Algebra Problem: Adding Water to 75% Pure Alcohol to Make 15% Mixture

[Omid]

How much water must be added to a gallon of alcohol 75% pure to make a mixture 15% pure ?


The book I'm reading has given this formula : 0.75/(n+1)= 0.15 while n is the number of gallons to be added.
I don't understand where the heck the formula comes from
Can you help me with this problem ?
Thanks

 

[Atheist]

1 * 0.75 = (n+1) * 0.15 seems like a more understandable form to me.

basically, above sais that the (absolute) amount of alcohol is the same in both mixtures.

 

[faust9]

That formula comes from this one--a more general form:

$$C_f(V_1+V_2+\cdots+V_n)=C_1V_1+C_2V_2+\cdots+C_nV_n$$

Where C is concentration and V is volume.

Now, using you original numbers I can do this:

$$15\%(1gal+V_2)=75\%(1gal)+0\%V_2$$

Which can be rewritten into the form you first asked about:

$$.15=\frac{.75}{1gal+V_2}$$

Hope this helped. Good luck.

 

[Omid]

$$15\%(1gal+V_2)=75\%(1gal)+0\%V_2$$



It clears things up, but what is $$c_2$$ ?
Thanks

 

[faust9]

C2 is 0% because pure water has 0% alcohol.