word problem?

Question

a radiator with a 6 gallon capacity is filled with 50% antifreeze solution. how much must be drained so that adding pure antifreeze gives a 60% antifreeze solution?

Answer 1

Let x gallon of the existing 50% solution be drained out and replaed by x gallon of pure antifreeze solution and a 60% solution is obtained
According to the problem
0.50(6-x)+x=0.60*6
or,3-0.50x+x=3.60
or,0.50x=3.60-3=0.60
or x=0.60/0.50=1.20
Therefore 1.20 gallon of the existing 50% solution is to be drained out and replaed with 1.20 gallons of pure antifreeze solution to make it 60%

Answer 2

1.2 gallons.

If 6 gallons is 50% solution, that is equivalent to 3 gallons of pure antifreeze.

What you want is 3.6 gallons of pure antifreeze within 6 gallons (60%)

When you drain, you'll be draining 50% solutiion. When you add, you'll be adding 100% solution.

Set up an equation:

3 - 0.5x + x = 3.6

[3 gallons of antifreeze - 50%(x ounces) + 100%(x ounces) solution = 3.6 gallons of antifreeze.]

Solve for x, and you'll get 1.2

Answer 3

here's the answer:
(0.5*6)-0.5x+x=(0.6*6)
3+0.5x=3.6
so x equals
x=1.2;
you must drain 1.2 gallons of the 50% solution and then add 1.2 gallons of 100% solution to make 6 gallons of 60% solution

to check:
if you have 6 gallons capacity then 60% of that will have to be 3.6gallons and that is to be of 100% antifreeze solution
if you drain 1.2 gallons of the 50% solution, that leaves you with 4.8 gallons of 50% solution.....
in the 4.8 gallons og 50% solution, 2.4 gallons of which is a 100% solution
add 2.4 gallons of 100% solution to 1.2 gallons of 100% solution and you get 3.6 gallons of 100% solution
3.6/6=0.6 which proves that you have now 60% antifreeze solution

Answer 4

5gallons