Question 154551
Let x=amount of pure antifreeze that needs to be added
Then 20-x=amount of 40% solution that needs to be used

Now we know that the amount of pure antifreeze in the 40% solution(0.40(20-x)) plus the amount of pure antifreeze added (x) has to equal the amount of pure antifreeze in the final solution (20*0.50) So our equation to solve is:

0.40(20-x)+x=20*0.50 get rid of parens and simplify

8-0.40x+x=10  subtract 8 from each side
8-8-0.40x+x=10-8  collect like terms

0.60x=2  divide both sides by 0.60
x=3.33 qts -----------amount of pure antifreeze that needs to be added

20-x=20-3.33=16.67 qts------------amount of 40% antifreeze used

CK
16*67*0.40+3.33=10
6.668+3.333=10
10~~~10


You do the chart

Hope this helps---ptaylor