Algebra Mixture Problems

[Jayce]

Problem : 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?

From the book, 2 equations were derived based from the problem :

Equation 1 -> x + y = 8 & Equation 2 -> 0.3333x + 0.2y = 0.25(8)

My question : What is the formula used to get the 0.3333 and 0.2? Why or how is the formula designed like that? (Note : My question is not how to answer the problem.)

My thoughts / analysis : To get 0.33333, the fraction must be 1/3. So from the first tank, we have 10 gallons of water and 5 gallons of pure alcohol.

So, 5 / (10 + 5) = 5 / 15 = 1/3 = 0.33333. If we used that on 2nd tank, we will get the 0.2

So, If we are to put it into words :Mixture of gal. of water and pure alcohol = x gallons of alcohol / (x gallons of alcohol + x gallons of water).

The question is why or how is that the formula for mixing the gallons of alcohol and water? In my mind it should be just an addition between the 2, water + alcohol = mixture.

Sorry If the question is silly but thanks in advance for those who can explain it clear. God bless.

 

[KlausST]

Hi,

What is the formula used to get the 0.3333 and 0.2?

0.33 and 0.2 is the ratio of alcohol in the whole mixture. Or more exact:
It is: volume_pure_alcohol / total_volume which is: volume_pure_alcohol / (volume_pure_alcohol + volume_water).

--> your calculation to get 0.33 and 0.2 is correct.

(indeed this is not absolutely correct. Because when you mix 10 gallons of water with 5 gallons of alcohol you don´t get exactly 15 gallons of mixture. You get a little below 15 gallons. But I assume we have to ignore this)

Klaus

 

[CataM]

My question : What is the formula used to get the 0.3333 and 0.2? Why or how is the formula designed like that? (Note : My question is not how to answer the problem.)

x --> represents gallons taken from tank A
y --> represents gallons taken from tank B

How much percent alcohol does whatever tank provide for each gallon ? In other words, what is the percent of alcohol in each tank ? --> Alcohol quantity in the tank divided by the total gallons in that tank.

Tank A --> 5 gallons of alcohol / 15 total gallons in tank A
Tank B --> ...

x+y=8 gallons total

Alcohol taken from tank A: it is x*(5/15) alcohol from tank A.

Alcohol from tank A + Alcohol from tank B =0.25 * 8 <=> x*(5/15) + y*(3/15)=0.25*8

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(indeed this is not absolutely correct. Because when you mix 10 gallons of water with 5 gallons of alcohol you don´t get exactly 15 gallons of mixture. You get a little below 15 gallons. But I assume we have to ignore this)

He did not mention anything about density, so yes, I assume alcohol density is the same as water density.

 

[KlausST]

Hi,

He did not mention anything about density, so yes, I assume alcohol density is the same as water density.

It´s not density of water or density of alcohol, it more about molecule size.
What I mean is this:
https://www.youtube.com/watch?v=84k206qaVRU

Klaus

 

[Jayce]

Thanks to the both of you. The explanations are clear~ it somewhat irritates me when something simple yet hard or can't understand.

Thanks for that link, an additional knowledge for me.

 

[Ratch]

Problem : 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?

From the book, 2 equations were derived based from the problem :

Equation 1 -> x + y = 8 & Equation 2 -> 0.3333x + 0.2y = 0.25(8)

My question : What is the formula used to get the 0.3333 and 0.2? Why or how is the formula designed like that? (Note : My question is not how to answer the problem.)

My thoughts / analysis : To get 0.33333, the fraction must be 1/3. So from the first tank, we have 10 gallons of water and 5 gallons of pure alcohol.

So, 5 / (10 + 5) = 5 / 15 = 1/3 = 0.33333. If we used that on 2nd tank, we will get the 0.2

So, If we are to put it into words :Mixture of gal. of water and pure alcohol = x gallons of alcohol / (x gallons of alcohol + x gallons of water).

The question is why or how is that the formula for mixing the gallons of alcohol and water? In my mind it should be just an addition between the 2, water + alcohol = mixture.

Sorry If the question is silly but thanks in advance for those who can explain it clear. God bless.

As you already understand, tank A contains 1/3 alcohol and tank B contains 1/5 alcohol both by volume. The third tank is supposed to contain 2 gallons of alcohol. If you drain x gallons from tank A, you can only drain (8-x) from tank B. So the equation is easily set up as:



It is easily solved for x=3. So, 3 gallons from tank A and 5 gallons from tank B.

Ratch