1. ## 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?

2. ## Re: Financial Planning

Originally Posted by harpazo
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.

3. ## Re: Financial Planning

Originally Posted by harpazo
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]

4. ## 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"!

5. ## 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?

Keerect!

7. ## 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.