Question

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?

Answer 1

First solution = 12 Liters

Second solution = 12 Liters

Third solution = 24 Liters

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