Question 1187114
.
When 1 kg of salt is added to a solution and water, the solution becomes 33 1/3 % salt by mass. 
When 1 kg of water is added to the new solution, the resulting solution is 30% by mass. 
The percentage of salt in the original solution is? Can someone please solve this question?
~~~~~~~~~~~~~~~~~~~



<pre>
Let x be the mass of salt (in kilograms), and let w be the mass of water (in kilograms) in the original solution.


Then from the problem's description, we have these equations for mass concentrations

    {{{(x+1)/(x+1+w)}}}    = {{{1/3}}},      (1)

    {{{(x+1)/(x+1+w+1)}}} = {{{3/10}}}.     (2)


From these equations, we have further

    3(x+1)   = x + w  + 1                   

    10*(x+1) = 3(x + w + 2)                   


Continue transformations

     3x +  3 = x + w + 1                    

    10x + 10 = 3x + 3w + 6                


The standard form equations

    2x -  w = -2,      (3)

    7x - 3w = -4.      (4)


Apply the Elimination method.  For it, multiply equation (3) by 3;  keep equation (4) as is

    6x - 3w = -6,      (3')

    7x - 3w = -4.      (4')


From equation (4'), subtract equation (3').  You will get

    7x - 6x = -4 - (-6),  or  x = 2.


Then from equation (3),  w = 2x + 2 = 2*2 + 2 = 6.



So, the original solution was 6 kg of water and 2 kg of salt,

and the original mass concentration of the solution was  {{{2/(2+6)}}} = {{{2/8}}} = {{{1/4}}} = 25%.    <U>ANSWER</U>
</pre>

Solved.