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?

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

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

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