# SOLUTION: Jane invested some amount at the rate of 12% simple interest and some other amount at the rate of 10% simple interest. She received yearly interest of 130.00. Randy also invested

Algebra ->  Algebra  -> Percentages: Solvers, Trainers, Word Problems and pie charts -> SOLUTION: Jane invested some amount at the rate of 12% simple interest and some other amount at the rate of 10% simple interest. She received yearly interest of 130.00. Randy also invested      Log On

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 Click here to see ALL problems on percentage Question 80556: Jane invested some amount at the rate of 12% simple interest and some other amount at the rate of 10% simple interest. She received yearly interest of 130.00. Randy also invested in the same scheme, but he interchanged the amounts invested, and received \$4 more as interest. How much amount did each of them invest at different rates?Answer by ankor@dixie-net.com(15663)   (Show Source): You can put this solution on YOUR website!Jane invested some amount at the rate of 12% simple interest and some other amount at the rate of 10% simple interest. She received yearly interest of \$130. Randy also invested in the same scheme, but he interchanged the amounts invested, and received \$4. more as interest. How much amount did each of them invest at different rates. : Let x = amt J invested at 12% Let y = amt J invested at 10% : Randy interchanged the amts so: x = amt R invested at 10% y = amt R invested at 12% : .12x + .10y = 130; J's earnings .10x + .12y = 134; R's earnings : Multiply 1st equation by 120 and 2nd equation by 100: 14.4x + 12y = 15600 10x + 12y = 13400 --------------------- subtracting eliminates y 4.4x + 0 = 2200 x = 2200/4.4 x = \$500 invested at 12% by J, \$500 invested at 10% by R : Find y: .12(500) + .10y = 130 60 + .1y = 130 .1y = 130 - 60 y = 70/.1 y = \$700 invested at 10% by J, \$700 invested at 12% by R : : Check our solutions using; .10x + .12y = 134: .1(500) + .12(700) = 50 + 84 = 134 : Make sense to you? Hope so.