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11 Feb 2020, 01:20
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
59% (01:06) correct
41%
(01:10)
wrong
based on 70
sessions
History
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Not Attempted Yet
11 Feb 2020, 01:42
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If x and y are integers, is x odd? Yes?
Or x is anything else? No?
(1) xy + y is odd
y(x+1) = odd
Odd•odd =odd
So y=odd , and (x+1) = odd
x= odd - 1 —> x = even (Sufficient)
(2) y is odd
We only know about y, what of x?
(Not sufficient)
Answer A
Posted from my mobile device
Or x is anything else? No?
(1) xy + y is odd
y(x+1) = odd
Odd•odd =odd
So y=odd , and (x+1) = odd
x= odd - 1 —> x = even (Sufficient)
(2) y is odd
We only know about y, what of x?
(Not sufficient)
Answer A
Posted from my mobile device
11 Feb 2020, 11:21
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Raxit85
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.
Visit https://www.mathrevolution.com/gmat/lesson for details.
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.
Condition 1) tells x is an even integer for the following reason.
Since xy+y = y(x+1) is an odd integer, both y and x+1 are odd integers.
Then x is an even integer and the answer is 'no'.
Since 'no' is also a unique answer by CMT (Common Mistake Type) 1, condition 1) is sufficient.
Condition 2)
Since we don't have any information about the variable x, condition 2) is not sufficient.
Therefore, A is the answer.
