document.write( "Question 1196477: A woman invest $10 000;part at 9% annual interests, and the rest at 14%. In each case the interest is compounded annually. The total annual income is $1275. How much is invested in each rate? \n" ); document.write( "
Algebra.Com's Answer #829310 by greenestamps(13200)![]() ![]() You can put this solution on YOUR website! \n" ); document.write( "(1) by standard algebra.... \n" ); document.write( "x = amount invested at 9% \n" ); document.write( "10000-x = amount invested at 14% \n" ); document.write( "The total interest was $1275: \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "ANSWER: x = $2500 at 9%; 10000-x = $7500 at 14% \n" ); document.write( "(2) using logical reasoning and mental arithmetic.... \n" ); document.write( "All $10,000 invested at 9% would have yielded $900 interest; all at 14% would have yielded $1400 interest. \n" ); document.write( "Look at the three interest amounts $900, $1275, and $1400 (on a number line, if it helps) and observe/calculate that $1275 is 3/4 of the way from $900 to $1400 (1275-900 = 375; 1400-900 = 500; 375/500 = 3/4). \n" ); document.write( "That means 3/4 of the $10,000 was invested at the higher rate. \n" ); document.write( "ANSWER: $7500 at 14%, the other $2500 at 9% \n" ); document.write( " \n" ); document.write( " |