He first solution contains 25 % acid, the second contains 35 % acid, and the third contains 55 % acid. she created 100 liters of a 45 % acid mixture, using all three solutions. the number of liters of 55 % solution used is 3 times the number of liters of 35 % solution used. how many liters of each solution was used?

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

b) mutualism

hope this : )

b.final