Statement 1: c*(d + 1) is even. Implies,
(1) c even, d even
or
(2) c even, d odd
or
(3) c odd, d odd
Not sufficient.
Statement 1: (c + 2)*(d + 4) is even. Implies,
(1) c even, d even
or
(2) c even, d odd
or
(3) c odd, d even
Not sufficient.
1 & 2 Together: Third possibility of statement 1 (c & d both odd) and third possibility of statement 2 ( c odd, d even) can't coexist. Only first two possibilities of both of the cases are possible. Thus d may be even or odd, but c is always even.
Sufficient.
The correct answer is C.
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Rahul@gurome
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Night reader
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Rahul, you mention that statements 1 and 2 contain different sets of variations, then you continue with opting C as two statements are sufficient to solve the problem. What was reasoning behind C?Rahul@gurome wrote:Statement 1: c*(d + 1) is even. Implies,
(1) c even, d even
or
(2) c even, d odd
or
(3) c odd, d odd
Not sufficient.
Statement 1: (c + 2)*(d + 4) is even. Implies,
(1) c even, d even
or
(2) c even, d odd
or
(3) c odd, d even
Not sufficient.
1 & 2 Together: Third possibility of statement 1 (c & d both odd) and third possibility of statement 2 ( c odd, d even) can't coexist. Only first two possibilities of both of the cases are possible. Thus d may be even or odd, but c is always even.
Sufficient.
The correct answer is C.
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Rahul@gurome
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Try to understand the logic that when two statements are combined then we must treat both of them as true. That may or may not result in answering the question, but we must treat both of them as TRUE.Night reader wrote:...
Rahul, you mention that statements 1 and 2 contain different sets of variations, then you continue with opting C as two statements are sufficient to solve the problem. What was reasoning behind C?
Now in this case, combining two statements we see that first two possibilities are identical. The third possibilities says that,
- (1) For statement 1, if c is odd then d must be odd.
(2) For statement 2, if c is odd then d must be even.
If this is still not clear, analyze the statements individually.
Case 1: c even
- (1) c*(d + 1) is always even for d even or odd
(2) (c + 2)*(d + 4) is always even for d even or odd
- (1) c*(d + 1) even if d is odd
(2) (c + 2)*(d + 4) is even if d is even
Hope this helps.
Rahul Lakhani
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Quant Expert
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