Question 29283: If 800 mL of a juice drink is 15% grape juice, how much grape juice should be added to make a drink that is 20% grape juice? Answer by venugopalramana(3286) (Show Source):
You can put this solution on YOUR website! SEE THE FOLLOWING EXAMPLE WHICH WILL HELP YOU TO GET AN INSIGHT INTO THESE PROBLEMS AND THE METHOD OF APPROACH TO SOLVE THEM
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How much pure alcohol must a nurse add to 8cc of a 70% solution to strengthen it to a 90% solution? Just need to know if this is right and if not what is right.
.90(8)+ 0(x) = .70(x+8)....NO ...FIRST WRITE DOWN WHAT IS X?I THINK X IS QTY.OF
PURE ALCOHOL TO BE ADDED TO 8 CC OF 70% ALCOHOL TO GET X+8 CC OF 90% SOLUTION.SO THE EQN. SHOULD BE X*100/100+8*70/100=(X+8)*90/100
7.2 = .70x+5.6.....X+5.6=0.9(X+8)
1.6 = 70x.......X-0.9X=7.2-5.6=1.6
2.29c = x.......0.1X=1.6
..................OR X=1.6/0.1=16 CC.
SEE THE FOLLOWING EXAMPLE WHICH WILL HELP YOU TO GET AN INSIGHT INTO THESE PROBLEMS AND THE METHOD OF APPROACH TO SOLVE THEM
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answer 15572 by venugopalramana(1008) About Me on 2006-02-26 07:15:31 (Show Source):
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.
