<PRE><B>Question 177765</B>
lets call the amount of 60% and 30% solutions a and b respectively
:
a+b=125.....eq 1
.6a+.3b=.36(125)..eq 2
:
lets rewrite eq 1 to a=125-b and plug that value into eq 2
:
.6(125-b)+.3b=.36(125)
:
75-.6b+.3b=45
:
-.3b=-30
:
{{{highlight(b=100)}}}ml of 30% solution
:
{{{a=125-b=125-100=highlight(25)}}}ml of 60% solution</PRE>