SOLUTION: You have 1 gallon of solution that is 20% salt and 80% water. You need 8 gallons of solution that is 2% salt and 98% water. How much water do you add to obtain this?
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Question 258411: You have 1 gallon of solution that is 20% salt and 80% water. You need 8 gallons of solution that is 2% salt and 98% water. How much water do you add to obtain this? Found 2 solutions by richwmiller, CharlesG2:Answer by richwmiller(17219) (Show Source):
You can put this solution on YOUR website! .8*1+x=8*.98
7.04 gallons
which works out in the equation but seems odd because 1 + 7.04 is more than 8
but if you add only 7 gallons it will tell that you are missing .04 gallons
Maybe the salt is absorbed into the solution?
.8*1+7=.98x
x=7.95918
You can put this solution on YOUR website! You have 1 gallon of solution that is 20% salt and 80% water. You need 8 gallons of solution that is 2% salt and 98% water. How much water do you add to obtain this?
1 gallon = 128 oz
0.20*128=25.6 oz salt (salt amount will not change)
0.02*(128+x)=25.6
2.56+0.02x=25.6
0.02x=23.04
x=1152 oz or 1152/128=9 gallons of water to be added
but this gives 10 gallons of prepared solution, and we wanted 8 gallons
0.8*128=102.4 oz of original solution
.20*102.4=20.48 oz salt (salt amount will not change)
0.02*(102.4+x)=20.48
2.048+0.02x=20.48
0.02x=18.432
x=921.6 oz or 921.6/128=7.2 gallons of water to be added
0.8 gallons original + 7.2 gallons of water = 8 gallons of prepared solution
so solution is either use 0.8 gallons of the original and prepare the 8 gallons
or use all of the original and have an extra 2 gallons of the prepared
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