A mixture of wine and water contains 80% wine. In 50 liters of such a mixture, how many liters of water are required to increase the percentage of water to 50%?
Solution
Correct Answer: Option B
Total mixture = 50 liters, then,
⇒ Amount of wine in the mixture = \(50 \times \frac{{80}}{{100}} = 40\ {\rm{litres}},\)
⇒ Amount of water in the mixture (50 - 40) = 10 liters,
Initially we have 40 liters wine and 10 liters water.
Now we are required to have 50% wine and 50% water in new mixture. So we have to have 40 liters of water, equal to the amount of pure wine (which is constant) available in the mixture. Thus we have to add up 30 liters (40 – 10 = 30) water in the original mixture.
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