document.write( "Question 1191434: Jolene invests her savings in two bank accounts, one paying 5 percent and the other paying 8 percent simple interest per year. She puts twice as much in the lower-yielding account because it is less risky. Her annual interest is 6498 dollars. How much did she invest at each rate?\r
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document.write( "Amount invested at 5 percent interest is $\r
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document.write( "Amount invested at 8 percent interest is \n" );
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Algebra.Com's Answer #823239 by greenestamps(13203)![]() ![]() You can put this solution on YOUR website! \n" ); document.write( "A standard straightforward algebraic solution.... \n" ); document.write( "let x = amount invested at 8% \n" ); document.write( "then 2x = amount invested at 5% \n" ); document.write( "The total interest was $6498: \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "ANSWER: x=$36,100 at 8%; 2x=$72,200 at 5%. \n" ); document.write( "I personally would use a slightly different path which keeps the numbers I need to work with a bit smaller -- allowing a mental solution if your mental arithmetic is good. \n" ); document.write( "Twice as much invested at 5% as at 8% means an average interest percentage rate of (2(5)+8)/3=18/3=6. $6498 interest at 6% means the total amount invested was $6498/.06 = $108,300. \n" ); document.write( "The amount invested at 8% was one-third of that, which is $36,100; the amount invested at 5% was twice that, or $72,200. \n" ); document.write( " \n" ); document.write( " |