Question

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

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 = 1700/17x = 100 ml   Answer.

Answer 2

Hi mate,

Solution :

=> Volume of alcohol in 400 ml solution

➾ 0.15 × 400 ml

=> Volume of alcohol in 400 ml solution

➾ 60 ml

=> Volume of others liquid

➾ 340 ml

=> Let x ml of alcohol is added to make it 32% strong

=> Then, volume of alcohol

➾ 60+x ml

=> Volume of other liquid

➾ 340 ml

=> Volume of solution

➾ 400 + x ml

➾ 0.32 × (400 + x) = 60 + x

➾ 128 + 0.32x = 60 + x

➾ 0.68x = 68

➾ 68x = 6800

➾ x = 100 ml ...

Answer : 100 ml pure alcohol must be added to 400ml of a 15% solution to make it's strength 32%

i hope it helps you.