document.write( "Question 279618: Investor Company loaned out a total of $36,000, part at 6% interest and part at 9% interest. They reported that the annual earnings from both investments were the same amount that would have been earned by the total loan if it had been invested at 8%. Find the amount loaned at each rate. \n" ); document.write( "
Algebra.Com's Answer #854164 by ikleyn(53742)\"\" \"About 
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\n" ); document.write( "Investor Company loaned out a total of $36,000, part at 6% interest and part at 9% interest.
\n" ); document.write( "They reported that the annual earnings from both investments were the same amount that would have been earned
\n" ); document.write( "by the total loan if it had been invested at 8%. Find the amount loaned at each rate.
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\n" ); document.write( "\n" ); document.write( "        The solution in the post by @mananth is incorrect,\r
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\n" ); document.write( "\n" ); document.write( "        I came to bring a correct and accurate solution.\r
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document.write( "Let x be the amount invested at 9% interest.\r\n" );
document.write( "Then (36000-x) dollars invested at 6%.\r\n" );
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document.write( "Write the total interest equation\r\n" );
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document.write( "    0.09x + 0.06*(36000-x) = 0.08*36000.\r\n" );
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document.write( "Simplify and find x\r\n" );
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document.write( "    0.09x + 0.06*36000 - 0.06x = 0.08*36000,\r\n" );
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document.write( "    0.09x - 0.06x = 0.08*36000 - 0.06*36000,\r\n" );
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document.write( "    0.03x = 0.02*36000\r\n" );
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document.write( "        x = \"%280.02%2A36000%29%2F0.03\" = 24000.\r\n" );
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document.write( "ANSWER.  $24000 was invested at 9% and the rest, 36000-24000 = 12000 dollars, was invested at 6%.\r\n" );
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document.write( "CHECK.   The total interest is  0.09*24000 + 0.06*12000 = 2880 dollars.\r\n" );
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document.write( "         Calculated by another way, it is  0.08*36000 = 2880 follars, the same amount.\r\n" );
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document.write( "         The solution is confirmed to be correct.\r\n" );
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\n" ); document.write( "\n" ); document.write( "Solved correctly.\r
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