Question

$5400 is invested, part of it at 11% and part of it at 9%. For a certain year, the total yield is is 546.00. How much was invested at each rate?

Answer 1

`$2400` invested at `9%`
`$3000` invested at `11%`

Explanation:

We can divide `$5400` by letting `5400-x` be the part invested at `11%` and the other `x` is invested at `9%`. We have the equation
`0.11(5400-x)+0.09x=546`. Solving for `x`. We have
`594-0.11x+0.09x=546`
#48=0.02x x = 2400#
Therefore we have
`$2400` invested at `9%`
`$3000` invested at `11%`