document.write( "Question 1000209: A man invests his savings in two accounts, one paying 6 percent and the other paying 10 percent simple interest per year.
\n" ); document.write( "He puts twice as much in the lower-yielding account because it is less risky. His annual interest is 3960 dollars.
\n" ); document.write( "How much did he invest at each rate?
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Algebra.Com's Answer #853967 by ikleyn(53742)\"\" \"About 
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\n" ); document.write( "A man invests his savings in two accounts, one paying 6 percent and the other paying 10 percent simple interest per year.
\n" ); document.write( "He puts twice as much in the lower-yielding account because it is less risky. His annual interest is 3960 dollars.
\n" ); document.write( "How much did he invest at each rate?
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\n" ); document.write( "\n" ); document.write( "        The answer in the post by @mananth, 10,000 at 10% and 20,000 at 6%, is incorrect\r
\n" ); document.write( "\n" ); document.write( "        I came to bring a correct solution.\r
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\n" ); document.write( "\n" ); document.write( "investment in 10% --------x
\n" ); document.write( "Investment in 6% ----------2x
\n" ); document.write( "10%x+6%(2x) = 3960
\n" ); document.write( "multiply by 100
\n" ); document.write( "10x + 12x = 396000
\n" ); document.write( "22x = 396000
\n" ); document.write( "x = 396000/22\r
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\n" ); document.write( "\n" ); document.write( "x=18000\r
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\n" ); document.write( "\n" ); document.write( "ANSWER.     18,000 at 10% and 36,000 at 6%\r
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\n" ); document.write( "\n" ); document.write( "CHECK.         0.1*18000 + 0.06*(2*18000) = 3960.         ! correct !\r
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\n" ); document.write( "\n" ); document.write( "Solved correctly.\r
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