Amixture contains 40 ounces of glycol and water and it is 10% glycol the mixture is to be strengthened to 25% by adding glycol how much glycol is in the original mixture

Reasoning solution:

40 oz of 10% glycol contains 40*0.1=4 oz of glycol, and 40-4=36 oz. of water.

To make a 25% glycol using 36 oz of water will make a total volume of 36*(4/3)=48 oz.
Volume of pure glycol to be added = 48-40=8 oz

Therefore 8 oz of pure glycol must be added to the mixture to make a 25% solution.

Algebraic solution:
Let x = volume of pure glycol required in oz.
then 
0.10*40+1.00*x = 0.25*(40+x)
expand
4.0+x=10+0.25x
solve
(1-0.25x)=10-4=6
x=6/0.75=8 oz

Therefore 8 oz of pure glycol must be added to the mixture to make a 25% solution.