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stonecold
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Consider two statements 03 Aug 2016, 21:41
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

55% (hard)

Question Stats:

62% (01:35) correct 38% (01:18) wrong based on 195 sessions

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Consider two statements
Statement 1 => If X is an integer and X=2p+1 where p is not an integer.
Statement 2 => If Y is an integer and Y=2q where q is not an integer

Which of the following correctly specifies the Even/Odd nature of X and Y respectively
1) Odd,Even
2) Odd,Odd
3) Even,Odd
4) Even,Even
5) Cannot be determined
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C
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DAllison2016
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Consider two statements 03 Aug 2016, 22:28
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stonecold
Consider two statements
Statement 1 => If X is an integer and X=2p+1 where p is not an integer.
Statement 2 => If Y is an integer and Y=2q where q is not an integer

\(2p\) and \(2q\) must be integers as \(X\) and \(Y\) are integers.

For \(2p\) to be an integer, \(p\) must be in the form \(\frac{n}{2}\) for some \(n\)

As \(p\) is not an integer, \(n\) must not be a multiple of 2. \(\implies n \text{ is odd}\)

\(p = \frac{\text{odd}}{2}\\\\
2p = \text{ odd}\\\\
2p + 1 = \text{ even}\)

Apply the same logic for \(q \implies q \text{ is odd}\)

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3) Even,Odd
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mbaprep2016
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Consider two statements 03 Aug 2016, 23:12
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answer will be E
x = integer
x = 2p+1 = integer here p can be any n/2 and n can be even or odd , matters alot 2 get cancelled out
if n were even x would odd
if n odd x would be even
Similarly other
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Consider two statements 05 Aug 2016, 07:46
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mbaprep2016
answer will be E
x = integer
x = 2p+1 = integer here p can be any n/2 and n can be even or odd , matters alot 2 get cancelled out
if n were even x would odd
if n odd x would be even
Similarly other

Hey mbaprep2016,

as per your statement, "here p can be any n/2 and n can be even or odd , matters alot 2 get cancelled out ", if n is even, 2 at the denominator would cancel the 2 at the numerator, which in turn would make p an integer. but it is given in the question that p is NOT an integer.

hence, E is not the right answer.
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Consider two statements 17 May 2025, 01:47
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Statement 1 => If X is an integer and X=2p+1 where p is not an integer.
From 1, X is an integer, p is not an integer. Therefore, for X to be integer, p denominator should be 2 and since p is not an integer, Nr of the number should be only odd. Therefore o+o= even

Statement 2 => If Y is an integer and Y=2q where q is not an integer
From 2, q is also an odd no. using above logic.

Method 2 but same logic:

p= X-1/2 therefore X-1 is not divisible by 2, X-1 is odd therefore X is even.

Similarly, q=Y/2. In this case Y is not divisible by 2. Hence, Odd
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