<PRE><B>Question 287786</B>
a certain amount of money is invested at 8%
 and PHP150,000 more than that amount is invested at 9%.
 The annual interest from the 9% investment exceeds the annual interest from 8% investment by PHP16,000. 
How much is invested at 9%? 
:
Not sure what PHP is but assume it will make no difference in the problem
Assume this is based on 1 yr
:
Let x = amt invested at 9%
then
(x-150000) = amt invested at 8%
:
Use the decimal form of the interest rates
:
Int at 9% - int at 8% = 16000
.09x - .08(x - 150000)  = 16000
:
.09x - .08x + 12000  = 16000
:
.01x = 16000 - 12000
x = {{{4000/.01}}}
x = 400,000 invested at 9%
:
:
Check solution by finding the interest at each rate
.09(400000) = 36000
.08(400000-150000) = 20000
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a difference of 16000</PRE>