SOLUTION: please help... bob invested $6100 in two funds paying 7% and 6% simple interest. the combined interest for both funds was $405. how much was invested at each rate? solution A

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: please help... bob invested $6100 in two funds paying 7% and 6% simple interest. the combined interest for both funds was $405. how much was invested at each rate? solution A      Log On


   

Question 86476: please help...
bob invested $6100 in two funds paying 7% and 6% simple interest. the combined interest for both funds was $405. how much was invested at each rate?
solution A is 35% alcohol. solution B is 60% alcohol. how many liters of each would be needed to obtain 10 liters of a mixture that is 40% alcohol?
Answer by checkley75(3666) About Me  (Show Source):
You can put this solution on YOUR website!
.07X+.06(6,100-X)=405
.07X+366-.6X=405
.01X=405-366
.01X=39
X=39/.01
X=3900 INVESTED @ 7%
6100-3900=2200 INVESTED @ 6%
PROOF
.07*3900+.06*2200=405
273+132=405
405=405