Question 51084:  Anne and Nancy use a metal alloy that is 21.22% copper to make jewelry. How many ounces of a 16% alloy must be mixed with a 25% alloy to form 100 ounces of the desired alloy? 
 Answer by venugopalramana(3286)      (Show Source): 
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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 
Y=60-20=40 
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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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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. 
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