SOLUTION: Help set up and solve
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.
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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?
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Answer by stanbon(75887) (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 amount invested by Jane at 12% be "x"
and amount invested by 10% be "y".
EQUATION:
1st: 0.12x +0.10y=130
1st: 12x+10y=13000
Amount invested by Randy at 12% is "y"
Amount invested by Randy at 10% is "x"
EQUATION
2nd: 0.10x+0.12y=134
2nd: 10x+12y=13400
Multiply 1st by 10 and 2nd by 12:
1st: 120x+100y=130000
2nd: 120x+144y=160800
Subtract 1st from 2nd to get:
44y=30800
y=700
Substitute this value into 1st or 2nd to solve for "x".
120x+100(700)=130000
120x=130000=60000
x=500
You now know x and y.
You can figure out how much is invested by Jane and by Randy
at the two raes.
Cheers,
Stan H.
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