SOLUTION: Pat invested a total of $3,000. Part of the money yields 10 percent interest per year, and the rest yields 8 percent interest per year. If the total yearly interest from this inv

Algebra ->  Equations -> SOLUTION: Pat invested a total of $3,000. Part of the money yields 10 percent interest per year, and the rest yields 8 percent interest per year. If the total yearly interest from this inv      Log On


   

Question 118385: Pat invested a total of $3,000. Part of the money yields 10 percent interest per year, and the rest yields 8 percent interest per year. If the total yearly interest from this investment is $256, how much did Pat invest at 10% and how much at 8%?
I assume this problem uses the I=P*RT equation, but with a twist. I have tried to come up with several formulas with no avail.
Answer by ptaylor(2198) About Me  (Show Source):
You can put this solution on YOUR website!

YOU ARE RIGHT ON!!!!!!!!!!!!!!!
I=PRT IS THE FORMULA THAT NEEDS TO BE USED AND WE'LL USE IT:
Let P=amount of money invested at 10%
Then (3000-P)=amount invested at 8%
Interest on the money invested at 10% for 1 year=P*0.10*1 (the formula)
Interest on the money invested at 8% fro 1 year=(3000-P)*0.08*1
Now we are told that the above interest adds up to $256, so our equation to solve is:
0.10P+0.08(3000-P)=256 get rid of parens
0.10P+240-0.08P=256 subtract 240 from both sides
0.10P+240-240-0.08P=256-240 collect like terms
0.02P=16 divide both sides by 0.02
P=$800-----------------------------------------amount invested at 10%
3000-P=3000-800=$2200----------------------------amount invested at 8%
CK
800*0.10+2200*0.08=256
80+176=256
256=256

Hope this helps---ptaylor