SOLUTION: How many oz. of pure chocolate must be added to 150 oz that is 60% chocolate to make a topping that is 75% chocolate?
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Question 34225: How many oz. of pure chocolate must be added to 150 oz that is 60% chocolate to make a topping that is 75% chocolate?
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
PLEASE SEE THE FOLLOWING EXAMPLE AND TRY.IF STILL IN DIFFICULTY,PLEASE COME BACK
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.
Mixture_Word_Problems/30587: The amount (by weight) of gold, silver and lead in three alloys of these metals are in ratios:
4:3:2 - Alloy 1
3:5:1 - Alloy 2
2:2:5 - Alloy 3
It is desired to make a fourth alloy containing equal amounts of gold, silver and lead. How many grams each alloy should be used for every 10 grams of the new alloy?
1 solutions
Answer 17262 by venugopalramana(1167) About Me on 2006-03-18 04:34:35 (Show Source):
The amount (by weight) of gold, silver and lead in three alloys of these metals are in ratios:
4:3:2 - Alloy 1
3:5:1 - Alloy 2
2:2:5 - Alloy 3
It is desired to make a fourth alloy containing equal amounts of gold, silver and lead. How many grams each alloy should be used for every 10 grams of the new alloy?
LET X GMS OF ALLOY1 ,Y GMS OF ALLOY2 AND 10-X-Y GMS OF ALLOY3 BE USED TO GET
X+Y+10-X-Y=10 GMS OF ALLOY 4
SO.................GOLD..............SIVER............LEAD IN THE MIX IS GIVEN BY
X GMS A1...........4X/9..............3X/9..........2X/9
Y GMS A2...........3Y/9..............5Y/9...........Y/9....
10-X-Y GMS A3....(20-2X-2Y)/9 ......(20-2X-2Y)/9...(50-5X-5Y)/9
-------------------------------------------------------------------------------
10 GMS A4.......(20+2X+Y)/9........(20+X+3Y)/9.....(50-3X-4Y)/9
THESE ARE ALL EQUAL...HENCE
20+2X+Y = 20+X+3Y...OR......................X-2Y=0..............I
20+2X+Y = 50-3X-4Y...OR...5X+5Y=30....OR....X+Y=6......II
EQN.II - EQN I...GIVES
X+Y-X+2Y=6......OR 3Y=6.....Y=2
SO X=6-Y=6-2=4
Z=10-4-2=4...
HENCE 4 GMS OF A1,2 GMS OF A2 AND 4 GMS OF A3 ARE TO BE ADDED TO GET 10 GMS OF A4.
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