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. Randy also invested in
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-> 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. Randy also invested in
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Question 144613: 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?
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%
J's total interest = $130
Also
x = amt R invested at 10%
y = amt R invested at 12%
R's total interest = $134
:
Two equations
.12x + .10y = 130 (J's investments)
.10x + .12y = 134 (R's investments)
:
Multiply the 1st equation by 1.2 and subtract the 2nd equation, find x
.144x + .12y = 156
.10x + .12y = 134
---------------------subtracting eliminates y
.044x = 22
x =
x = $500; J has invested at 12% and R has invested at 10%
:
Find y using (.12x + .10y = 130)
.12(500) + .10y = 130
60 = .10y = 130
.10y = 130 - 60
.1y = 70
y =
y = $700; J has invested at 10% and R has invested at 12%
:
:
Check solution using (.10x + .12y = 134)
.10(500) + .12(700) =
50 + 84 = 134; confirms our solutions
