Question

Tank A contains a mixture of 10 gallons water and 5 gallons pure alcohol. Tank B has 12 gallons water and 3 gallons alcohol. How many gallons should be taken from each tank and combined in order to obtain an 8 gallon solution containing 25% alcohol by volume?

Answer 1

The mixture in Tank A is 5/(5+10) = 1/3 pure alcohol. The mixture in Tank B is 3/(3+12) = 1/5 pure alcohol. We want to combine these to make a mixture that is 25% = 1/4 pure alcohol.

Let a represent the volume in gallons of solution taken from Tank A. Then the volume taken from Tank B will be (8 -a) because we are making a total of 8 gallons of solutions. The alcohol content of the mixture is
.. (1/3)a +(1/5)(8 -a) = (1/4)*8
.. 5a + 3(8 -a) = 30 . . . . . . . . multiply by 15
.. 2a = 6 . . . . . . . . . . . . . . . . . subtract 24, collect terms
.. a = 3

3 gallons should be taken from Tank A; 5 gallons should be taken from Tank B.

Answer 2

Tank A contains 15 gallons of mixture, from what 5 gallons is alcohol,
that is 1/3
Tank B contains 15 gallons of mixture, from what 3 gallons is alcohol,
that is 1/5

According to this, tank A contains 100/3 % alcohol and tank B contains 20% alcohol.

Let's named variables:

gallons from tank A will be the x and from tank B will be the y

Equations are:  x+y=8 and (100/3)x+20y= 8 * 25

From first we expressed x=8-y, than replace in the second

(100/3) ( 8-y) + 20y = 200 the whole equation multiply with 3 and get

100(8-y) + 60y = 600 => 800 -100y+60y=600 => 800-40y=600 =>

800-600=40y => 40y=200 => y=5

than we replace this result in

x= 8-5=3 => x=3

It will be taken from tank A- 3 gallons and from tank B- 5 gallons.

Good luck!!!