Question 997311
FOLLOWING INSTRUCTIONS:
Removing {{{3}}} liters from {{{8}}} liters leaves {{{8-3=5}}} liters.
If the solution water content was {{{x}}} liters if water per liter of original solution,
those {{{5}}} liters of solution now contain
{{{5x}}} liters of water, and {{{5-5x=5(1-x)}}} liters of milk.
Adding {{{3}}}liters of water, you end up with a total of
{{{5+3=8}}} liters of new, diluted solution, containing
{{{5-5x=5(1-x)}}} liters of milk.
The expected milk concentration (as a volume/volume fraction) is
{{{5(1-x)/8}}}

THE WAY THE ASSISTANT DID IT:
If the solution water content was {{{x}}} liters if water per liter of original solution,
in the {{{8}}} liters of original milk solution there was initially
{{{8x}}} liters of water, and {{{8-8x=8(1-x)}}} liters of the milk.
After adding {{{3}}} liters of water, the resulting solution had a volume of
{{{8+3=11}}} liters of waters, containing {{{8-8x=8(1-x)}}} liters of the milk.
The resulting milk concentration (as a volume/volume fraction) is
{{{8(1-x)/11}}} , and does not change on removing {{{3}}} liters of solution.


The ratio of expected concentration of the milk to actual concentration of the milk is
{{{5(1-x)/8}}}{{{":"}}}{{{8(1-x)/11}}}={{{5/8}}}{{{":"}}}{{{8/11}}}={{{5*11}}}{{{":"}}}{{{8*8}}}={{{55}}}{{{":"}}}{{{64}}} .