document.write( "Question 1172239: Solve the problems in one variable.\r
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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. \n" );
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Algebra.Com's Answer #797216 by greenestamps(13200)![]() ![]() You can put this solution on YOUR website! \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( " \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 \n" ); document.write( " \n" ); document.write( " |