The diluted wine contains only 8 liters of wine and the rest is water. A new mixture whose concentration is 30%, is to be formed by replacing wine. How many liters of mixture shall be replaced with pure wine if there was initially 32 liters of water in the mixture ?
Correct Answer: 5
Wine Water
8L 32L
1 : 4
20 % 80% (original ratio)
30 % 70% (required ratio)
In ths case, the percentage of water being reduced when the mixture is being replaced with wine.
so the ratio of left quantity to the initial quantity is 7:8
Therefore , 78=1-K40
=> K = 5 Lit