document.write( "Question 1172239: Solve the problems in one variable.\r
\n" ); document.write( "\n" ); document.write( "A company decided to invest P100, 000 which was partly invested on a bank account paying 8% per annum and the remainder was invested in stocks paying 11% per annum. If the total yearly interest from the two investments is P10, 100, how much was invested in each? 3 points.
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Algebra.Com's Answer #797216 by greenestamps(13200)\"\" \"About 
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\n" ); document.write( "x = amount invested at 8%
\n" ); document.write( "100000-x = amount invested at 11%

\n" ); document.write( "The total interest is 10100:

\n" ); document.write( "\".08%28x%29%2B.11%28100000-x%29+=+10100\"

\n" ); document.write( "Solve using basic algebra....

\n" ); document.write( "Doing the above is a good exercise in solving problems using formal algebra.

\n" ); document.write( "For a quick and easy way to answer a \"mixture\" problem like this, you can do this:

\n" ); document.write( "(1) The 100,000 all invested at 8% would earn 8000 interest; all at 11% would earn 11,000 interest; the actual interest is 10,100.
\n" ); document.write( "(2) View the three numbers on a number line and use simple calculations to determine that 10,100 is 7/10 of the way from 8000 to 11,000.
\n" ); document.write( "(3) That means 7/10 of the total 100,000 was invested at the higher rate.

\n" ); document.write( "ANSWER: 7/10 of the total 100,000, or 70,000 was invested at 11%; the other 30,000 at 8%.

\n" ); document.write( "CHECK: .11(70,000)+.08(30,000) = 7700+2400 = 10,100

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