1. Financial Planning

Wendy, 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?

2. Originally Posted by symmetry
Wendy, 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 = part of the $50,000 that is to be invested for 15 percent per year. So, 50,000 -x = the part to be invested for 7 percent per year. Interest = principal * rate x(0.15) +(50,000 -x)(0.07) = 6000 0.15x +3500 -0.07x = 6000 0.15x -0.07x = 6000 -3500 0.08x = 2500 x = 2500/0.08 x = 31,250 dollars. Hence, 50,000 -x = 50,000 -31,250 = 18,750 dollars. Therefore, to earn$6000 in interest per year, $31,250 should be invested for 15 percent per year, and$18,750 for 7 percent per year. -------answer.

3. ok

Is there a way to answer this question without setting up an equation and solving for x?

4. Originally Posted by symmetry
Is there a way to answer this question without setting up an equation and solving for x?
Umm, if we believe in "Nothing is impossible", then probably there is, are, other ways to solve this kind of problem without using equations, and/or without using unknowns.

But, sorry, I do not know of any.

5. ok

Question:

What portion of the question indicated that x should be subtracted from 5000?

This sort of question is more involved that just the simple I = pr, right?

6. Originally Posted by symmetry
Question:

Why portion of the question indicated that x should be subtracted from 5000?

This sort of question is more involved that just the simple I = pr, right?
There are two investments. The monet at hand is $50,000. So if x is one of the two investments, then the other must be ($50,000 -x)

Or, like this.
Two investments = \$50,000 initially.
Let x = one of them
And Y = the other.

x +y = 50,000
So,
y = 50,000 -x.

Therefore, the two investments are x and y, or, x and (50,000 -x).

In using the x and (50,000 -x) only one unknown, the x, is involved. And playing with one unknown, instead of two unknowns, on the same game/problem, is, I think, simpler or easier.
Of course, with two unknowns, the game will transform to a single unknown eventually or momentarily. One unknown at a time. Cannot solve for the two unknowns in the same instance.