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If x, y and z are positive integers, is x  y odd?
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09 Mar 2011, 11:29
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If x, y and z are positive integers, is x  y odd? (1) x=z^2 (2) y=(z1)^2
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Re: 172 OG DS
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09 Mar 2011, 11:40
Baten80 wrote: If x, y and z are positive integers, is x  y odd? (1) x=z^2 (2) y=(z1)^2
Can this problem be solve by plunging number? If x, y and z are positive integers, is x  y odd?(1) x=z^2. No info about y. Not sufficient. (2) y=(z1)^2. No info about x. Not sufficient. (1)+(2) Subtract (2) from (1): \(xy=z^2(z^22z+1)=2z1=evenodd=odd\). Sufficient. Answer: C.
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Re: 172 OG DS
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09 Mar 2011, 21:59
(1) and (2) are clearly not enough on their own. So from (1) and (2) together: z^2  z^2 + 2z  1 = 2z1 which is odd, so answer is C.
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Re: 172 OG DS
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10 Mar 2011, 11:32
Another way to solve this problem:
(1) Two cases: (i) If x = odd => z = odd (ii) If x = even => z = even
(2) Two cases: (i) if y = odd => z = even (and vice versa) (ii) if y = even => Z = odd (and vice versa)
(1) & (2) combined, again two cases: (i) If x = odd => y = even (ii) If x = even => y = odd
As subtracting an odd number from an even number and subtracting an even number from an odd number always results in an odd number it follows that C is the correct solution.



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Re: 172 OG DS
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10 Mar 2011, 21:34
Baten80 wrote: If x, y and z are positive integers, is x  y odd? (1) x=z^2 (2) y=(z1)^2
Can this problem be solve by plunging number? Yes, you can plug in numbers. Generally, in even odd questions, plugging numbers works. (mind you, generally, not always) Break it down in the following way: If x, y and z are positive integers, is x  y odd? Question: Is one of x and y even and one odd? (because x  y will be odd only if one of them is even and one is odd) (1) x=z^2 If z is even, x is even. If z is odd, x is odd. No info about y. or if z = 2, x is 4. If z = 1, x is 1. (2) y=(z1)^2 If z is even, y is odd. If z is odd, y is even. No info about x. or If z = 2, y = 1. If z = 1, y is 0 (even number). Together: If z is even, x is even and y is odd. If z is odd, x is odd and y is even. One is always odd, other is always even. or If z is 2, x = 4 and y = 1 If z = 1, x = 1 and y = 0 Answer (C)
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Re: 172 OG DS
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11 Mar 2011, 18:09
Rephrasing the question  Is x odd and y even or x even and y odd? Or Is x and y squares of consecutive integers respectively?
1. Insufficient. y is unknown 2. Insufficient. x is unknown
combine 1) + 2) z and z1 are consecutive integers. Hence sufficient.



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Re: 172 OG DS
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12 Mar 2011, 18:39
1. Not sufficient.
x = z^2 , no information about y
xy can be even or odd lets say x y z are 16 3 4 respectively => xy = 13 odd x y z are 16 6 4 respectively => xy = 12 even
2. Not sufficient
no x , similar to the above example , depending upon x , xy can be odd or even . Not sufficient.
together
x = z^2 , y = (z1)^2
=> xy = 2z1 .i.e is odd. Hence answer is C.



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If x, y and z are positive integers, is x  y odd?
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08 Mar 2015, 12:51
Another approach might be the following which doesn't involve calculating:
(1) x = z^2 Not suff. (as explained) (2) y = (z1)^2 Not suff. (as explained)
(1) & (2) x = z^2 states that x is some number, which for instance could be even. An even number to any power stays even. So if z^2 is even, x is even.
Now (2) states that (z1)^2. Hence, as z^2 is even, (z1)^2 has to be odd. Again, an odd number to any power stays odd. Therefore, x has to be even (in the example) and y has to be odd.
Of course, it also works for odd x and even y.
Does that make sense?



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If x, y and z are positive integers, is x  y odd?
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05 Dec 2015, 11:21
I think most of the gmat problems could have elegant solutions like the one given below. Though this one is not a very difficult question, i think it has implications. Test writers could make it into a difficult question by changing the answer choices as follows: 1) x=z^p & 2)y=(z1)^q...where p and q are pos int or 1) x=z^p & 2)y=(z3)^q...where p and q are pos int or 1) x=z^p & 2)y=(zm)^q...where m=odd integer What do math experts think about it? gmat1220 wrote: Rephrasing the question  Is x odd and y even or x even and y odd? Or Is x and y squares of consecutive integers respectively?
1. Insufficient. y is unknown 2. Insufficient. x is unknown
combine 1) + 2) z and z1 are consecutive integers. Hence sufficient.
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Re: If x, y and z are positive integers, is x  y odd?
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06 Dec 2015, 11:29
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution. If x, y and z are positive integers, is x  y odd? (1) x=z^2 (2) y=(z1)^2 There are 3 variables (x,y,z) but only 2 equations are given by the 2 conditions, so there is high chance (E) will become the answer. Looking at the conditions together, x=z^2, y=(z1)^2, and as x=odd and y=even or x=even and y=odd, the answer to the question always becomes 'yes' and the answer seems like (C) However, this is a question with commonly made mistakes, so just to make sure, if the conditions are examined separately, we cannot know the value of y from condition 1, and condition 2 is not sufficient as well, so the answer becomes (C). For cases where we need 3 more equations, such as original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 80% chance that E is the answer (especially about 90% of 2 by 2 questions where there are more than 3 variables), while C has 15% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since E is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, C or D.
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Re: If x, y and z are positive integers, is x  y odd?
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13 Jan 2017, 20:41
Question asks if either of x and y is even and other is odd.
Independently neither of two statements answer the question, so insufficient.
Together, if z is even: x:z^2 is even and y:(Z1)^2 is odd Same is the case, if z is odd, hence sufficient.
Ans: C



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Re: If x, y and z are positive integers, is x  y odd?
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01 Apr 2017, 10:30
Baten80 wrote: If x, y and z are positive integers, is x  y odd?
(1) x=z^2 (2) y=(z1)^2 x,y,x>0 (They are positive and zero is neither positive nor negative) xy= Odd Possible if one of the two integers is even. (1) x=z^2 The even/odd nature of x depends on even/odd nature of z. And we have no idea about the even/odd nature of z. Insufficient. BCE (2) y=(z1)^2 Even/odd nature of y depends on even/odd nature of z. Insufficient. Both the statements: x=z^2 ..........eq.1 y=(z1)^2 .........eq.2 Substitute eq. 1 in eq. 2 y=(x^21)^2 If x is odd, y is even. If x is even, y is odd. So one of the two is even and the other is odd. C



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Re: If x, y and z are positive integers, is x  y odd?
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20 Apr 2018, 08:48
Bunuel wrote: Baten80 wrote: If x, y and z are positive integers, is x  y odd? (1) x=z^2 (2) y=(z1)^2
Can this problem be solve by plunging number? If x, y and z are positive integers, is x  y odd?(1) x=z^2. No info about y. Not sufficient. (2) y=(z1)^2. No info about x. Not sufficient. (1)+(2) Subtract (2) from (1): \(xy=z^2(z^22z+1)=2z1=evenodd=odd\). Sufficient. Answer: C. Bunuel how did you get this > \((z^22z+1)\) can you please explain ?



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Re: If x, y and z are positive integers, is x  y odd?
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20 Apr 2018, 10:56
dave13 wrote: Bunuel wrote: Baten80 wrote: If x, y and z are positive integers, is x  y odd? (1) x=z^2 (2) y=(z1)^2
Can this problem be solve by plunging number? If x, y and z are positive integers, is x  y odd?(1) x=z^2. No info about y. Not sufficient. (2) y=(z1)^2. No info about x. Not sufficient. (1)+(2) Subtract (2) from (1): \(xy=z^2(z^22z+1)=2z1=evenodd=odd\). Sufficient. Answer: C. Bunuel how did you get this > \((z^22z+1)\) can you please explain ? Hello I think Bunuel has just squared (z1). Since y = (z1)^2 so it becomes = z^2  2z + 1



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Re: If x, y and z are positive integers, is x  y odd?
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20 Apr 2018, 11:38
amanvermagmat many thanks for clarification just started practicing OG DS questions



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Re: If x, y and z are positive integers, is x  y odd?
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08 May 2018, 11:12
Baten80 wrote: If x, y and z are positive integers, is x  y odd?
(1) x=z² (2) y=(z  1)² Here's an algebraic approach: Target question: Is xy odd?Given: x, y, and z are positive integers Statement 1: x = z² There's no information about y, so there's no way to determine whether or not xy is odd.Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT Statement 2: y = (z  1)² There's no information about x, so there's no way to determine whether or not xy is odd.Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT Statements 1 and 2 combined Statement 1: x = z² Statement 2: y = (z1)² Subtract equations to get: xy = z²  (z1)² Expand to get: xy = z²  [z²  2z + 1] Simplify to get: xy = 2z  1 Since z is a positive integer, we know that 2z is EVEN, which means 2z1 is ODD. If 2z1 is ODD, we can conclude that xy is definitely ODDSince we can answer the target question with certainty, the combined statements are SUFFICIENT Answer: C Cheers, Brent
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Re: If x, y and z are positive integers, is x  y odd?
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