Answer:
Volume of first solution = 20 Liters
Volume of second solution = 20 Liters
Volume of third solution = 60 Liters
Explanation:
Let the volume second solution used= y
Let the volume of third solution used = 3y
Let the volume of first solution used = 100 - (y +3y)
= 100- 4y
The volume of acid in the first solution ( V₁) = 25% of (100-4y)
= 25-y
The volume of acid in the second solution(V₂) =35% of y
= 0.35y
The volume acid in the third solution (V₃) = 55% of 3y
=1.65y
The Volume of acid in Mixture (V₄) = 45% of 100
= 45
From the conservation of volume:
V₄ = V₁ + V₂ + V₃
45 = (25-y )+ 0.35y + 1.65y
45 =25 -y + 0.35y + 1.65y
45 -25 = y
y = 20
So the volume of first solution used = 100- 4y
= 100- 4(20)
= 20 Liters
So the volume of second solution used = y
= 20 Liters
So the volume of third solution used = 3y
= 3(20)
= 60 Liters
