LM wrote:
bmwhype2 wrote:
Two solutions of acid were mixed to obtain 10 liters of a new solution. Before they were mixed, the first solution contained .8 liters of acid whole te second contained .6liters of acid. If the percentage of acid in the first solution was twice that in the second, what was the volume of the first solution?
1st: X liter & Acid = 0.8L
2nd: Y liter & Acid = 0.6L
Given:
X+Y=10L.........( i )
(0.8/X)*100 = 2*[(0.6/Y)*100]
Y=(1.5)*X
Therefore, ( i ) can be re-written as
X+1.5X =10
X= 4Liter
Responding to a pm:
Quote:
Could you please solve this using Weighted averages concept . And also , i haven't understood the method used to solve this in the original post.
[ 0.8/A = 2* (0.6/B) ] --- This part.
First solution had .8 liters of acid and second had .6 liters. Total the 10 liters solution must have had .8 + .6 = 1.4 liters of acid in 10 liters of solution which is 14% (average %age)
If %age of acid in second solution is x, %age of acid in first solution is 2x. Say volume of first solution is w which means volume of second solution is (10 - w)
Let's see how x and w connect:
.8 = 2x% of w = 2xw/100
xw = 40
Weighted Average Formula:
Cavg = (C1*w1 + C2*w2)/(w1 + w2)
14 = [2x*w + x*(10 - w)]/10
140 = [40 + 10x]
x = 10
Since xw = 40, w = 4
So volume of first solution is 4 liters.
Thanks Karishma for beautiful explanation.