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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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The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify conditions if necessary.
The question xy + 1 > x + y is equivalent to (x - 1)(y - 1)>0 for the following reason.
xy + 1 > x + y
=> xy + 1 - x - y > 0
=> xy - x - y + 1 > 0 (rearranging the equation)
=> x(y - 1) - 1(y - 1) > 0
=> (x - 1)(y - 1) > 0
=> x > 1, y > 1 or x < 1, y < 1
Condition 1)
Condition 1) tells us that 0 ≤ x < 1 and 0 ≤ y < 1. This tells us that x < 1, and y < 1, which fits our modified condition. Therefore, the answer is unique, 'yes,' and the condition is sufficient.
Condition 2)
If x = -1, and y = -1, then xy + 1 = (-1)(-1) + 1 = 1 + 1 = 2, x + y = (-1) + (-1) = -2. In this case, xy + 1 is greater than x + y and the answer is 'yes'.
If x = 2, and y = -3, then xy + 1 = (2)(-3) + 1 = -6 + 1 = -5, x + y = 2 + (-3) = -1. In this case, xy + 1 is less than x + y and the answer is 'no'.
Since condition 2) does not yield a unique solution, it is not sufficient.
Therefore, A is the answer.
Answer: A
