the annual interest on an 18,000 investment exceeds the interest earned on a 9000 investment by 648. the 18,000 is invested at a 0.6% higher rate of interest than the 9000. what is the interest rate of each investment
Let x=annual interest rate of 18000 investment. Then 18000*x - 9000(x-0.006)=648 Solve for x 18000x-9000x+9000*0.006=648 9000x=648-54=594 x=594/9000=0.066 Answer: the annual interest rate of the 18000 investment is 6.6% the annual interest rate of the 9000 investment is 6.0%
