Hi ! I have a problem in this question::::::

How much pure alcohol must be added to 400 ml of a 15% solution to make its 32%?

Plz ans fast. tomorrow is my exam....

Let x = the required number of milliliters of pure alcohol.

Then we have:

400(15%) + x(100%) = (400+x)(32%)

400(0.15) + x(1) = (400+x)(0.32)

Simplify and solve for x.

60 + x = 128 + 0.32x

60 + 0.68x = 128

0.68x = 68

x = 100 ml of pure alcohol is required.

Check:

400(0.15) + 100 = (400+100)(0.32)

60 + 100 = 500(0.32)

160 = 160