SOLUTION: Isabel invested $8000. part in an account paying 6% interest and part in bonds paying 8%. If she has reversed the amounts invested, she would have received $100 less. How much did

Algebra ->  Percentage-and-ratio-word-problems -> SOLUTION: Isabel invested $8000. part in an account paying 6% interest and part in bonds paying 8%. If she has reversed the amounts invested, she would have received $100 less. How much did       Log On


   

Question 1142975: Isabel invested $8000. part in an account paying 6% interest and part in bonds paying 8%. If she has reversed the amounts invested, she would have received $100 less. How much did she invest in cash?
Answer by ikleyn(52832) About Me  (Show Source):
You can put this solution on YOUR website!
.
Let x = invested at 6%.

Then the amount invested at 8% is  (8000 -x).


The total interest in this case is  0.06x + 0.08*(8000-x).



In the reversed case, the total interest is  0.08x + 0.06*(8000-x).


The difference is $100:


    0.06x + 0.08*(8000-x) = 0.08x + 0.06*(8000-x) + 100.


Simplify and solve for x:


    0.06x + 640 - 0.08x = 0.08x + 480 - 0.06x + 100


    640 - 0.02x = 0.02x + 580

    640 - 580 = 0.02x + 0.02x

    60        = 0.04x

     x        = 60%2F0.04 = 1500.


ANSWER.  $1500 invested at 6% and the rest 8000-1500 = 6500 dollars invested at 8%.


CHECK.  0.06*1500 + 0.08*6500 = 610 dollars.

        0.06*6500 + 0.08*1500 = 510 dollars.

        The difference is exactly $100  ---->   hence, the solution is correct !