Question 260651

Let x=amount of 100% antifreeze needed
Then 55-x=amount of 10% antifreeze mixtures needed

Now we know that the amount of pure antifreeze used (x) plus the amount of pure antifreeze in the 10% antifreeze mixtures (0.10(55-x)) has to equal the amount of pure antifreeze in the final mixture (0.20*55).  So our equation to solve is:

x+0.10(55-x)=0.20*55  simplify
x+5.5-0.10x=11 subtract 5.5 from each side
x+5.5-5.5-0.10x=11-5.5  collect like terms
0.9x=5.5  divide each side by 0.9
x=6.1 gal-------------------------amount of 100% antifreeze needed
CK

6.1+0.10(48.9)=11
6.1+4.9=11
11=11

Hope this helps---ptaylor