Question 1190822
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A man invested part of P10, 000 at 6% and the rest at 7%. 
The annual income from the 7% investment was P35 less than seven times 
the annual income from the 6 % investment. How much did he invest at each rate?
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<pre>
Let x be invested at 7%;  then the rest,  (10000-x) is invested at 6%.


The annual interest of the 7% investment is  0.07x.

The annual interest of the 6% investment is  0.06*(10000-x).


The problem says that


    0.07x = 7*0.06*(10000-x) - 35.


From this equation


    x = {{{(7*0.06*10000-35)/(0.07 + 7*0.06)}}} = 8500.


<U>ANSWER</U>.  P 8500 invested at 7% and the rest, 10000-8500 = 1500 invested at 6%.


<U>CHECK</U>.   0.07*8500 = 595;   0.06*1500 = 90;   7*90 - 595 = 630 - 595 = 35,  the difference.    ! Correct !
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Solved.