Question 30976: Joe has $40,000 to invest. He invests part of it at 5%, one-fourth of this amount at 6%, and the rest of the money at 7%. His total annual interest income is $2,530. Find the amount invested at each rate.
Answer by blubunny01(20)   (Show Source):
You can put this solution on YOUR website! Let's first define three variables to keep track of the amounts invested at each rate.
x = amount invested at 5%
y = amount invested at 6%
z = amount invested at 7%
This problem has three unknown values - let's establish the relationships between these unknowns.
We know that the amount invested at 6% (y) is 1/4 the amount invested at 5% (x). Therefore, we can say:
Then, whatever is not invested at 5% or 6% will be invested at 7%. The total amount of money Joe is investing is $40,000. So, we can say:
 (to find the "rest of the money" invested at 7%)
or...
 (substituting the y equation we created earlier)
We know his total annual interest income is $2,530. To find the income, we just take the amounts invested at each rate and multiply them by their respective rates (i.e. since x is the amount invested at 5%, we find the interest income by multiplying 0.05x). This means:
or...
Now, solve for x:
 [simplify equation]
[simplify equation]
 [more simplifying the equation]
 [combine like terms]
 [combine like terms]
 [divide both sides by -0.0225 to isolate x]
Going back to our original equations:
x = amount invested at 5%
y = amount invested at 6% = (1/4)x
z = amount invested at 7% = 40,000 - x - y
We have x = 12,000.
y = (1/4)(12,000) = 3,000.
z = 40,000 - 12,000 - 3,000 = 25,000.
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