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Math Help - TV or Not TV finding the interest

  1. #1
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    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?
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  2. #2
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    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 .= .(1.01)^n

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

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

    Take logs: . . . . . .nln(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 .= .(1.015)^n

    . . . . . . . . . . . . . (1.015)^n .= .1

    . . . . . . . . . . . . . . (1.015)^n .= .4

    Take logs: . . . . . .nln(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.

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