Expert: Sherry Wallin - 12/23/2008

Question
QUESTION: Hello:

Here is my question:  An investor has $30,000 to invest at simple interest, one portion at 2% and the other at 5%.
How much must he invest in each account to earn a total return of 4% or $1,200?

Answers: 2% at $10,000 and 5% at $20,000

I want to try to determine the answers by using the following:
If all of the $30,000 is invested at 2%, this will equal $600. He is now short $600. (Remember the interest must be $1,200.) Any cash removed from the 2% account and invested in the 5% account will earn an extra 3%.  How much does he need to invest at the remaining 3% to earn the shortfall of $600?

Try to determine the answer by using a simple non-algebra calculation if possible. I already have a calculation using algebra.
I think division may help determine the answer, but I'm not sure.

I thank you for any helpful reply.

ANSWER: Hi Kenneth~
    You can use algebra to find a "non-algebraic" way to your answer. Since you are earning an additional 3% of some number and it has to equal $600, write .03x=600 and thus 600/.03 is the division you need to calculate "non-algebraically" the amount you need to remove from the amount invested at 2% to the account earning 5%. By the way, that amount is $20,000 as you've already indicated  above.

Math Prof

---------- FOLLOW-UP ----------

QUESTION: Hello Math Prof.:

I want to thank you for your reply and willingness to help!

If the situation is reversed and the full amount is invested at 5%, the amount is $1,500. This amount is $300 too much, $,1,500 - $1,200 = $300.
2% X ? = $300 does not produce the correct amount of $10,000. Can you determine the correct amount using the same calculation in your first answer/reply?

I thank you for your assistance.

ANSWER: Hi Kenneth~
    The reason you don't get $10,000 is because you're not using a difference (change) of 3%. (You used 2%). So try it with 3%.

Math Prof

---------- FOLLOW-UP ----------

QUESTION: Hello Math Prof:

I will explain beginning with your first solution:

"Since you are earning an additional 3% of some number and it has to equal $600, write .03x=600 and thus 600/.03 is the division you need to calculate "non-algebraically" the amount you need to remove from the amount invested at 2% to the account earning 5%. By the way, that amount is $20,000 as you've already indicated above."

Disregard my second question your reply. I do not think that it applies to the following question.

Now, suppose the shortfall of $600 is divided by the full amount in the 5% account, that is, divide $600 by 5% instead of 3%, as you explained in your first solution.
$600 divided by 3% = $20,000, the answer. However, $600 divided by 5% = $12,000.

How would the amount of $20,000 be determined? Is it possible?

I hope that I have explained better, and I apologize for any confusion. Most of all, I thank you for your help and assistance regarding my questions.


Answer
Hi Kenneth~
   It is still unclear why you would want to divide by 5%. Essentially you are going to be earning 2% on the first $18,000. The $12,000 that you get by dividing the shortfall of $600 by 5% is how much more you would need to invest at 7% (the 2% already being earned and the 5% you are wanting to divide by to get a total of $1200 interest). This all depends on your goals. It would help if you would tell me what you are trying to accomplish. If your goal is to get $1200 interest on $30,000 then this is just one way to accomplish that goal. (You already know that investing $10,000 at 5% and 20,000 at 2% gives you a total of $1200 interest or an overall return of 4%). If your goal is to use $30,000 in the least risky way and to get $1200 interest then there might be a better way. I assume the 2% is less risky than the 5% or a 7%. Are you learning about financial planning?

Math Prof