Question 159080
*See this one, thank you.
Let {{{a}}}, amount in Liters to be drained
{{{b}}}, amount of 100% A/F to be added
Therefore,
{{{highlight(15(0.40)-a(0.40)+b(1.00)=15(0.50))}}} ------> working eqn
</pre<font size=4><b>The above eqn means the 15L capacity with 40% A/F is drained with {{{a}}} liters amount with the same mixture of %40 A/F of course.Then you're adding 100% A/F in {{{b}}} liters amount that will result to the same capacity of 15L with 50% A/F</pre>
Also, take note:
{{{15L-a(L)+b(L)=15L}}}
</pre><font size=4><b>The cooling system 15L has taken out {{{a(liters)}}} amount plus {{{b(liters)}}} amount equals the same capacity of 15L<pre>.
Continuing,
{{{cross(15L)-a(L)+b(L)=cross(15L)}}}
{{{a(L)=b(L)}}} ---> substitute in working eqn:
{{{15(0.40)-a(0.40)+highlight(a)(1.00)=15(0.50)}}}
{{{6+0.60a=7.5}}}
{{{0.60a=7.5-6=1.5}}}
{{{cross(0.60)a/cross(0.60)=cross(1.5)2.5/cross(0.60)}}}
{{{highlight(a=2.5L)}}} --------> amount to be drained = amount to be added
In doubt? go back working eqn:
{{{15(0.40)-2.5(0.40)+2.5(0.40)=15(0.50)}}}
{{{6-1+2.5=7.5}}}
{{{7.5=7.5}}}
thank you,
Jojo