Question

$5400 is invested, part of it at 12% and part of it at 9%. For a certain year, the total yield is 576.00. How much was invested at each rate?

Answer 1

`$3000` was invested at `12%` and `$2400` was invested at `9%`

Explanation:

Let the money invested at `12%` be `$x`. Interest on it for a year will be `12/100 xx x`.

Then money invested at `9%` would be `$(5400-x)` and interest on it for a year will be `9/100 xx (5400-x)`.

As total yield is `576`, we have

`(12x)/100+(9(5400-x))/100=576` - multiplying both sides by `100`

or `12x+9(5400-x)=57600`

or `12x+48600-9x=57600`

or `3x=57600-48600=9000`

i.e. `x=9000/3=3000` and `5400-x=2400`

Hence, `$3000` was invested at `12%` and `$2400` was invested at `9%`