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16 Nov 2020, 01:15
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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:
75%
(hard)
Question Stats:
50% (02:06) correct
50%
(02:15)
wrong
based on 92
sessions
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If x and y are integers, is x even?
(1) (x + 2)(y^2 + 7) is even.
(2) (x^3 + 8)(y^2 -4) is even.
(1) (x + 2)(y^2 + 7) is even.
(2) (x^3 + 8)(y^2 -4) is even.
16 Nov 2020, 02:14
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Statement 1: (x + 2)(y^2 + 7) is even.
=> Either (or both) of the entities in pair, must be even.
Assuming, (y^2 + 7) is even
=> y must be odd. But, (x + 2) could be even or odd, and so x could also be even or odd. NOT SUFFICIENT
Statement 2: (x^3 + 8)(y^2 -4) is even.
=> Either (or both) of the entities in pair, must be even.
Assuming, (y^2 -4) is even
=> y must be even. But, (x^3 + 8) could be even or odd, and so x could also be even or odd. NOT SUFFICIENT
Let's check if they are together sufficient? (Now, our answer could be either- C or E)
Assuming, (y^2 + 7), one of the pairs of statement 1, is EVEN
=> y^2 must be odd. (7 is odd + [must be odd] = Even)
=> y must be odd. (square root of an odd number would be an odd number)
=> (y^2 -4), which is other pair of statement 2, must also be odd, since y is odd.
=> (x^3 + 8), pair of statement 2, must be even. (since statement 2 is even- given)
=> x^3 must be even.
=> x must be EVEN
Assuming, (y^2 + 7), one of the pairs of statement 1, is ODD
=> (x + 2) must be EVEN. (Since statement 1 is even- given)
=> x must be EVEN.
Let's check for statement 2 as well, if this assumption holds true in the first place:
If (y^2 + 7) is Odd, it implies y must be even.
So, the other pair of statement 2, (y^2 -4) must also be even. This means that its other pair (x^3 + 8) could be even or odd. => x can take even value to conform with the assumption here.
Hence, x must be EVEN while the other 'entity of y' could be even or odd, if both statements are taken together.
=> Both together sufficient.
Correct Ans- C IMO
PS- Please award Kudos if you find this post helpful. It's very encouraging to answer more questions. Thanks
=> Either (or both) of the entities in pair, must be even.
Assuming, (y^2 + 7) is even
=> y must be odd. But, (x + 2) could be even or odd, and so x could also be even or odd. NOT SUFFICIENT
Statement 2: (x^3 + 8)(y^2 -4) is even.
=> Either (or both) of the entities in pair, must be even.
Assuming, (y^2 -4) is even
=> y must be even. But, (x^3 + 8) could be even or odd, and so x could also be even or odd. NOT SUFFICIENT
Let's check if they are together sufficient? (Now, our answer could be either- C or E)
Assuming, (y^2 + 7), one of the pairs of statement 1, is EVEN
=> y^2 must be odd. (7 is odd + [must be odd] = Even)
=> y must be odd. (square root of an odd number would be an odd number)
=> (y^2 -4), which is other pair of statement 2, must also be odd, since y is odd.
=> (x^3 + 8), pair of statement 2, must be even. (since statement 2 is even- given)
=> x^3 must be even.
=> x must be EVEN
Assuming, (y^2 + 7), one of the pairs of statement 1, is ODD
=> (x + 2) must be EVEN. (Since statement 1 is even- given)
=> x must be EVEN.
Let's check for statement 2 as well, if this assumption holds true in the first place:
If (y^2 + 7) is Odd, it implies y must be even.
So, the other pair of statement 2, (y^2 -4) must also be even. This means that its other pair (x^3 + 8) could be even or odd. => x can take even value to conform with the assumption here.
Hence, x must be EVEN while the other 'entity of y' could be even or odd, if both statements are taken together.
=> Both together sufficient.
Correct Ans- C IMO
Bunuel
PS- Please award Kudos if you find this post helpful. It's very encouraging to answer more questions. Thanks
16 Nov 2020, 08:00
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Bunuel
The exponents are just distractions here. x+2 and x^3 + 8 are the same: they're both even or both odd. y^2 + 7 and y^2 - 4 are different: one is even and the other is odd. So we really have a question like this:
If a and b are integers, is a even?
1. ab is even
2. a(b+1) is even
Neither Statement could be sufficient alone, since each can be true when a is odd, as long as b is the right type of number. But using both Statements, since either b or b+1 is odd, we'll only get an even product both times if a itself is even. So the answer is C.
