Question 74968
Let us suppose that 'x' ml of 30% alcohol be mixed with 'y' ml of 15% alcohol to form 150 ml of 20% alcohol.


150 ml of 20% alcohol consists {{{20*150/100}}} = 30 ml of alcohol and (150 - 30) = 120 ml of water.


Similarly, 'x' ml of 30% alcohol contains {{{30*x/100}}} = 0.3x ml and (x - 0.3x) = 0.7x ml of water.
Also, 'y' ml of 30% alcohol contains {{{15*y/100}}} = 0.15y ml and (y - 0.15y) = 0.85y ml of water.


Thus total volume of alcohol in final mixture = (0.3x + 0.15y) ml and total volume of water in final mixture = (0.7x + 0.85y) ml.


But final mixture contains 30 ml alcohol and 120 ml water.


So, {{{0.3x + 0.15y = 30}}}______________(1)
and {{{0.7x + 0.85y = 120}}}______________(2)


Multiplying (1) by 7 and (2) by 3 and subtracting (1) from (2)
{{{2.55y - 0.85y = 360 - 210}}}
{{{1.7y = 170}}}
{{{y = 170/1.7 = 100}}}


So, {{{0.3x + 0.15*100 = 30}}} [from (1)]
{{{0.3x + 15 = 30}}}
{{{0.3x = 30 - 15 = 15}}}
{{{x = 15/0.3 = 50}}}


Hence, 50 ml of 30% alcohol should be mixed with 100 ml of 15% alcohol.