Question 178565
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We know {{{x}}}=ml of 80%
Let {{{y}}}=ml of 15%
Then, {{{x+y=600ml}}}---------------------> EQN 1
Also, when mixed:
{{{0.80(x)+0.15(y)=600(0.40)}}}-----------> EQN 2
Goin back EQN 1 we get,
{{{x=600-y}}} ----------------------------> EQN 3, subst. in EQN 2:
{{{0.80(600-y)+0.15y=240}}}
{{{480-0.80y+0.15y=240}}}
{{{480-240=0.80y-0.15y}}}
{{{240=0.65y}}}---------->{{{cross(240)369.23/cross(0.65)=cross(0.65)y/cross(0.65)}}}
{{{y=369.23ml}}} needed for 15% sol'n.
Going back EQN 3:
{{{x=600-369.23=highlight(230.77ml)}}}, needed for the 80% sol'n.
In doubt? Go back EQN 2:
{{{0.80(230.77)+0.15(369.23)=600(0.40)}}}
{{{184.616+55.3845=240ml}}}
{{{240ml=240ml}}}
Thank you,
Jojo</pre>