Question

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

Answer 1

Ans: 2727

Solution 1
Let total quantity of the mixture = 70 litre
then quantity of acid = 70×80100=5670×80100=56 litre

After the replacement, quantity of acid = 70×47=4070×47=40 litre

This means,  56 - 40 = 16 litre of acid is removed from the original mixture.

Suppose xx litre of the mixture was removed and replaced by water
then x×80100=16x×80100=16
=> xx= 20 litre

Therefore, required part of the mixture = 2070=272070=27

Solution 2 (using rule of alligation)
part of acid in the original mixture = 80100=4580100=45
part of acid in the final solution = 4747

4545         0
  \      /
     4747       
  /      \
47−0=4747−0=47    45−47=83545−47=835


Ratio = 47:835=20:8=5:247:835=20:8=5:2
=>  Quantity of the original mixture : Quantity of the water = 5 : 2

Therefore, 2727 of the mixture is removed and replaced with water.