SOLUTION: $6300 is invested, part of it at 10% and part of it at 8%. For a certain year, the total yield is $566.00. How much was invested at each rate?
How much was invested at 10%?
H
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Coordinate Systems and Linear Equations
-> SOLUTION: $6300 is invested, part of it at 10% and part of it at 8%. For a certain year, the total yield is $566.00. How much was invested at each rate?
How much was invested at 10%?
H
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Question 885372: $6300 is invested, part of it at 10% and part of it at 8%. For a certain year, the total yield is $566.00. How much was invested at each rate?
How much was invested at 10%?
How much was invested at 8%?
You can put this solution on YOUR website! $6300 is invested, part of it at 10% and part of it at 8%. For a certain year, the total yield is $566.00. How much was invested at each rate?
How much was invested at 10%?
How much was invested at 8%?
solution
let $ X at 10% and Rs $6300-x at 8%
X(10)/100 +(6300-x)8/100 =$566.00 by LCM
10X -8x +50400/100 =566
10x-8x +50400 = 56600
2x =56600-50400 = 6200
x=6200/2 = $3100
other 6300-3100 = $3200
answer $3100 at 10%
$3200 at 8%
