Question 1168702
Target is 60% antifreeze.
60% is closer to 50% and farther from 100%.
MOST of the antifreeze should come from the lower concentration 50% antifreeze.



{{{(100-60)/(100-50)}}} of the  10L is of the 50% antifreeze.
{{{4/5}}} of 10L is of the 50% antifreeze.


8L of 50% and 2L of 100%.


--

If you would do the solution with algebra and hold off on all computation until the very end, you would see that same fraction (or 1 minus it).


v, how much 100%
10-v, how much 50%
{{{100v+50(10-v)=10*60}}}----to account for pure antifreeze itself
-
{{{100v+50*10-50v=10*60}}}
{{{100v-50v=10*60-10*50}}}
{{{(100-50)v=10(60-50)}}}

{{{v=10((60-50)/(100-50))}}}------see the one-minus-the-other fraction form; depends on which antifreeze the variable is assigned.


{{{(60-50)/(100-50)=10/50=1/5}}}