Question

A mixture contains 80% acid and rest water. Part of the mixture that should be removed and replaced by same amount of water to make the ratio of acid and water 4 : 3 is:

Answer 1

Answer:

2/7

Explanation:

Let Original Volume of Mixture = x litre and it's to be replaced by y litre of water.

Ratio of Acid & Water in mixture = 80% : 20% = 4:1 =5

volume of the Acid in x - y litre of Mixture = 4/5 (x - y) litre;

and volume of water in x - y litre of Mixture = 1/5 (x - y) Litre

But volume of water will be increased by y litre = 1/5 (x - y) litre + y Litre. ==> (x + 4y)/5

Now new ratio of Acid to Water is given 4 : 3

Thus 4/5 (x - y) : (x + 4y)/5 = 4:3

Solving we get (4x - 4y) / (x + 4y) = 4 : 3

==> 16x - 12y = 4x + 16y

==> 12x = 28y

==> y/x = 12/28 = 2/7