A mixture of sugar and water contains sugar and water in the ratio of 3: 2. Another mixture of sugar and water contains sugar and water in the ratio 2: 5. In what ratio should the two mixtures be mixed so that the resulting mixture contains equal proportion of sugar and water?
A. 2 : 1
B. 3 : 1
C. 3 : 2
D. 4 : 1
Explanation: Let \(x\) and \(y\) be the amounts of Mixture 1 and Mixture 2, respectively. Mixture 1 (Ratio 3:2, Total 5): Sugar concentration = \(\frac{3}{5}\). Water concentration = \(\frac{2}{5}\). Mixture 2 (Ratio 2:5, Total 7): Sugar concentration = \(\frac{2}{7}\). Water concentration = \(\frac{5}{7}\). Resulting mixture needs equal proportion of sugar and water (Ratio 1:1). Sugar concentration = \(\frac{1}{2}\). The total amount of sugar in the new mixture is: \(\frac{3}{5}x + \frac{2}{7}y\). The total amount of mixture is \(x+y\). The concentration of sugar in the new mixture must be \(\frac{1}{2}\). \(\frac{\frac{3}{5}x + \frac{2}{7}y}{x+y} = \frac{1}{2}\). \(2 (\frac{3}{5}x + \frac{2}{7}y) = x+y\). \(\frac{6}{5}x + \frac{4}{7}y = x+y\). \(\frac{6}{5}x - x = y - \frac{4}{7}y\). \(\frac{1}{5}x = \frac{3}{7}y\). The required ratio \(\frac{x}{y} = \frac{\frac{3}{7}}{\frac{1}{5}} = \frac{3}{7} \times \frac{5}{1} = \frac{15}{7}\). Since $15:7$ is not an option, the answer is **None of these**.
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