SOLUTION: How can $69,000 be​ invested, part at 12% annual simple interest and the remainder at 11% annual simple​ interest, so that the interest earned by the two accounts will be​ eq

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Question 1193974: How can $69,000 be​ invested, part at 12% annual simple interest and the remainder at 11% annual simple​ interest, so that the interest earned by the two accounts will be​ equal?
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52776)   (Show Source): You can put this solution on YOUR website!
.
How can $69,000 be​ invested, part at 12% annual simple interest and the remainder at 11% annual simple​ interest,
so that the interest earned by the two accounts will be​ equal?
~~~~~~~~~~~~~~ 

x dollars invested at 12%;  the rest (69000-x) dollars invested at 11%.


Write equation as you read the problem

    0.12x = 0.11*(69000-x).


Simplify and find x

    0.12x = 0.11*69000 - 0.11x

    0.12x + 0.11x = 0.11*69000

         0.23x    = 7590

             x    = 7590/0.23 = 33000.


ANSWER.  $33000 invested at 12%;  the rest  69000-33000 = 36000 invested at 11%.


CHECK.  0.12*33000 = 3960 dollars, the annual interest from the 12% investment.

        0.11*36000 = 3960 dollars, the annual interest from the 11% investment.

        the interest is the same, so the answer is correct.

Solved.



Answer by greenestamps(13198)   (Show Source): You can put this solution on YOUR website!


The solution from tutor @ikleyn is a typical formal algebraic solution; you should understand it and know how to do it.

But a bit of logical reasoning gives us an easier way to solve the problem.

If the amounts of interest from the investments at 12% and 11% are equal, then the amounts invested in the two accounts must be in the ratio 11:12. So

11x = amount invested at 12%
12x = amount invested at 11%

The total amount invested was $69,000:

11x+12x=69000
23x=69000
x=3000

ANSWER: The amounts of interest are equal if 11x=$33,000 is invested at 12% and 12x=$36,000 is invested at 11%.
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