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?

So here's how we determine hom gallons should be taken from each tank:

Tank A:

contains 15 gal of mixture, the 5 gals are alcohol which is considered as 1/3

Tank B:

contains 15 gal of mixture, the 3 gals is alcohol which is considered as 1/5

Tank A contains 100/3% alcohol

Tank B contains 20% alcohol

Let:

x = gallons from Tank A

y = gallons from Tank B

Equation (s):

x + y = 8

Substitute:

(100/3) x + 20y = 8 * 25

Express the x = 8 - y

Replace (100/3) (8-y) + 20 y = 200

Multiply the whole equation by 3 and get 100 (8-y) + 60y = 600

800 - 100y + 60y = 600

800-40y = 600

800-600 = 40y

40y = 200

y = 5

Replace the result to:

x = 8-5 = 3

x = 3

The answer is:

From Tank A will be taken 3 gallons and from Tank B, 5 gallons

Fraction of alcohol in Tank A is 5 / (5+10) = 1/3

Fraction of alcohol in Tank B is 3 / (3+12) = 1/5

Fraction of final alcohol required = 1/4 (i. e. 25%)

Let 'x' be the volume of solution taken from Tank A (in gallons).

Now, since final volume of alcohol is 8 gallons.

Therefore, volume taken from Tank B will be (8 - x).

The alcohol content of the mixture can be estimated as follows,

(1/3) x + (1/5) (8 - x) = (1/4) * 8

∴, 5x + 3 (8 - x) = 30

∴ 2x = 6

∴ x = 3

Thus, 3 gallons should be taken from Tank A and 5 gallons should be taken from Tank B so that resultant 8 gallon solution contain 25% alcohol by volume