SOLUTION: An investor has part of her money in an account that pays 9% annual interest, and the rest in an account that pays 11% annual interest. If she has $8000 less in the higher paying a
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Question 638642: An investor has part of her money in an account that pays 9% annual interest, and the rest in an account that pays 11% annual interest. If she has $8000 less in the higher paying account than in the lower paying one and her total annual interest income is $2010, how much does she have in each account? Answer by mananth(16946) (Show Source):
You can put this solution on YOUR website! 9%--- x
11%---- x-8000
Interest = 2010
9%*x+11%(x-8000)= 2010
multiply by 100
9x+11(x-8000)= 201000
9x+11x-88000=201000
20x= 201000+88000
x= (201000+88000)/20
x=14450
Investment at 9%= $14,450
Investment at 11%= $6,450
