A mixture 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.