document.write( "Question 980087: A & B take equal amount of loan. A pays 8% Interest two times per year while b pays 12% Interest three times per year. After 17 years, B pays 1400 more than that of A. What amount they take the loan at starting? \n" ); document.write( "
Algebra.Com's Answer #601294 by Boreal(15235)![]() ![]() You can put this solution on YOUR website! A=Ao{1+r/n}^nt\r \n" ); document.write( "\n" ); document.write( "{1+.08/2}^34+1200={1+.12/3}}^51\r \n" ); document.write( "\n" ); document.write( "1.04^34+ 1200= {1/04}^51 \n" ); document.write( "3.7943 +1200=7.3909\r \n" ); document.write( "\n" ); document.write( "In other words, \n" ); document.write( "A for A=3.7943*Ao \n" ); document.write( "A for B=7.3909*Ao \n" ); document.write( "3.7943Ao+1400=7.3909Ao, since I need to add $1400 to A to match B. \n" ); document.write( "1400=3.5967Ao \n" ); document.write( "Ao=$389.25\r \n" ); document.write( "\n" ); document.write( "389.25{1.04}^34=$1476.94 \n" ); document.write( "389.25(1.04)^51=2876.93, $1400 more within rounding error.\r \n" ); document.write( "\n" ); document.write( "The answer is $389.25 \n" ); document.write( " |