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Thread: Two mistakes in my tutorial sheet?

  1. July 7th 2008, 10:31 PM #1
    Finmathlearner
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    Two mistakes in my tutorial sheet?

    I don't get the answers on the sheet. but I think I am right

    13. A man purchases a car on January 1, 1979 by paying a deposit of $3000 and agreeing to pay $500 every 3 months for the next 5 years. Interest is included at the rate of 10% p.a. payable quarterly. He makes no payments at all until April 1, 1980 at which time under pressure of repossession he agrees to fulfil his obligations by regular quarterly payments for the rest of the 5 year period starting immediately. What should the payments be?

    14. A couple pays a total of $1,000 yearly into an investment fund for their children’s education. The fund pays 8% pa compounded quarterly. How much will be in the fund after 18 years?


    Answers on the sheet


    13. $659.04; 14. $38,347.58

    My answers
    13) everything happens 4 quarters later than it should so the 500 payments become 500*(1.025)^4= 551.9064453

    14) 1000* (( 1.02)^4)^18-1)/(1.02^4-1)=38,348.39
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  2. July 8th 2008, 09:46 AM #2
    jonah
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    From January 1, 1979 to January 1, 1980 is 1 year.
    The price of the car is = a down payment/deposit of 3000 + remaining balance of <br /> 500 \cdot \frac{{1 - \left( {1 + \tfrac{{.10}}<br /> {4}} \right)^{ - \left( {5*4} \right)} }}<br /> {{\left( {\tfrac{{.10}}<br /> {4}} \right)}}<br /> <br />
    Since the man did not bother to make any of the $500 quarterly payments until April 1, 1980, the accumulated amount (future value) of his remaining balance on January 1,1980 is
    <br /> \left[ {500 \cdot \frac{{1 - \left( {1 + \tfrac{{.10}}<br /> {4}} \right)^{ - \left( {5*4} \right)} }}<br /> {{\left( {\tfrac{{.10}}<br /> {4}} \right)}}} \right]\left( {1 + \tfrac{{.10}}<br /> {4}} \right)^{\left( {1*4} \right)} <br /> <br />
    Accordingly, “the rest of the 5 year period” is 5-1.
    The new quarterly payments can then be determined by solving for R in the following equation:
    <br /> \left[ {500 \cdot \frac{{1 - \left( {1 + \tfrac{{.10}}<br /> {4}} \right)^{ - \left( {5*4} \right)} }}<br /> {{\left( {\tfrac{{.10}}<br /> {4}} \right)}}} \right]\left( {1 + \tfrac{{.10}}<br /> {4}} \right)^{\left( {1*4} \right)} = R \cdot \frac{{1 - \left( {1 + \tfrac{{.10}}<br /> {4}} \right)^{ - \left[ {\left( {5 - 1} \right)*4} \right]} }}<br /> {{\left( {\tfrac{{.10}}<br /> {4}} \right)}}<br /> <br />
    By my reckoning, the answer on you tutorial sheet is correct.
    Last edited by jonah; July 8th 2008 at 11:08 AM. Reason: disorientation with time reference
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  3. July 8th 2008, 10:39 AM #3
    jonah
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    14) 1000* (( 1.02)^4)^18-1)/(1.02^4-1)=38,348.39
    That is quite correct. There’s a slight typo though. There should be three open parentheses after 1000*.
    It should be 1000* ((( 1.02)^4)^18-1)/(1.02^4-1)=38,348.39.
    Also, $38,348.39 is correct only if the $1,000 payments/deposits are made at the end of the year.
    If the $1,000 payments/deposits are made at the beginning of the year, you’d have to multiply your equation by (1.02)^4. Thus,
    1,000 \cdot \frac{{\left[ {\left( {1.02} \right)^4 } \right]^{18} - 1}}<br /> {{\left( {1.02} \right)^4 - 1}} \cdot \left( {1.02} \right)^4 \approx 41,509.5274<br />
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  4. July 8th 2008, 11:25 AM #4
    Soroban
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    Hello, Finmathlearner!


    14) A couple pays a total of $1,000 yearly into a fund for their children’s education.
    The fund pays 8% p.a. compounded quarterly.
    How much will be in the fund after 18 years?

    Answers on the sheet: .$38,347.58

    My work: . 1000\cdot \frac{(1.02^4)^{18} - 1}{1.02^4-1} \:= \: 38,348.39

    I agree with your formula and your answer!
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  5. July 8th 2008, 12:54 PM #5
    Finmathlearner
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    Quote Originally Posted by jonah View Post
    From January 1, 1979 to January 1, 1980 is 1 year.
    The price of the car is = a down payment/deposit of 3000 + remaining balance of <br /> 500 \cdot \frac{{1 - \left( {1 + \tfrac{{.10}}<br /> {4}} \right)^{ - \left( {5*4} \right)} }}<br /> {{\left( {\tfrac{{.10}}<br /> {4}} \right)}}<br /> <br />
    Since the man did not bother to make any of the $500 quarterly payments until April 1, 1980, the accumulated amount (future value) of his remaining balance on January 1,1980 is
    <br /> \left[ {500 \cdot \frac{{1 - \left( {1 + \tfrac{{.10}}<br /> {4}} \right)^{ - \left( {5*4} \right)} }}<br /> {{\left( {\tfrac{{.10}}<br /> {4}} \right)}}} \right]\left( {1 + \tfrac{{.10}}<br /> {4}} \right)^{\left( {1*4} \right)} <br /> <br />
    Accordingly, “the rest of the 5 year period” is 5-1.
    The new quarterly payments can then be determined by solving for R in the following equation:
    <br /> \left[ {500 \cdot \frac{{1 - \left( {1 + \tfrac{{.10}}<br /> {4}} \right)^{ - \left( {5*4} \right)} }}<br /> {{\left( {\tfrac{{.10}}<br /> {4}} \right)}}} \right]\left( {1 + \tfrac{{.10}}<br /> {4}} \right)^{\left( {1*4} \right)} = R \cdot \frac{{1 - \left( {1 + \tfrac{{.10}}<br /> {4}} \right)^{ - \left[ {\left( {5 - 1} \right)*4} \right]} }}<br /> {{\left( {\tfrac{{.10}}<br /> {4}} \right)}}<br /> <br />
    By my reckoning, the answer on you tutorial sheet is correct.
    Thanks for that I did not read the question carefully I interpreted "rest of the 5 year period" as time as "pay off the loan in 5 years"
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  6. July 8th 2008, 12:55 PM #6
    Finmathlearner
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    Quote Originally Posted by Soroban View Post
    Hello, Finmathlearner!



    I agree with your formula and your answer!
    Thanks for you and jonah for your help
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  7. July 9th 2008, 09:04 AM #7
    jonah
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    13. A man purchases a car on January 1, 1979 by paying a deposit of $3000 and agreeing to pay $500 every 3 months for the next 5 years. Interest is included at the rate of 10% p.a. payable quarterly. He makes no payments at all until April 1, 1980 at which time under pressure of repossession he agrees to fulfil his obligations by regular quarterly payments for the rest of the 5 year period starting immediately. What should the payments be?
    Let’s look at this again from an annuity due perspective.
    From January 1, 1979 to April 1, 1980 is approximately 1.25 years.
    The price of the car is = a down payment/deposit of 3000 + remaining balance of 20 (=5*4) payments of $500 at the end of each 3 months for 5 years as represented by 500 \cdot \frac{{1 - \left( {1 + \tfrac{{.10}}<br /> {4}} \right)^{ - \left( {5*4} \right)} }}<br /> {{\left( {\tfrac{{.10}}<br /> {4}} \right)}}. Since the man did not bother to pay any of the first four $500 quarterly payments until April 1, 1980, the accumulated amount (future value) of his remaining balance on April 1,1980 is \left[ {500 \cdot \frac{{1 - \left( {1 + \tfrac{{.10}}<br /> {4}} \right)^{ - \left( {5*4} \right)} }}<br /> {{\left( {\tfrac{{.10}}<br /> {4}} \right)}}} \right]\left( {1 + \tfrac{{.10}}<br /> {4}} \right)^{\left( {1.25*4} \right)}
    The original agreement called for 20 end of quarter payments. The man failed to pay the first four end of quarter payments of $500. (We assumed the quarterly payments are made at the end of each 3 months because there was a down payment already.) He will then have to make up by making 16 (= 20 – 4) payments (the 1st on April 1,1980 and the 16th at the end of 3.75 years after April 1, 1980) to amortize the accumulated remaining balance on April 1, 1980. 3.75 years because “the rest of the 5 year period” is 5 - 1.25 = 3.75 years. If these 16 payments are to be of equal amount and since the first new unknown payment is due immediately on April 1, 1980, these payments form an annuity due of 16 payments whose present value on April 1, 1980 is \left[ {500 \cdot \frac{{1 - \left( {1 + \tfrac{{.10}}<br /> {4}} \right)^{ - \left( {5*4} \right)} }}<br /> {{\left( {\tfrac{{.10}}<br /> {4}} \right)}}} \right]\left( {1 + \tfrac{{.10}}<br /> {4}} \right)^{\left( {1.25*4} \right)} <br />
    These 16 payments can then be determined by solving for R_1 <br /> in the following equation:
    \begin{gathered}<br /> \left[ {500 \cdot \frac{{1 - \left( {1 + \tfrac{{.10}}<br /> {4}} \right)^{ - \left( {5*4} \right)} }}<br /> {{\left( {\tfrac{{.10}}<br /> {4}} \right)}}} \right]\left( {1 + \tfrac{{.10}}<br /> {4}} \right)^{\left( {1.25*4} \right)} = R_1 + R_1 \cdot \frac{{1 - \left( {1 + \tfrac{{.10}}<br /> {4}} \right)^{ - 15} }}<br /> {{\left( {\tfrac{{.10}}<br /> {4}} \right)}} \hfill \\<br /> \Leftrightarrow \hfill \\<br /> R_1 \approx \$ 659.0392485 \hfill \\ <br /> \end{gathered} <br />
    As you can see, although this approach is just as effective as the ordinary annuity perspective approach, it takes a bit of orientation with a number line. Cheers.
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