SOLUTION: How much water must be added to 20.0 mL of a 60.0% slaine solution in order to make a 45.0% saline solution. Please help solve. Thank You!

Question 261856: How much water must be added to 20.0 mL of a 60.0% slaine solution in order to make a 45.0% saline solution. Please help solve. Thank You!
Found 2 solutions by checkley77, Alan3354:
Answer by checkley77(12844) (Show Source): You can put this solution on YOUR website!
.60*20=.45(20+X)
12=9+.45X
.45X=12-9
.45X=3
X=3/.45
X=6.667 ML OPF WATER IS USED.
PROOF:
.60*20=.45(20+6.667)
12=.45*26.667
12=12
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
How much water must be added to 20.0 mL of a 60.0% slaine solution in order to make a 45.0% saline solution.
-------------------
0.6*20 = 12 ml of salt
For 12 ml to be 45%, the total is 12/0.45
= 80/3 ml total
Add 20/3 ml of water

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