . . . . . . . ln(1.015) You will be paying$100 per month for 93 months
. . a total of $9,300. You will pay$4,300 in interest charges.

Even with the \$1000 TV set, you will play more than at Bank A.

Take the loan from Bank A.