# $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? ##### 1 Answer Nov 19, 2016 $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 $\frac{12}{100} \times x$.

Then money invested at 9% would be $(5400-x) and interest on it for a year will be $\frac{9}{100} \times \left(5400 - x\right)$. As total yield is $576$, we have $\frac{12 x}{100} + \frac{9 \left(5400 - x\right)}{100} = 576$- multiplying both sides by $100$or $12 x + 9 \left(5400 - x\right) = 57600$or $12 x + 48600 - 9 x = 57600$or $3 x = 57600 - 48600 = 9000$i.e. $x = \frac{9000}{3} = 3000$and $5400 - x = 2400$Hence, $3000 was invested at 12% and \$2400 was invested at 9%