document.write( "Question 1171078: Astro Boy won $22 000 at a weight lifting competition. He decided to invest his money
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document.write( "in two different accounts. One fund earns 8.5% interest, the other earns 4% interest. If
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document.write( "Astro Boy earned $1 510 in interest in one year in total, how much was invested in each
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document.write( "account.
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Algebra.Com's Answer #796014 by greenestamps(13208) You can put this solution on YOUR website! \n" ); document.write( "A setup for a formal algebraic solution.... \n" ); document.write( "x = amount invested at 8.5% \n" ); document.write( "22000-x = amount invested at 4% \n" ); document.write( "The total interest was $1510: \n" ); document.write( " \n" ); document.write( "Solve using basic algebra... though the decimals make the calculations a bit unpleasant. \n" ); document.write( "I leave it to you to finish the problem by that path. \n" ); document.write( "Here is a method that gets you to the answer with what I think is far less effort.... \n" ); document.write( "(1) $22000 all invested at 8.5% would yield $1870 interest; all at 4% would yield $880 interest. \n" ); document.write( "(2) Look at the three interest amounts on a number line -- 880, 1510, and 1870 -- and determine with simple calculations that $1510 is 630/990 = 7/11 of the way from $880 to $1870. \n" ); document.write( "(3) That means 7/11 of the total was invested at the higher rate. \n" ); document.write( "ANSWER: 7/11 of $22,000, or $14,000, was invested at 8.5%; the other $8000 at 4%. \n" ); document.write( "CHECK: .085(14000)+.04(8000) = 1190+320 = 1510 \n" ); document.write( " \n" ); document.write( " |