Question 419077
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Dave Horn invested half of his money at 5%, one-tird of his money at4%, and the rest at 3.5%. 
If his total annual investment income is $530, how much had he invested?
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        @mananth incorrectly interpreted the problem and incorrectly treated it.

        Therefore, his solution and his answer both are incorrect.

        I came to bring a correct solution.



<pre>
percent    part   
  5        1/2
  4        1/3
  3.5      the rest = 1 - 1/2 - 1/3 = 1 - 5/6 = 1/6


Let x be the total investment.

Write an equation for the total interest

    {{{0.05*(x/2)}}} + {{{0.04*(x/3)}}} + {{{0.035*(x/6)}}} = 530 dollars.


It is your setup equation.  Now you should find the unknown x from it.
For it, first multiply this equation by 6 to rid the denominators.  You will get

    0.05*(3x) + 0.04*(2x) + 0.035x = 6*530,

    0.15x + 0.08x + 0.035x = 3180.

            0.265x         = 3180,

                 x         = 3180/0.265 = 12000.


<U>ANSWER</U>.  $6000 was invested at 5%;  $4000 was invested at 4%,  and  $2000 was invested at 3.5%.


<U>CHECK</U>.  0.05*6000 + 0.04*4000 + 0.035*2000 = 530  dollars.   ! correct !
</pre>

Solved correctly.