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If x and y are integers and xy^2 is a positive odd integer

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If x and y are integers and xy^2 is a positive odd integer [#permalink] New post 01 Apr 2009, 02:25
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A
B
C
D
E

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76% (01:49) correct 24% (01:02) wrong based on 212 sessions
If x and y are integers and xy^2 is a positive odd integer, which of the following must be true?

Ⅰ. xy is positive.
Ⅱ. xy is odd.
Ⅲ. x + y is even.

(A) Ⅰ only
(B) Ⅱ only
(C) Ⅲ only
(D) Ⅰ and Ⅱ
(E) Ⅱ and Ⅲ
[Reveal] Spoiler: OA
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Re: Numbers again............ [#permalink] New post 01 Apr 2009, 03:16
Ⅱ. xy is odd.
Ⅲ. x + y is even.


both are correct
because x ad y are both ODD
but y could be -ve ODD
hence I is not CORRECT

good Q
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Re: Numbers again............ [#permalink] New post 26 Aug 2009, 07:52
I think the answer should be E

As (xy)^2 is odd that means xy should be odd only.

Since xy is odd that mean both x & y are odd . When you add two odd numbers, sum should be even.
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Re: Numbers again............ [#permalink] New post 27 Aug 2009, 11:01
xy^2 is a positive odd integer

1) Only way for this to hold true is that if x and y^2 are both odd. As ExO and ExE are even

2) Further Y must be odd. The only way y^2 to be odd is OxO.

Therefore we know both X and Y are odd

i) Not inferrable since we are squaring Y which eliminates +/- inference
ii)As noted in premise 1 OxO is always odd so true
iii) For addition the number property rule is this O+O = E E+E = E and O+E = O
Therefore always even and true

E
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Re: Numbers again............ [#permalink] New post 05 Oct 2009, 21:20
Thought it might help..i was confused initially

just to share a finding. zero is an even number.
Hence, even considering x=1 and y=-1 will yield x+y as zero, which is even.
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Re: Numbers again............ [#permalink] New post 19 Dec 2009, 18:58
one of x or y may be negative hence xy may be negative and even then (xy)^2 is positive odd integer.

Let x=2a and y=2b both be even no.and x+y even hence xy=4ab (xy)^2 = 16(ab)^2 is even hence C is wrong

odd odd*odd =odd
hence (xy)^2 is odd implies xy is odd Hence B is correct

Hence ans is B
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Re: Numbers again............ [#permalink] New post 17 Jan 2010, 18:52
The answer is E.

I.- xy is positive. It's not possible to determine whether xy^2 could be -x*-y (i.e -3*-9 = +27) or x*y (i.e 3*9 = +27).
II.- xy is odd. Anytime an ODD number is multiplied by an EVEN number the answer will be an EVEN number. Since xy^2 is ODD it implies that xy is ODD (.i.e. 27^2 =729 an ODD number).
III.- x + y is even. Always an ODD number plus another ODD number is equal to a EVEN number (.i.e. 3 + 9 =12).
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Re: Numbers again............ [#permalink] New post 18 Jan 2010, 09:20
XY^2 is +ve

1. we cannot say this holds true as square of a number is always +ve.

2. Since XY^2 is ODD , both X n Y should be odd, because product of numbers is odd only when all of them are odd

That means XY is odd = Product of odd numbers is odd.

3. X+Y is even >>> Sum of odd numbers = even

Hence E
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Re: Numbers again............ [#permalink] New post 20 Dec 2010, 11:01
I guess answer is B. III option is not always right.Reason:-

When x=3 and y=-3 xy^2=27(odd integer) but x+y=0 and that is not an odd integer.Hence answer will be B.

Let me have your inputs on the same.
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Re: Numbers again............ [#permalink] New post 20 Dec 2010, 11:15
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Eshika wrote:
I guess answer is B. III option is not always right.Reason:-

When x=3 and y=-3 xy^2=27(odd integer) but x+y=0 and that is not an odd integer.Hence answer will be B.

Let me have your inputs on the same.


If x and y are integers and xy^2 is a positive odd integer, which of the following must be true?
I. xy is positive.
II. xy is odd.
III. x + y is even.

A. I only
B. II only
C. III only
D. I and II
E. II and III

As x and y are integers and xy^2 is a positive odd integer then both x and y must be odd.

Now, let's consider each option:
I. xy is positive --> not necessarily true as x can be positive odd number and y can be negative odd number.

II. xy is odd --> as both x and y are odd then xy=odd*odd=odd, hence this statement is always true.

III. x + y is even --> as both x and y are odd then x+y=odd+odd=even, hence this statement is always true.

Answer: E (II and III)

As for your example: if x=3 and y=-3 then x+y=3-3=0=even, so statemet III is still satisfied.

Hope it's clear.
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Re: Numbers again............ [#permalink] New post 20 Dec 2010, 11:31
But Bunuel I guess 0 is neither odd nor even number. Thats why I ruled out option III.Let me know your thoughts.
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Re: Numbers again............ [#permalink] New post 20 Dec 2010, 11:35
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Eshika wrote:
But Bunuel I guess 0 is neither odd nor even number. Thats why I ruled out option III.Let me know your thoughts.


Zero is an even integer.

An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder.
An even number is an integer of the form n=2k, where k is an integer.

So for k=0 --> n=2*0=0.

For more on number properties check: math-number-theory-88376.html

Hope it's clear.
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Re: Numbers again............ [#permalink] New post 20 Dec 2010, 11:42
Bunuel wrote:
Eshika wrote:
But Bunuel I guess 0 is neither odd nor even number. Thats why I ruled out option III.Let me know your thoughts.


Zero is an even integer.

An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder.
An even number is an integer of the form n=2k, where k is an integer.

So for k=0 --> n=2*0=0.

For more on number properties check: math-number-theory-88376.html

Hope it's clear.



Alright...Thanks :)
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Re: Numbers again............ [#permalink] New post 20 Dec 2010, 13:01
I think you might have confused positive/negative with even/odd. Zero is an even integer that's neither positive nor negative

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Re: Numbers again............ [#permalink] New post 21 Dec 2010, 13:22
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milind1979 wrote:
16. If x and y are integers and xy^2 is a positive odd integer, which of the following must be true?
Ⅰ. xy is positive.
Ⅱ. xy is odd.
Ⅲ. x + y is even.
(A) Ⅰ only
(B) Ⅱ only
(C) Ⅲ only
(D) Ⅰ and Ⅱ
(E) Ⅱ and Ⅲ


xy^2 is a positive odd integer

The moment you come across such information, first of all, think what it implies in this question... If you do, getting to your answer will be quick and easy... Given that x and y are integers,

xy^2 is positive implies that x is positive (Since y^2 is never negative so pos = pos*pos). y is either positive or negative (neither of them is 0)

xy^2 is odd means both x and y are odd. If either one of them were even, xy^2 would have been even.

So we get the following: x - positive odd; y - odd

Now run through the statements to get your answer.
Ⅰ. xy is positive. - Not necessary. If y is negative, xy is negative
Ⅱ. xy is odd. - Necessary since both x and y are odd.
Ⅲ. x + y is even. - Odd + Odd = Even hence necessary

Answer (E)
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Re: Numbers again............ [#permalink] New post 15 Jan 2012, 04:10
Im only getting 45 consistently in maths ...how to improve

The problematic area is DS especially Equalities

Anybody have good notes on that...
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Re: Numbers again............ [#permalink] New post 15 Jan 2012, 04:24
Expert's post
Eshika wrote:
Im only getting 45 consistently in maths ...how to improve

The problematic area is DS especially Equalities

Anybody have good notes on that...


Search for hundreds of question with solutions by tags: viewforumtags.php

DS questions on inequalities: search.php?search_id=tag&tag_id=184
PS questions on inequalities: search.php?search_id=tag&tag_id=189
Hardest DS inequality questions with detailed solutions: inequality-and-absolute-value-questions-from-my-collection-86939.html

Hope it helps.

Finally please post general math questions/queries in the main Math forum. Thanks.
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Re: If x and y are integers and xy^2 is a positive odd integer [#permalink] New post 15 Jan 2012, 21:09
Thanks Bunuel..okay will do that
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Re: If x and y are integers and xy^2 is a positive odd integer [#permalink] New post 15 Jan 2012, 21:14
If x and y are integers and xy^2 is a positive odd integer, which of the following must be true?
Ⅰ. xy is positive.
Ⅱ. xy is odd.
Ⅲ. x + y is even.
(A) Ⅰ only
(B) Ⅱ only
(C) Ⅲ only
(D) Ⅰ and Ⅱ
(E) Ⅱ and Ⅲ

xy^2 is odd implies both x and y are Odd

Ⅰ. xy is positive -> y can be positive or negative
Ⅱ. xy is odd -> true, can be positive or negative depending on y.
Ⅲ. x + y is even -> true.

Option E
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Re: If x and y are integers and xy^2 is a positive odd integer [#permalink] New post 22 May 2013, 03:00
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Re: If x and y are integers and xy^2 is a positive odd integer   [#permalink] 22 May 2013, 03:00
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