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# For how many ordered pairs (x, y) that are solutions of the

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For how many ordered pairs (x, y) that are solutions of the [#permalink]  14 Jan 2011, 15:40
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59% (02:26) correct 41% (01:24) wrong based on 747 sessions
$$2x + y = 12$$
$$|y| \leq 12$$

For how many ordered pairs (x, y) that are solutions of the system above are x and y both integers?

A. 7
B. 10
C. 12
D. 13
E. 14
[Reveal] Spoiler: OA
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Re: Quant Rev. #152 [#permalink]  14 Jan 2011, 16:28
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tonebeeze wrote:
152.

$$2x + y = 12$$
$$|y| \leq 12$$

For how many ordered pairs (x, y) that are solutions of the system above are x and y both integers?

a. 7
b. 10
c. 12
d. 13
e. 14

The solution of $$|y| \leq 12$$ is straight forward.
$$-12 \leq y \leq 12$$
(If you are not comfortable with this, check out my blog post:
http://www.veritasprep.com/blog/2011/01/quarter-wit-quarter-wisdom-do-what-dumbledore-did/

If both x and y have to be integers, y should be an integer and hence can take any value from the set {-12, -11, -10 ... 10, 11, 12} i.e. any one of 25 values (these are 25 values -12 to -1 (12 values), 0, 1 to 12 (another 12 values)) 13 of them are even and 12 of them are odd.

$$2x + y = 12$$
Every time y is even, x will be integer. e.g. y = 12, x = 0 (because x = (12 - even)/2 will be an integer)
Every time y is odd, x will be non-integer e.g. y = 1, x = 5.5 (because x = (12 - odd)/2 will not be an integer)

Therefore, for 13 values, x and y both will be integers.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Math Forum Moderator Joined: 20 Dec 2010 Posts: 2027 Followers: 142 Kudos [?]: 1209 [2] , given: 376 Re: 152. Algebra Absolute value [#permalink] 10 Mar 2011, 10:15 2 This post received KUDOS Sol: |y| <= 12 Means; -12<=y<=12 2x + y = 12 x = (12-y)/2 x will be integers for y=even; because even-even = even and even is always divisible by 2. We need to find out how many even integers are there between -12 and 12 ((12-(-12))/2)+1 = (24/2)+1 = 12+1 = 13 Ans: "D" _________________ Math Expert Joined: 02 Sep 2009 Posts: 29100 Followers: 4722 Kudos [?]: 49622 [4] , given: 7400 Re: 152. Algebra Absolute value [#permalink] 10 Mar 2011, 10:19 4 This post received KUDOS Expert's post Baten80 wrote: 2x + y = 12 |y| <= 12 152. For how many ordered pairs (x , y) that are solutions of the system above are x and y both integers? A. 7 B. 10 C. 12 D. 13 E. 14 Given: $$-12\leq{y}\leq{12}$$ and $$2x+y=12$$ --> $$y=12-2x=2(6-x)=even$$, (as $$x$$ must be an integer). Now, there are 13 even numbers in the range from -12 to 12, inclusive each of which will give an integer value of $$x$$. Answer: D. _________________ VP Status: There is always something new !! Affiliations: PMI,QAI Global,eXampleCG Joined: 08 May 2009 Posts: 1356 Followers: 14 Kudos [?]: 177 [1] , given: 10 Re: 152. Algebra Absolute value [#permalink] 15 Jun 2011, 23:00 1 This post received KUDOS -12<= y <=12 gives 0<=x <=12 thus 13 values in total. _________________ Visit -- http://www.sustainable-sphere.com/ Promote Green Business,Sustainable Living and Green Earth !! Manager Joined: 26 Jul 2011 Posts: 130 Location: India WE: Marketing (Manufacturing) Followers: 1 Kudos [?]: 67 [0], given: 15 Re: Quant Rev. #152 [#permalink] 06 Sep 2012, 23:22 Hi Karishma Using the number properties this indeed is very convenient to solve. I was wondering can we substitute y = 12 - 2x in the inequality and solve for the possible values of x. Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 5850 Location: Pune, India Followers: 1478 Kudos [?]: 7931 [2] , given: 190 Re: Quant Rev. #152 [#permalink] 09 Sep 2012, 20:21 2 This post received KUDOS Expert's post ratinarace wrote: Hi Karishma Using the number properties this indeed is very convenient to solve. I was wondering can we substitute y = 12 - 2x in the inequality and solve for the possible values of x. Certainly and it is quick too. y = 12 - 2x Whenever x is an integer, y will be an integer. So if we can solve for integral values of x, the number of values we get will be the number of solutions. $$|y| \leq 12$$ $$|12 - 2x| \leq 12$$ $$|x - 6| \leq 6$$ From 6, x should be at a distance less than or equal to 6. So x will lie from 0 to 12 i.e. 13 values. (Check http://www.veritasprep.com/blog/2011/01 ... edore-did/ if this is not clear) There are 13 solutions. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: For how many ordered pairs (x, y) that are solutions of the [#permalink]  10 Sep 2012, 09:36
Thanks Karishma..Wonderful explaination
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Re: For how many ordered pairs (x, y) that are solutions of the [#permalink]  25 Sep 2012, 01:50
Expert's post
Responding to a pm:
Changing the sign within the mod has no impact on anything outside the mod.

$$|6 - x| \leq 12$$ is same as
$$|x - 6| \leq 12$$

Think about it: Whether you write |x| or |-x|, it is the same.
|6| = |-6|

So for every value of x,
|x - 6| = |6 - x|
So you don't need to flip the inequality sign.

|x - 6| and - |x - 6| are of course different. If you change |x - 6| to - |x - 6|, you will need to flip the inequality sign.
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Re: For how many ordered pairs (x, y) that are solutions of the [#permalink]  04 Jul 2013, 00:44
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Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

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DS Abolute Values Questions to practice: search.php?search_id=tag&tag_id=37
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Hard set on Abolute Values: inequality-and-absolute-value-questions-from-my-collection-86939.html

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Re: For how many ordered pairs (x, y) that are solutions of the [#permalink]  05 Jul 2013, 02:54
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$$y=12-2x=2*(6-x).$$
Since $$|y| \leq 12 , -12 \leq y \leq 12$$ . Substituting for y from above, $$-6 \leq (6-x) \leq 6.$$. This reduces to $$x \geq 0$$ and $$x \leq 12.$$ Including 0 and 12 there are thus 13 integer solutions.
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Re: For how many ordered pairs (x, y) that are solutions of the [#permalink]  09 Jul 2013, 15:34
1
KUDOS
2x+y=12
|y|<=12

For how many ordered pairs (x, y) that are solutions of the system above are x and y both integers?

y=12-2x
|y|<=12
|12-2x| <= 12
12 - 2x <= 12
-2x <= 0
x>=0

-(12-2x) <= 12
-12+2x <= 12
2x <= 24
x<=12

13 solutions between 0 and 12 inclusive.
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Re: For how many ordered pairs (x, y) that are solutions of the [#permalink]  08 Aug 2014, 06:41
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: For how many ordered pairs (x, y) that are solutions of the [#permalink]  24 Nov 2014, 21:20
Hi Moderators,

Does this qualify as a 700 level question, I think it should be in the lower range?

Thanks
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Re: For how many ordered pairs (x, y) that are solutions of the [#permalink]  25 Nov 2014, 03:47
Expert's post
aj0809 wrote:
Hi Moderators,

Does this qualify as a 700 level question, I think it should be in the lower range?

Thanks

User statistics on the question say it's 700.
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Re: For how many ordered pairs (x, y) that are solutions of the   [#permalink] 25 Nov 2014, 03:47
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