SOLUTION: The manager of a cleaning supply company wants to make 12 gallons of mildew stain remover that is 25% chlorine. To make the new cleaner, he'll use a 5% chlorine liquid and an 85%

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: The manager of a cleaning supply company wants to make 12 gallons of mildew stain remover that is 25% chlorine. To make the new cleaner, he'll use a 5% chlorine liquid and an 85%       Log On


   

Question 52992: The manager of a cleaning supply company wants to make 12 gallons of mildew stain remover that is 25% chlorine. To make the new cleaner, he'll use a 5% chlorine liquid and an 85% chlorine concentrate. How much of each should he use to make 12 gallons of a 25% chlorine mixture?
Find the equation(s) and solve using substitution or elimination.
Not out of a text book!
Answer by venugopalramana(3286) About Me  (Show Source):
You can put this solution on YOUR website!
see the following and try.if still i difficulty please come back
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A chemist needs 10 liters of a 50% salt solution. All
she has available is a 20% salt solution and a 70%
salt solution. How much of each of the two solutions
should she mix to obtain her desired solution?
LET X LITRES OF 20% SOLUTION BE REQUIRED...SINCE TOTAL
IS 10 LITRES ,WE HAVE TO ADD 10-X LITRES OF 70%
SOLUTION..SO TAKING A SALT BALANCE
SALT IN X LITRES OF 20 % SOLUTION......=X*20/100=0.2X
SALT IN 10-X LITRES OF 10 % SOLUTION..
=(10-X)*70/100=7-0.7X
SALT IN 10 LITRES OF FINAL 50 % SOLUTION =10*50/100=5
HENCE
0.2X+7-0.7X=5
0.7X-0.2X=7-5=2
0.5X=2
X=2/0.5=20/5=4 LITRES OF 20 % SWOLUTION AND 10-4=6
LITRES OF 70% ARE NEEDED TO BE MIXED TO GET 10 LITRES
OF 50% SOLUTION
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SEE THIS EXAMPLE WHICH IS SIMILAR TO YOUR PROBLEM AND
TRY.IF YOU STILL HAVE DIFFICULTY PLEASE COME BACK
In a chemistry class, 7 liters of a 4% silver iodide
solution must be mixed with a 10% solution to get a 6%
solution. How many liters of a 10% solution are
needed?
A. 4.5L LET X LITRES OF 10% SOLUTION BE NEEDED.IT HAS
X*10/100 LITRES OF SILVER IODIDE.
B. 2.5L 7 LITRES OF 4% SILVER IODIDE SOLUTION HAS
7*4/100=28/100 LITRES OF SILVER IODIDE
C. 7.0L TOTAL SOLUTION =X+7 LITRES��.IT HAS TOTAL
10X/100+28/100=(10X+28)/100 LITRES OF SILVER IODIDE
D. 3.5L FINAL CONCENTRATION =6 %�SO
6/100=(10X+28)/(X+7)*100..OR�6=(10X+28)/(X+7)...CROSS
MULTIPLYING..
6(X+7)=(10X+28)�OR�6X+42=10X+28..OR�.10X-6X=42-28=14�OR�.4X=28..OR..X=7
LITRES.C IS THE ANSWER
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