Question 1191699
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Here is a non-traditional non-algebraic way to solve any two-part "mixture" problem like this.<br>
All P140,000 invested at 9% would yield P12,600 interest; all at 11% would yield P15,400 interest; the actual interest was P13,900.
Consider the three interest amounts 12600, 13900, and 15400 on a number line and observe/calculate that 13900 is 1300/2800 = 13/28 of the way from 12600 to 15400.
That means 13/28 of the total was invested at the higher rate.<br>
(13/28)(140000)=13(5000)=65000<br>
ANSWER: P65000 at 11%; P75000 at 9%<br>
CHECK: .11(65000)+.09(75000)=7150+6750=13900<br>