document.write( "Question 87452: Bob invested $ 30,00 part at 10% and part at 1%. If the total interest at the end of the year is $1,560 how much did he invest at 10% \n" ); document.write( "
Algebra.Com's Answer #63373 by checkley75(3666)![]() ![]() ![]() You can put this solution on YOUR website! I'LL ASSUME THE AMOUNT TO BE INVESTED IS $30,000 THEN: \n" ); document.write( ".10X+.01(30,000-X)=1,560 \n" ); document.write( ".10X+300-.01X=1,560 \n" ); document.write( ".09X=1,560-300 \n" ); document.write( ".09X=1,260 \n" ); document.write( "X=1,260/.09 \n" ); document.write( "X=14,000 INVESTED @ 10%. \n" ); document.write( "30,000-14,000=16,000 INVESTED @ 1% \n" ); document.write( "PROOF \n" ); document.write( ".10*14,000+.01*1,600=1,560 \n" ); document.write( "1,400+160=1,560 \n" ); document.write( "1,560=1,560\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |