How much pure anti freeze must be added to 10 gal of 10% antifreeze solution to increase its concentration to 60 % anti freeze? If X represents the number of pure antifreeze solution, what is the correct statement of the problem?

Answer 1

Add up the amounts of anti-freeze in the parts and the mixture. They must be the same. If x gallons of 100% anti-freeze are added, then

.10*10 + 1.00x = .60(10+x)

Answer 2

original 10 gal is
1 gal af, 9 gal water
so
1 gal + x gal = 0.60 ( 10 + x)

1 + x = 6 + .6 x
.4 x = 5
x = 12.5

Answer 3

thanks,Steve and Anonymous,but i do not understand.please,i will need you to elucidate a little bit more.if you could give me the steps explaining them,that would not only provide the answer but also teach me how to do it and the logic behind it.
thanks anyway.i have an idea,but it is not gelling for me

Answer 4

so far this what i clearly understand
10 gallons of 10% antifreezee solution
so 1 gallon has 10% antifreeze and 90%water.
so it's 1 gallon of antifreeze and 9 gallons of water which make up the 10 gallons total
but i do not know how to proceed from here