Question 312145
Let A be the amount of 40% solution currently in the radiator. You add B liters of pure antifreeze (100% solution).
1.{{{40(A)+100(B)=50(A+B)}}}
2.{{{A+B=6}}}
From eq. 2,
{{{A=6-B}}}
Substitute into eq. 1,
{{{40(6-B)+100(B)=50(6-B+B)}}}
{{{240-40B+100B=300}}}
{{{60B=60}}}
{{{highlight(B=1)}}}
Then from eq. 2,
{{{A=6-B=6-1}}}
{{{highlight(A=5)}}}
There are currently 5 liters of 40% solution. Adding 1 liter will fill the radiator and make a 50% solution.