$5400 is invested, part of it at 12% and part of it at 9%. For a certain year, the total yield is 576.00. How much was invested at each rate?

1 Answer

#$3000# was invested at #12%# and #$2400# was invested at #9%#

Explanation:

Let the money invested at #12%# be #$x#. Interest on it for a year will be #12/100 xx x#.

Then money invested at #9%# would be #$(5400-x)# and interest on it for a year will be #9/100 xx (5400-x)#.

As total yield is #576#, we have

#(12x)/100+(9(5400-x))/100=576# - multiplying both sides by #100#

or #12x+9(5400-x)=57600#

or #12x+48600-9x=57600#

or #3x=57600-48600=9000#

i.e. #x=9000/3=3000# and #5400-x=2400#

Hence, #$3000# was invested at #12%# and #$2400# was invested at #9%#