80 liters of a mixture of alcohol and water contains 60% alcohol. If 'X' liters of water is added to this mixture, then the percentage of alcohol in the new mixture becomes 40%. What is the value of X?

40

Initial solution,

Alcohol : Water

    3    :   2            ..(1)

After adding X liters water, Final solution,

Alcohol : Water

    2    :    3           ..(2)

⇒ Making ratio of Alcohol in both solution equal, we multiply (1) with 2 and (2) with 3.

⇒ (3 : 2) x 2 = 6 : 4,   and   (2 : 3) x 3 = 6 : 9,

⇒ Difference between After water and initial water ratio = 9 - 4 = 5 units.

⇒ Quantity of initial solution = 80 liters,

⇒ (6 + 4) units = 80,

⇒ 10 units = 80,

⇒ 1 unit = 8 liters,

Therefore, amount of water 'X' added is = 5 x 8 = 40 liters.