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

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(2198) (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

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