SOLUTION: A 25% SOLUTION OF ALCOHOL IS TO BE MIXED WITH A 40% SOLUTION TO GET 50l OF A FINAL MIXTURE THAT IS 30% alcohol. How much of each of the original solution should be used?

Question 54768: A 25% SOLUTION OF ALCOHOL IS TO BE MIXED WITH A 40% SOLUTION TO GET 50l OF A FINAL MIXTURE THAT IS 30% alcohol. How much of each of the original solution should be used?
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
SEE THE FOLLOWING EXAMPLE AND TRY
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PLEASE HELP ASAP: Ziggy's famous yogurt blends regular
yogurt that is 3% fat with its no fat yogurt to obtain
low fat yogurt that is 1% fat. How many pounds of
regular and how many pounds of non-fat yogurt should
be mixed to obtain 60 pounds of lowfat yogurt.
PLEASE HELP ASAP. thank you
THESE ARE MATERIAL BALANCE PROBLEMS.THE PRINCIPLE IS
TO APPLY
TOTAL OF ALL INPUTS =TOTAL OF ALL OUTPUTS..
THIS PRINCIPLE CAN BE APPLIED TO TOTAL MIXTURE AS A
WHOLE AS WELL AS INDIVIDUAL COMPONENTS OF THE
MIXTURE.LET US SEE THE APPLICATION USING YOUR PROBLEM.
HERE THE MIXTURE COMPRISES 2 INPUTS-REGULAR YOGURT
(RY) & NO FAT YOGURT (NFY)
AND ONE OUT PUT-LOW FAT YOGURT (LFY).THE COMPONENT OF
IMPORTANCE IN THE MIXTURE IS FAT CONTENT.SO WE TAKE 2
BALANCES HERE ..ONE FOR THE TOTAL MIXTURE AND ANOTHER
FOR COMPONENT OF FAT IN THE MIXTURE.
I..TOTAL BALANCE...
INPUTS
1.QTY.OF.RY=X POUNDS
2.QTY OF NFY=Y POUNDS
OUT PUT
1.QTY.OF LFY=60 POUNDS
SO APPLYING
TOTAL OF ALL INPUTS =TOTAL OF ALL OUTPUTS.....WE GET
X+Y=60.............................I
II..COMPONENT BALANCE..HERE IT IS FAT .
INPUTS
1.QTY.OF FAT IN RY=X*3/100=3X/100 POUNDS
2.QTY OF FAT IN NFY=Y*0/100=0 POUNDS
OUT PUT
1.QTY.OF FAT IN LFY=60*1/100=60/100 POUNDS
SO APPLYING
TOTAL OF ALL INPUTS =TOTAL OF ALL OUTPUTS.....WE GET
3X/100 + 0=60/100.............................II
3X=60
X=20 POUNDS. OF REGULAR YOGURT
Y=60-20=40 POUNDS OF NO FAT YOGURT.

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