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Thread: Financial Planning

  1. #1
    Super Member harpazo's Avatar
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    Financial Planning

    My biggest problem continues to be setting up the right equation based on information given in a word problem.

    Betsy, a recent retiree, requires 6000 dollars per year in extra income. She has 50,000 dollars to invest and can invest in B-rated bonds paying 15 percent per year or in a certificate of deposit paying 7 percent per year. How much money should be invested in each to realize exactly 6000 dollars in interest per year?

    Let x = money to be invested in each

    (50,000-x)(0.15) + (50,000)(0.07) = 6000

    Is this the correct set up?
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  2. #2
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    Re: Financial Planning

    Quote Originally Posted by harpazo View Post
    My biggest problem continues to be setting up the right equation based on information given in a word problem.

    Betsy, a recent retiree, requires 6000 dollars per year in extra income. She has 50,000 dollars to invest and can invest in B-rated bonds paying 15 percent per year or in a certificate of deposit paying 7 percent per year. How much money should be invested in each to realize exactly 6000 dollars in interest per year?

    Let x = money to be invested in each

    (50,000-x)(0.15) + (50,000)(0.07) = 6000

    Is this the correct set up?
    What does 50,000-x represent? What does 0.15 represent? What does 50,000 represent? How about 0.07? After you write the equation, ask yourself what everything represents. It may help to create a table:

    $$\begin{array}{c|cc} & \text{Amt Invested} & \text{Interest Rate} \\ \hline \text{B-rated bond} & 50,000-x & 0.15 \\ \text{CoD} & 50,000 & 0.07 \\ \hline \text{Total} & 100,000-x & (50,000-x)(0.15)+(50,000)(0.07)\end{array}$$

    You want the total to invest to be 50,000, not 100,000-x. So, something is not quite right. Let's change the amount invested in certificates of deposit to $x$ instead:


    $$\begin{array}{c|cc} & \text{Amt Invested} & \text{Interest Rate} \\ \hline \text{B-rated bond} & 50,000-x & 0.15 \\ \text{CoD} & x & 0.07 \\ \hline \text{Total} & 50,000 & (50,000-x)(0.15)+(x)(0.07)\end{array}$$

    That looks better.
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  3. #3
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    Re: Financial Planning

    Quote Originally Posted by harpazo View Post
    My biggest problem continues to be setting up the right equation based on information given in a word problem.

    Betsy, a recent retiree, requires 6000 dollars per year in extra income. She has 50,000 dollars to invest and can invest in B-rated bonds paying 15 percent per year or in a certificate of deposit paying 7 percent per year. How much money should be invested in each to realize exactly 6000 dollars in interest per year?

    Let x = money to be invested in each
    There's your first mistake! You will probably want to invest different amounts in each. The can't both be "x"!

    (50,000-x)(0.15) + (50,000)(0.07) = 6000
    And here is your second! If you invest all 50000 at 7% you have nothing left to invest at 15%!

    Let x be the amount you invest at 7%. Then you have 50000- x to invest at 15%. (50000- x)(0.15)+ x(0.07)= 6000. 50000(0.15)= 7500 so this is the same as 7500- 0.15x+ 0.07x= 7500- 0.08x= 6000. -0.08x= -1500. x= 1500/0.08= ?.

    Is this the correct set up?[/QUOTE]
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  4. #4
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    Re: Financial Planning

    0 @ 7% = 0
    40,000 @ 15% = 6,000

    Why invest anything at 7%?
    Plus problem does not prohibit $0

    Problem should be SIMPLY worded this way:
    50,000 is invested at 7% and 15%.
    Interest over 1 year totals 6,000.
    How much was invested at each rate?

    I'll never understand why teachers make up problems
    through inventing a long story...no wonder these
    are called "word problems"!
    Last edited by DenisB; Nov 2nd 2018 at 10:00 AM.
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  5. #5
    Super Member harpazo's Avatar
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    Re: Financial Planning

    x= 1500/0.08

    x = 18, 750

    At 15 percent:

    50,000 - 18,750 = 31,250

    At 7 percent:

    18,750

    Yes?
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  6. #6
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    Re: Financial Planning

    Keerect!
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  7. #7
    Super Member harpazo's Avatar
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    Re: Financial Planning

    I will surely post addition word problems. I will explain my logic leading to a correct or incorrect equation. Translating words to algebraic language is probably the most important math skill to master. In years gone by, I recall taking state exams for jobs in different fields. My inability to solve word problems kept me away from success. However, I managed to pass the ASVAB with a 55 percent in 1995 and the LAST with a score of 224 (passing score is 220) between 2000 and 2003.
    Last edited by harpazo; Nov 3rd 2018 at 06:13 AM.
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