Two vessels contain milk and water in the ratio of 7:3 and 2:3 respect...
Answer – a) 2:1 Explanation : Let the ratio be k:1 then in first mixture, milk = 7k/10 and water = 3k/10 and in second mixture, milk = 2/5 and water = 3/5 [7k/10 + 2/5]/[3k/10 3/5] = 3/2 K = 2, so ratio will be 2:1
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Two vessels contain milk and water in the ratio of 7:3 and 2:3 respect...
To solve this problem, we need to consider the ratios of milk and water in each vessel and determine the ratio in which the contents of both vessels must be mixed to obtain a new mixture with a desired milk-water ratio.
Let's consider the first vessel, which contains milk and water in the ratio of 7:3. This means that for every 7 units of milk, there are 3 units of water. Similarly, for the second vessel, the ratio of milk to water is 2:3.
Now, let's assume that we mix x units from the first vessel and y units from the second vessel to obtain the new mixture. Therefore, the total amount of milk in the new mixture would be 7x/10 + 2y/5, and the total amount of water would be 3x/10 + 3y/5.
According to the problem, we want the ratio of milk to water in the new mixture to be 3:2. This can be expressed as (7x/10 + 2y/5) : (3x/10 + 3y/5) = 3/2.
To solve this equation, we can cross-multiply and simplify:
2(7x/10 + 2y/5) = 3(3x/10 + 3y/5)
14x/10 + 4y/5 = 9x/10 + 9y/5
14x + 8y = 9x + 18y
5x = 10y
x/y = 2/1
Therefore, the ratio in which the contents of both vessels must be mixed to obtain a new mixture with a milk-to-water ratio of 3:2 is 2:1. Option (A) is the correct answer.