The first solution contains 15 % acid, the second contains 25 % acid, and the third contains 70 % acid. she created 48 liters of a 45 % acid mixture, using all three solutions. the number of liters of 70 % solution used is 2 times the number of liters of 25 % solution used. how many liters of each solution should be used?

First solution = 12 Liters

Second solution = 12 Liters

Third solution =  24 Liters

Step-by-step explanation:

Let the  number of  liters of 25% acid =  y

The number of  liters of 70 % acid = 2y

The number of  liters of  15 % acid = 48-3y

Volume of  acid in  first solution=  0.15 (48-3y)

Volume of  acid in  second solution=  0.25y

Volume of  acid in  third solution=  0.7 (2y)

Volume of acid in mixture  =  0.45 x48

                                             = 21.6 liters

Assuming no loss of volume,

0.15 (48-3y) + 0.25y + 0.7 (2y) = 21.6

7.2 -0.45y + 0.25y +1.4y = 21.6

7.2+ 1.2y = 21.6

1.2y = 21.6 - 7.2

1.2 y = 14.4

y = 14.4/1.2

y = 12 liters

Substituting the value of y above:

Volume of first solution =  48-3y

                                       =  48- 3(12)

                                       =  48-36

                                       = 12 liters

Volume of second solution =  y

                                              = 12 liters

Volume of third   solution =  2y

                                            = 2 (12)

                                            = 24 liters

option (b) is correct

step-by-step explanation:

to find the solution of this system, we need to convert it into reduced row-echelon form. we have,

after applying row operations r2 -> r2/2 and r3 -> r3/3, we get,

answer:   c ?

step-by-step explanation: