A sum of $3000 was invested, part at 6% and the rest at 9%. How much was invested at each rate if the interest on each investment is the same?

1 Answer
Sep 7, 2015

Answer:

Amount invested at 6% yield# = 1800#
Amount invested at 9% yield# = 1200#

Explanation:

Amount invested at 6% yield# = x#
Let the yield be #=y#
Then # x xx 6/100 = y#
# (6x)/100 = y# -----------------------(1)

Amount invested at 9% yield # = 3000 - x#
Let the Yield be # = y# [ since yield from both investment is the same)
Then
#(3000-x)9/100=y# -----------------(2)

Substitute #y = (6x)/100# in equation (2)
#(3000-x)9/100=(6x)/100# [multiplying both sides by 100

#(3000-x)9=(6x)#
#27000-9x -6x=0#
#-15x=-27000#
#x=27000/15 = 1800#
Amount invested at 6% yield# = 1800#
Amount invested at 9% yield# = 3000-1800=1200#