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| 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): 
You can put this solution on YOUR website! SEE THE FOLLOWING EXAMPLES AND TRY.IF STILL IN DIFFICULTY PLEASE COME BACK -------------------------------------------------------
 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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