SOLUTION: A chemist has two large containers of salt solution. The concentration of the salt is different in the two containers. She blends 80mL of the first solution with 320 mL of the se

Question 356917: A chemist has two large containers of salt solution. The concentration of the salt is different in the two containers. She blends 80mL of the first solution with 320 mL of the second solution to obtain a solution with concentration 56g/mL. She blends 300 mL of the first solution with 200 mL of the second solution to obtain a solution with concentration 48g/mL. What are the concentrations of the original solutions.
Found 2 solutions by mananth, robertb:
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
A chemist has two large containers of salt solution. The concentration of the salt is different in the two containers. She blends 80mL of the first solution with 320 mL of the second solution to obtain a solution with concentration 56g/mL. She blends 300 mL of the first solution with 200 mL of the second solution to obtain a solution with concentration 48g/mL. What are the concentrations of the original solutions.
Solution A concentration be x%
solution B concentration be y%
..
0.8x+3.2y=400*0.56
0.8x+3.2y= 224
multiply by 10
8x+32y=2240....................1
3x+2y= 240......................2
multiply by -16 and add to first equation
-48x-32y=-3840
add
-40x=-1600
x= 40% solution A
plug value of x in equation 2
120+2y=240
2y=120
y=60% solution B
...
m.ananth@hotmail.ca
Answer by robertb(5830) (Show Source): You can put this solution on YOUR website!
Let x = concentration, in g/mL of the first salt solution.
Let y = concentration, in g/mL of the second salt solution.
Then the pertinent equations are:
80x + 320y = 56*400 = 22,400, and
300x + 200y = 48*500 = 24,000.
solving the system simultaneously, we get x = 40 g/mL, and y = 60 g/mL.

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