self doubt

Question

a vessel  contains mixture of milk and water in which  water  concentration is 30 percent  .if 20 percent water and 30 percent milk is taken  out and replaced with equal  quantity  of water , then quantity  of water increases by what percent of initial  of water in the mixture?

Comments

Let initial amount of water is 3x, the amount of milk is 7x

30 % milk is taken out, so required milk = 7x -(30*7x)100 = 4.9x

Remaining quantity which is water  = 5.1x

% increase in water = (5.1 x - 3 x)/3 x  * 100
                               = 70 %

---

How will 21 come ?

---

Ok, thanks

Answer 1

Let total mixture = 100 ltr

30 ltr Water + 70 ltr Milk = 100 ltr Mixture

20% of Water = 6 ltr

30% of Milk = 21 ltr

Water Left=24 ltr

Milk Left=49 ltr

Add 27 ltr Water, so water=24+27=51 ltr

% Increase=$((51-30)\div30)\ast100 = 70$%