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 were initially 32 liters of water in the mixture?
Wine: Water 8lt: 32lt. 1:4 20%:80% (original ratio) 30%:70% (required ratio) In this 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 = 7:8 Let the replaced amount of water with pure wine be X lt. As per the question (40-X)/40 = 7/8 320-8X=280 40 = 8X 5 = X So, 5 lt. Of pure wine replaced water in the mixture
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