Question 357302
20 grams of 10% solution.
x grams of 6% solution.
want y grams of 8% solution.


usually you will come up with 2 or more equations that need to be solved simultaneously.


in this particular problem you start off with 20 grams of 10% solution.


you are going to add x grams of 6% solution.


you want to make a total of y grams of 8% solution.


the first equation would be:


20 + x = y


the second equation would be (20*.10) + (x*.06) = (y*.08)


The first equation is telling you how many grams of total solution.


The second equation is telling you how many grams of alcohol.


if you know how to solve simultaneous equations, the rest is just an application of the rules for solving them.


I'll use substitution.


in the first equation, you know that y = 20 + x.


you can substitute for y in the second equation to get:


(20*.10) + (x*.06) = ( (20+x) *.08)


you have now reduced 1 equation in 2 unknowns to 1 equation in 1 unknown which can now be solved for the unknown.


simplify your equation to get:


2 + .06*x = 1.6 + .08*x


subtract 1.6 from both sides of the equation to get:


.4 + .06*x = .08*x


subtract .06*x from both sides of the equation to get:


.4 = .02*x


divide both sides of the equation by .02 to get:


20 = x


that's the same as x = 20.


substitute for x in the first equation to get:


y = 20 + x = 40


the total grams in the final solution should be 40 grams.


you have  x = 20 and y = 40


substitute for  and y in the second equation to get:


(20*.10) + (x*.06) = (y*.08) becomes:


(20*.1) + (20*.06) = (40*.08)


simplify this to get:


2 + 1.2 = 3.2 which becomes:


3.2 = 3.2 which is true, so the values of x and y should be good.


you will need 20 grams of total solution.


you will require 3.2 grams of alcohol.


2 of them will come from 20 grams of the first solution (10%).


1.2 of them will come from 20 grams of the second solution (6%)