Question

A tank contains 20 gallons of a mixture of alcohol and water which is 40% alcohol by volume. How much of the mixture should be removed and replaced by an equal volume of water so that the resulting solution will be 25% alcohol by volume?​

Answer 1

Step-by-step explanation:

Let x = amt of water to be added,

and

Let x = amt of mixture to be removed

:

:

Water is a 0% solution

:

The equation:

.40(20-x) + .0(x) = .25(20)

or just

.40(20-x) = .25(20)

:

8 - .4x = 5

:

-.4x = 5 - 8

;

-.4x = -3

x = 

x = +7.5 gal mixture removed and 7.5 gal water added

;

:

Check in an "amt of alcohol" equation

.4(20-7.5) = .25(20)

.4(12.5) = 5

5 = 5