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Question 73
Two bottles A and B contain diluted acid. In bottle A, the amount of water is double the amount of acid while in bottle B, the amount of acid is 3 times that of water. How much mixture(in litres) should be taken from each bottle A and B respectively in order to prepare 5 liters diluted acid containing equal amount of acid and water?
Solution
Therefore, Required ratio = $$\dfrac{1}{4} :Â \dfrac{1}{6} = 6 : 4 = 3 :Â 2$$
Acid : Water in Bottle A = 1 : 2
Acid in Bottle A = $$\dfrac{1}{3}$$
Acid : Water in Bottle B = 3 : 1
Acid in Bottle B = $$\dfrac{3}{4}$$
Acid : Water in Required solution = 1 :Â 1
Acid in required solution = $$\dfrac{1}{2}$$
Therefore, Required ratio = $$\dfrac{1}{4} :Â \dfrac{1}{6} = 6 : 4 = 3 :Â 2$$
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