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Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
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Since we have 2 variables (x and y) and 0 equations, C is most likely the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.

Conditions 1) & 2)

5x(5 + 2 \(\sqrt{5}\) ) - 5\(\sqrt{5}\)y(3 - 2\(\sqrt{5}\))
= 25x + 10\(\sqrt{5}\)x - 15\(\sqrt{5}\)y + 50y (multiplying through the brackets)
= 25x + 50y + 10\(\sqrt{5}\)x - 15\(\sqrt{5}\)y (rearranging the terms)
= 25(x + 2y) + 5\(\sqrt{5}\)(2x - 3y) (taking out common factors)

We must have 2x – 3y = 0 in order for 5\(\sqrt{5}\)(2x-3y) to be zero (and therefore cancel out the square root or irrational number because condition 1 states that the answer is a rational number) when x and y are rational numbers.
Thus, we have 2x = 3y or x/y = 3/2.

Since both conditions together yield a unique solution, they are sufficient.

Therefore, C is the answer.
Answer: C

Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B, or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D, or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.