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M22-33

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M22-33  [#permalink]

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New post 16 Sep 2014, 01:17
1
2
00:00
A
B
C
D
E

Difficulty:

  75% (hard)

Question Stats:

46% (01:32) correct 54% (01:00) wrong based on 116 sessions

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Re M22-33  [#permalink]

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New post 16 Sep 2014, 01:17
1
1
Official Solution:

If \(x\) and \(y\) are even integers, which of the following must also be an even integer?

A. \(x^y\)
B. \(2\frac{x}{y}\)
C. \(\frac{x - y}{x + y}\)
D. \(\frac{x^2 - y^2}{2}\)
E. \((x + 1)(y - 1)\)


\(\frac{x^2 - y^2}{2} = \frac{(x - y)(x + y)}{2} = \frac{even*even}{2} = even*integer = even\).

\(x^y\) is 1 if \(y = 0\) and \(x\) is positive.

\(2*\frac{x}{y}\) is 1 if \(x = 2\) and \(y = 4\).

\(\frac{x - y}{x + y}\) might not be an integer at all.

\((x + 1)(y - 1)\) is always odd.


Answer: D
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Re: M22-33  [#permalink]

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New post 18 Apr 2015, 16:16
Here, if x=y (Question didn't say that x and y are different, so we can consider this use case as well), then option (D) becomes 0/2 =0, which is not an even integer. Can you please let me know your thoughts on this?
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Re: M22-33  [#permalink]

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New post 19 Apr 2015, 03:47
reddyMBA wrote:
Here, if x=y (Question didn't say that x and y are different, so we can consider this use case as well), then option (D) becomes 0/2 =0, which is not an even integer. Can you please let me know your thoughts on this?


You should brush up fundamentals before attempting questions.

ZERO:

1. 0 is an integer.

2. 0 is an even integer. An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even.

3. 0 is neither positive nor negative integer (the only one of this kind).

4. 0 is divisible by EVERY integer except 0 itself.

Check more here: tips-and-hints-for-specific-quant-topics-with-examples-172096.html#p1371030
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Re: M22-33  [#permalink]

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New post 15 Aug 2017, 16:10
Option 2 in the question looks like a Mixed fraction and not 2*(x/y)
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Re: M22-33  [#permalink]

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New post 07 Feb 2018, 06:53
Bunuel wrote:
Official Solution:

If \(x\) and \(y\) are even integers, which of the following must also be an even integer?

A. \(x^y\)
B. \(2\frac{x}{y}\)
C. \(\frac{x - y}{x + y}\)
D. \(\frac{x^2 - y^2}{2}\)
E. \((x + 1)(y - 1)\)


\(\frac{x^2 - y^2}{2} = \frac{(x - y)(x + y)}{2} = \frac{even*even}{2} = even*integer = even\).

\(x^y\) is 1 if \(y = 0\) and \(x\) is positive.

\(2*\frac{x}{y}\) is 1 if \(x = 2\) and \(y = 4\).

\(\frac{x - y}{x + y}\) might not be an integer at all.

\((x + 1)(y - 1)\) is always odd.


Answer: D

Hello. How can I know the difficulty level of this or any other quant question (part of gmat club test)?

Sent from my Redmi Note 4 using GMAT Club Forum mobile app
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Re: M22-33  [#permalink]

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New post 07 Feb 2018, 06:57
HarshBazooka wrote:
Bunuel wrote:
Official Solution:

If \(x\) and \(y\) are even integers, which of the following must also be an even integer?

A. \(x^y\)
B. \(2\frac{x}{y}\)
C. \(\frac{x - y}{x + y}\)
D. \(\frac{x^2 - y^2}{2}\)
E. \((x + 1)(y - 1)\)


\(\frac{x^2 - y^2}{2} = \frac{(x - y)(x + y)}{2} = \frac{even*even}{2} = even*integer = even\).

\(x^y\) is 1 if \(y = 0\) and \(x\) is positive.

\(2*\frac{x}{y}\) is 1 if \(x = 2\) and \(y = 4\).

\(\frac{x - y}{x + y}\) might not be an integer at all.

\((x + 1)(y - 1)\) is always odd.


Answer: D

Hello. How can I know the difficulty level of this or any other quant question (part of gmat club test)?

Sent from my Redmi Note 4 using GMAT Club Forum mobile app


You can check the stats in the original post:
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2018-02-07_1755.png

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Re: M22-33  [#permalink]

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New post 07 Feb 2018, 11:42
I used test cases and got down to two answer choices: A and D. I couldn't find an example of two even integers that disproved A?

Any Advice?
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Re: M22-33  [#permalink]

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New post 07 Feb 2018, 11:55
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Re: M22-33  [#permalink]

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New post 08 Feb 2018, 20:51
After initial analysis , answer choices narrowed down to A and D. The option "0" was the caveat in the choice A which made the only possible option is D.

Thanks Bunuel for clear explanation as always
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Re M22-33  [#permalink]

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New post 17 Nov 2018, 08:03
I think this is a high-quality question and I agree with explanation.
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M22-33  [#permalink]

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New post 13 Jan 2019, 03:06
Hi Bunuel,

Using the same logic( i.e. 0 is neither even nor odd) shouldn't A also satisfy the conditions?

x^y would always be even as y can never be 0(as 0 is not an even number).

Thanks,
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New post 13 Jan 2019, 03:11
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New post 13 Jan 2019, 03:35
Thanks, Bunuel. That was rather a stupid query from my end.
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Re M22-33  [#permalink]

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New post 16 Mar 2019, 02:35
I think this is a high-quality question and I agree with explanation.
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Re M22-33   [#permalink] 16 Mar 2019, 02:35
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