SOLUTION: The annual interest on a $14,000 investment exceeds the interest earned on a $7000 investment by $294. The $14,000 is invested at a 0.6% higher rate of interest than the $7000. W

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Question 1196412: The annual interest on a $14,000 investment exceeds the interest earned on a $7000 investment by $294. The $14,000 is invested at a 0.6% higher rate of interest than the $7000. What is the interest rate of each investment?
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39615) About Me  (Show Source):
You can put this solution on YOUR website!
If simple interest, one year, r the rate as decimal for the part invested for $7000
Then amount invested at the 14000 is r+0.06.

Interest from the $14000 MINUS interest from the $7000 EQUALS $294.

14000%28r%2B0.006%29-7000r=294
Solve this for r.

Answer by ikleyn(52769) About Me  (Show Source):
You can put this solution on YOUR website!
.
The annual interest on a $14,000 investment exceeds
the interest earned on a $7000 investment by $294.
The $14,000 is invested at a 0.6% higher rate of interest than the $7000.
What is the interest rate of each investment?
~~~~~~~~~~~~~~ 

Let x be the annual interest rate of the $7000 investment (in decimal form).

Then the annual interest rate of the $14000 investment is (x+0.006), 
according to the problem.


Having it, we write an equation expressing the difference of two annual
interest amounts


    14000*(x+0.006) - 7000x = 294  dollars.


Simplify it and find x

    14000x + 14000*0.006 - 7000x = 294

    7000x = 294 - 14000*0.006

    7000x = 210

        x = 210%2F7000 = 0.03.


ANSWER.  $7000 invested at 3%;  $14000 invested at 3.6%.

Solved.