SOLUTION: Suppose we have two solutions of sulphuric acid in water. The first is 40% strong and second is 60% strong. We mix the two solutions, add 5 kg of pure water and obtain a 20% soluti
Question 1038623: Suppose we have two solutions of sulphuric acid in water. The first is 40% strong and second is 60% strong. We mix the two solutions, add 5 kg of pure water and obtain a 20% solution. If instead of adding 5 kg of pure water, we were to add 5 kg of an 80% solution, we would get a 70% solution. How much of the 40% solution and 60% solution do we have?
a) 2 kg : 1 kg b) 1kg : 2 kg c) 2 kg : 3 kg d) 3 kg : 1kg Answer by jorel555(1290) (Show Source): You can put this solution on YOUR website! Let m be the 60% solution, n be the 40% solution. Then:
.6m+.4n=.2(m+n+5), so
.4m+.2n=1
Also:
.6m+.4n+.8(5)=.7(m+n+5)
.1m+.3n=.5
So:
.4m+1.2n=2
.4m+.2n=1
Subtracting, we get:
1n=1 kg of 40% solution
m=2 kg of 60% solution!!!!!!!!!!!!!!