Question 1007377
x amount of juice in first vat, y amount of juice in second vat, to start.



The transfer process:
0.9x in first vat;
0.10x+y in second vat;
{{{(9/10)x=3((1/10)x+y)}}}


How much change in the second vat?
{{{3((1/10)x+y)-y=(3/10)x+3y-y=3x/10+2y}}}


The comparison of the change to the original quantity of second vat
{{{(3x/10+2y)/y}}}
{{{highight_green(3x/(10y)+2)}}}, as a fraction for the increase, and you would multiply by 100 to make this a percentage.


Any way to know the value?


Still the same amount of juice contained in the two vats as before the transfer process.
{{{x+y=3x/10+2y}}}
{{{10x+10y=3x+20y}}}
{{{7x=10y}}}
{{{x=(10/7)y}}}
Substitute this into the fraction increase expression:



{{{3((10/7)y)/(10y)+2}}}
{{{(30/7)y/(10y)+2}}}
{{{30/70+2}}}
{{{3/7+2}}}
{{{highlight(2&3/7)}}}---------the FRACTION increase.
You can change this to percentage if you want.


The second  vat increased in volume held, two and three-sevenths times.