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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 each condition has 2 equations, D is most likely the answer. So, we should consider each condition on its own first.
Condition 1)
We have x + y = 7 and x - y = 1.
Adding the 2 equations together gives us (x + y) + (x - y) = 7 + 1, 2x = 8, or x = 4.
When we substitute x with 4 in the first equation, we have 4 + y = 7 or y = 3.
Therefore, we have xy = 4*3 = 12.
Since condition 1) yields a unique solution, it is sufficient.
Condition 2)
Case 1: x ≥ y
Then we have 2x + 3y - 13 = x and 3x - y - 6 = y.
Then we have x + 3y - 13 = 0 and 3x - 2y - 6 = 0.
When we subtract three times the first equation from the second equation, we have (3x - 2y - 6) - 3(x + 3y - 13) = 3x - 2y - 6 - 3x - 9y + 39 = -11y + 33 = 0 or y = 3.
When we substitute y with 3 in the first equation, we have x + 3*3 - 13 = 0 or x = 4.
Then, we have xy = 4*3=12.
Case 2: x < y
Then we have 2x + 3y - 13 = y and 3x - y - 6 = x.
Then we have 2x + 2y - 13 = 0 and 2x - y - 6 = 0.
When we subtract the second equation from the first equation, we have (2x - y - 6) - (2x + 2y - 13) = 2x - y - 6 - 2x - 2y + 13 = -3y + 7 = 0 or y = 7/3.
When we substitute y with 7/3 in the second equation, we have 2x - 7/3 - 6 = 2x - 7/3 - 18/3 = 2x - 25/3 = 0, 2x = 25/3, or x = 25/6.
However, we have x > y in this case so we don't have a solution in this case.
Therefore, we have a unique solution of x = 4 and y = 3.
Since condition 2) yields a unique solution, it is sufficient.
Therefore, D is the answer.
Answer: D
This question is a CMT 4(B) question: condition 1) is easy to work with, and condition 2) is difficult to work with. For CMT 4(B) questions, D is most likely the answer.
If the original condition includes "1 variable", or "2 variables and 1 equation", or "3 variables and 2 equations" etc., one more equation is required to answer the question. If each of conditions 1) and 2) provide an additional equation, there is a 59% chance that D is the answer, a 38% chance that A or B is the answer, and a 3% chance that the answer is C or E. Thus, answer D (conditions 1) and 2), when applied separately, are sufficient to answer the question) is most likely, but there may be cases where the answer is A, B, C or E.
