document.write( "Question 1129933: sally has money in two accounts. one account pays 5% interest, whereas the other pays 11% annual interest. if she has $400 more invested at 11% than she does at 5% and her total interest for a year is $172, how much does she have in each account? \n" ); document.write( "

Algebra.Com's Answer #746548 by ikleyn(52794) $\"\"$ $\"About$

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document.write( "Let x be the amount invested at 11%, in dollars.\r\n" );
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document.write( "Then the amount invested at 5% is (x-400) dollars.\r\n" );
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document.write( "The condition says\r\n" );
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document.write( "    interest + interest     = total interest,  or\r\n" );
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document.write( "    0.11x    + 0.05*(x-400) = 172\r\n" );
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document.write( "    0.11x + 0.05x - 0.05*400 = 172\r\n" );
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document.write( "    0.16x = 172 + 0.05*400\r\n" );
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document.write( "    x =  = 1200 dollars.\r\n" );
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document.write( "Answer.  1200 invested at  11%,  and  1200 - 400 = 800 dollars invested at 5%.\r\n" );
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document.write( "Check.  0.11*1200 + 0.05*800 = 172 dollars.   ! Correct !\r\n" );
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