Question

How much pure alcohol must be added to 400ml of 15percent solution to make its strength 32 percent

Answer 1

Solution:-
Let the required number of milliliters of alcohols be x ml.
Given 15 % of 400 ml should be added with 100 % of pure alcohol.
Hence 400(15 %) + x(100 %) = (400+x) (32 %)
⇒ 400 × 15/100 + x = (400 + x) × (32/100)
⇒ 60 + x = 128 + (8x/25)
⇒ x - 8x/25 = 128 - 60
⇒ 17x/25 = 68
⇒ x = (68 × 25)/17
⇒ x = 1700/17
x = 100 ml Answer.

Answer 2

Let x = the required number of milliliters of pure alcohol. 
Then we have:
400(15%) + x(100%) = (400+x)(32%) Change the percents to decimals. 
400(0.15) + x(1) = (400+x)(0.32) Simplify and solve for x. 
60 + x = 128 + 0.32x Subtract 0.32x from both sides.
60 + 0.68x = 128 Subtract 60 from both sides.
0.68x = 68 Divide both sides by 0.68
x = 100 ml of pure alcohol is required. 
Check: 
400(0.15) + 100 = (400+100)(0.32)
60 + 100 = 500(0.32)
160 = 160 Check ok.