# SOLUTION: A broker has invested \$17,500 in two mutual funds, one earning 10% annual interest, and the other earning 12%. After 1 year, his combined interest is \$1,941. how much was invested

Algebra ->  Algebra  -> Percentages: Solvers, Trainers, Word Problems and pie charts -> SOLUTION: A broker has invested \$17,500 in two mutual funds, one earning 10% annual interest, and the other earning 12%. After 1 year, his combined interest is \$1,941. how much was invested      Log On

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 Question 102271: A broker has invested \$17,500 in two mutual funds, one earning 10% annual interest, and the other earning 12%. After 1 year, his combined interest is \$1,941. how much was invested at each rate? Fill up the given table, set up and equation and solve. Mutual fund 1 Mutual fund 2 Answer by ptaylor(2048)   (Show Source): You can put this solution on YOUR website!Interest(I)=Principal(P) times Rate(R) times Time(T) or I=PRT Let P=amount invested at 12% (1) Interest earned at 12%=P*0.12*1=0.12P Then (17,500-P)=amount invested at 10% (2) Interest earned at 10%=(\$17,500-P)*0.10*1=0.10(\$17,500-P) Now we are told that (1)+(2)=\$1,941, so our equation to solve is: 0.12P+0.10(\$17,500-P)=\$1,941 get rid of parens 0.12P+\$1,750-0.10P=\$1,941 subtract \$1,750 from both sides 0.12P+\$1,750-\$1,750-0.10P=\$1,941-\$1,750 collect like terms 0.02P=\$191 divide both sides by 0.02 P=\$9,550-------------------------------amount invested at 12% (\$17,500-P)=\$17,500-\$9,550)=\$7,950----------amount invested at 10% CK \$9550*0.12+\$7950*0.10=\$1941 \$1146+\$795=\$1941 \$1941=\$1941 Hope this helps----ptaylor