Question 1192358
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a painter wanted to mix 2 liters of blue paint with 3 liters of yellow paint to make 5 liters of green paint. 
however, by mistake, he used 3 liters of blue and 2 liters of yellow so that he made the wrong shade of green. 
what is the smallest amount of this green paint that he must throw away so that, 
using the rest of his green paint, and some extra blue or yellow paint he could make 5 liters of paint 
of the correct shade of green?
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After first (mistaken) mix, the painter has too much blue paint and too little yellow paint in the mixture, 
so, OBVIOUSLY, he should add extra yellow paint at some moment.


After first (mistaken) mix, the painter has 5 liters of paint with 2 liters of yellow paint in it, 
which contents is 40% = 0.4 of yellow in the total of 5L volume.


He poors out an unknown amount of V liters of the mixed paint and adds V liters of yellow paint.
After it, the painter has 0.4*(5-V)+V liters of yellow paints in the 5 liters total volume.


He wants the ratio {{{yellow_paint/total_5_liters}}}  be  {{{3/5}}} = 0.6, i.e.

    0.4*(5-V) + V = 0.6*5.


Simplify and find V

    2 - 0.4V + V = 3

    -0.4V + V    = 3 - 2

        0.6V     = 1

           V     = 1/0.6 = {{{1/((6/10))}}} = {{{10/6}}} = {{{5/3}}}  liters.


<U>ANSWER</U>.  {{{5/3}}} = 1.666... liters of the original mistaken mixture should be poored out and replaced with  {{{5/3}}} = 1.666... liters of yellow paint.
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Solved.