Question 32255: Two kilograms of Columbian coffee are mixed with 1 kilogram of Brazilian coffee to produce a mixture selling for $7.00 a kilogram. If 3 kilograms of Columbian coffee were mixed with 2 kilograms of Brazilian coffee, the mixture would sell for $6.90 a kilogram. What is the value of a kilogram of each type of cofee?
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! SEE THE FOLLOWING,APPLY THE SAME METHOD AND YOU SHOULD GET THE ANSWER.IF STILL IN DIFFICULTY PLEASE COME BACK
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Mixture_Word_Problems/30725: How many gallons of pure alcohol solution should be added to 63 gal of 32% alcohol solution to make a 37% solution? Thank you.
1 solutions
Answer 17458 by venugopalramana(1167) About Me on 2006-03-19 22:24:14 (Show Source):
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/30727: How many gallons of 16% alcohol solution should be added to 30 gal of a 45% alcohol solution to make a 21% solution? Please help me set this up.
1 solutions
Answer 17453 by venugopalramana(1167) About Me on 2006-03-19 21:54:56 (Show Source):
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
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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.
Money_Word_Problems/29402: The nut store sells walnuts for 4.00 a pound and cashews for 7.00 a pound how many pounds of cashews should be mixed with 10 pounds of walnuts to obtain a mixture that sells for 5.50 a pound?
1 solutions
Answer 16262 by venugopalramana(1088) About Me on 2006-03-07 05:47:32 (Show Source):
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.
logarithm/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?
1 solutions
Answer 16163 by venugopalramana(1088) About Me on 2006-03-06 11:02:24 (Show Source):
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.
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