Lorna invests $4000, with some at 6% annual interest and some at 11% annual interest. If she receives a total of $580 from these investments at the end of a year, how much was invested at the 6% interest rate?	Choices
A) $160 B) $1,600 C) $800 D) $4,400 E) $5,840
We have that 'x' invested at 6%   We also have that 4000-x' invested at 11%   Now then 580/4000 is approx 10%   Therefore, -4x+4000-x=0   5x=4000   x=800 approximately. Probably higher   Only $800 matches
<result>C</result>
Lorna invests $4000, with some at 6% annual interest and some at 11% annual interest. If she receives a total of $580 from these investments at the end of a year, how much was invested at the 6% interest rate?