Question
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?Solution
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
More Mixture Questions
- Simplify the following expression:
16 + [17 - (8 + 11) + 6 - 3] ÷ 0.2 18/2 of 3/9 of 2/6 of 69690 = ?
- Simplify:
(13 X 11) + (19 X 3) = 400% of √?
8 × 12 + 110 ÷ 5 = 72 + ?
√529 * 5 – 15% of 220 + ? = 120% of 160
18(1/3) + 9(2/3) – 10(1/3) = 1(2/3) + ?
Solve: 3/4÷2/3
140% of 1270 + 60% of 2085 = 1881 + ‘?’% of 287
Find the simplified value of the given expression