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.
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.
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
To solve this problem, we can use a system of equations. Let's denote the amount invested at 8% as 'x' and the amount invested at 6% as '16000 - x' (since the total investment is $16000).
We know that the interest earned from the 8% investment can be calculated as 0.08x (since 8% is equivalent to 0.08), and the interest earned from the 6% investment is 0.06(16000 - x) (since 6% is equivalent to 0.06).
According to the problem, the total interest earned is $1180. Therefore, we can set up the following equation:
0.08x + 0.06(16000 - x) = 1180
Let's solve this equation:
0.08x + 0.06(16000 - x) = 1180
0.08x + 960 - 0.06x = 1180
0.02x + 960 = 1180
0.02x = 220
x = 220 / 0.02
x = 11000
Therefore, Mr. Mitchell invested $11000 at 8% and the remaining amount ($16000 - $11000 = $5000) at 6%.