SOLUTION: A chemist is mixing three chemicals to get 15 liters of a 6% solution. How much of each of 3%, 7%, and 9% solutions should he use if he must use twice as much of the 3% as the 9% s

Question 512137: A chemist is mixing three chemicals to get 15 liters of a 6% solution. How much of each of 3%, 7%, and 9% solutions should he use if he must use twice as much of the 3% as the 9% solution?
Found 2 solutions by Maths68, josmiceli:
Answer by Maths68(1474) (Show Source):
You can put this solution on YOUR website!
Solution A
Amount = x
Concentration =9% = 0.09
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Solution B
Amount = 2x
Concentration =3% = 0.03
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Solution C
Amount = 15-3x
Concentration =7% = 0.07
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Resultant Solution
Amount =15
Concentration = 6%=0.06
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[Amount Solution A * Concentration A] + [Amount Solution B * Concentration of B]+ [Amount Solution B * Concentration of B] = Amount of Resultant * Concentration of resultant
(0.09)x+(0.03)2x+(0.07)(15-3x)=(0.06)(15)
0.09x+0.06x+1.05-0.21x=0.9
0.09x+0.06x-0.21x=0.9-1.05
0.15-0.21x=-0.15
-0.06x/0.06=-0.15/-0.06
x=2.5
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Solution A
Amount = x = 2.5 liters
Concentration =9% = 0.09
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Solution B
Amount = 2x =2(2.5)= 5 liters
Concentration =3% = 0.03
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Solution C
Amount = 15-3x =15-3(2.5)=15-7.5=7.5 liters
Concentration =7% = 0.07
================================================================================

Result
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He has to mix 5 liters of 3%, 7.5 liters of 7%, and 2.5 liters of 9% solutions to get 15 liters of a 6% solution

Answer by josmiceli(9680) (Show Source):
You can put this solution on YOUR website!
Let = liters of 3% solution needed
Let = liters of 7% solution needed
Let = liters of 9% solution needed
given:
(1)
(2)
(3)
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There are 3 equations and 3 unknowns, so it's solvable
(2)
(2)
(2)
Multiply both sides of (1) by
and subtract (2) from (1)
(1)
(2)
(4)
Substitute (3) into (4)
(4)
(4)
(4)
and, since
(3)
(3)
and
(1)
(1)
(1)
(1)
5 liters of 3% solution are needed
7.5 liters of 7% solution are needed
2.5 liters of 9% solution are needed
check answer:
(2)
(2)
(2)
(2)
OK