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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.
Since we have 2 variables (x and y) and 0 equations, C is most likely to be 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) together:
\(10|x|=|y|\)
\(=> y = ±10x\)
Case 1: \(y = 10x\)
\(\frac{x}{(x+2y)} = \frac{x}{(x+20x)} = \frac{x}{21x} = \frac{1}{21}< 1\)
Case 2: \(y = -10x\)
\(\frac{x}{(x+2y)} = \frac{x}{(x-20x)} = \frac{x}{(-19)x} = -(\frac{1}{19}) < 1\)
Both conditions together are sufficient.
Since this question is an inequality question (one of the key question areas), CMT (Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B.
Condition 1):
\(10|x|=|y|\)
\(=> y = ±10x\)
Case 1: \(y = 10x\)
\(\frac{x}{(x+2y)} = \frac{x}{(x+20x)} = \frac{x}{21x} = \frac{1}{21} < 1\)
Case 2: \(y = -10x\)
\(\frac{x}{(x+2y)} = \frac{x}{(x-20x)} = \frac{x}{(-19)x} = -(\frac{1}{19}) < 1\)
Thus, condition 1) only is sufficient.
Condition 2):
Since it gives us no information about y, condition 2) is not sufficient.
Therefore, A is the answer.
Answer: A
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