Question

A chemist has three different acid solutions. The first acid solution contains 20% acid, the second contains 35% and the third contains 80%. He wants to use all three solutions to obtain a mixture of 54 liters containing 45% acid, using 2 times as much of the 80% solution as the 35% solution. How many liters of each solution should be used?

Answer 1

the volume of First acid solution =24 liters

the volume of second acid solution = 10 liters

the volume of third acid solution = 20 liters

Explanation:

Let,

the volume of First acid solution be 'x'

the volume of second acid solution be 'y'

the volume of third acid solution be 'z'

Data provided in the question:

Acid in first acid solution = 20% = 0.2

Acid in Second acid solution = 35% = 0.35

Acid in third acid solution = 80% = 0.8

Acid in the final solution = 45% = 0.45

Volume of the final solution = 54 liters

According to question

0.2x + 0.35y + 0.8z = 0.45 × 54

or

0.2x + 0.35y + 0.8z = 24.3 ......................(1) [concentration = Percentage × volume ]

z = 2y .............(2)

also,

x + y + z = 54 ...........(3) [Total volume = sum of individual solutions]

substituting 2 in 3,

x + y + 2y = 54

x + 3y = 54

or

x = 54 - 3y ...........(4)

substituting 2 and 4 in equation 1

0.2(54 - 3y) + 0.35y + 0.8(2y) = 24.3

or

10.8 - 0.6y + 0.35y + 1.6y = 24.3

or

1.35y = 13.5

or

y = 10 liters

substituting the value of y in 2, we get

z = 2(10)

or

z = 20 liters

and,

substituting the value of y in 4, we get

x = 54 - 3(10)

or

x = 24 liters

Hence,

the volume of First acid solution =24 liters

the volume of second acid solution = 10 liters

the volume of third acid solution = 20 liters