Asked by Yakima
Mr. Micthell invested $16000, part at 8%, and the remainder at 6%. If the total yearly interest from these investments was $1180, find the amount invested at each rate.
Answers
Answered by
Henry
Invested:
$X @ 8%.
$(16000-X) @ 6%.
X*0.08*1 + (16000-X)*0.06 = $1180.
0.08x + 960 - 0.06x = 1180,
0.02x = 1180 - 960 = 220,
X = 220 / 0.02 = $11,000.
16000-X = 16000 - 11000 = $5,000.
$X @ 8%.
$(16000-X) @ 6%.
X*0.08*1 + (16000-X)*0.06 = $1180.
0.08x + 960 - 0.06x = 1180,
0.02x = 1180 - 960 = 220,
X = 220 / 0.02 = $11,000.
16000-X = 16000 - 11000 = $5,000.
Answered by
Sam Shelly
Invested:
$x@ 8%
$(16000-x)*0.06=$1180
0.08x+960-0.06x=1180
0.02x=1180-960=220
x=220/0.02=$11,000
16000-x=16000-11000=$5,000
$x@ 8%
$(16000-x)*0.06=$1180
0.08x+960-0.06x=1180
0.02x=1180-960=220
x=220/0.02=$11,000
16000-x=16000-11000=$5,000
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