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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]

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14 Jan 2011, 16:40
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62% (01:23) correct 38% (01:30) wrong based on 1846 sessions

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

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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14 Jan 2011, 17:28
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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 Kudos [?]: 16894 [23], given: 230 Math Forum Moderator Joined: 20 Dec 2010 Posts: 1970 Kudos [?]: 2005 [3], given: 376 Re: 152. Algebra Absolute value [#permalink] ### Show Tags 10 Mar 2011, 11:15 3 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" _________________ Kudos [?]: 2005 [3], given: 376 Math Expert Joined: 02 Sep 2009 Posts: 41640 Kudos [?]: 124181 [5], given: 12075 Re: 152. Algebra Absolute value [#permalink] ### Show Tags 10 Mar 2011, 11:19 5 This post received KUDOS Expert's post 1 This post was BOOKMARKED 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. _________________ Kudos [?]: 124181 [5], given: 12075 VP Status: There is always something new !! Affiliations: PMI,QAI Global,eXampleCG Joined: 08 May 2009 Posts: 1295 Kudos [?]: 274 [2], given: 10 Re: 152. Algebra Absolute value [#permalink] ### Show Tags 16 Jun 2011, 00:00 2 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 !! Kudos [?]: 274 [2], given: 10 Manager Joined: 26 Jul 2011 Posts: 121 Kudos [?]: 131 [0], given: 16 Location: India WE: Marketing (Manufacturing) Re: Quant Rev. #152 [#permalink] ### Show Tags 07 Sep 2012, 00:22 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. Kudos [?]: 131 [0], given: 16 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7608 Kudos [?]: 16894 [3], given: 230 Location: Pune, India Re: Quant Rev. #152 [#permalink] ### Show Tags 09 Sep 2012, 21:21 3 This post received KUDOS Expert's post 1 This post was BOOKMARKED 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]

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25 Sep 2012, 02:50
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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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Kudos [?]: 16894 [0], given: 230 Math Expert Joined: 02 Sep 2009 Posts: 41640 Kudos [?]: 124181 [0], given: 12075 Re: For how many ordered pairs (x, y) that are solutions of the [#permalink] ### Show Tags 04 Jul 2013, 01:44 Bumping for review and further discussion*. Get a kudos point for an alternative solution! *New project from GMAT Club!!! Check HERE Theory on Abolute Values: math-absolute-value-modulus-86462.html DS Abolute Values Questions to practice: search.php?search_id=tag&tag_id=37 PS Abolute Values Questions to practice: search.php?search_id=tag&tag_id=58 Hard set on Abolute Values: inequality-and-absolute-value-questions-from-my-collection-86939.html _________________ Kudos [?]: 124181 [0], given: 12075 Intern Joined: 05 May 2013 Posts: 27 Kudos [?]: 23 [3], given: 5 GMAT 1: 730 Q50 V39 GRE 1: 1480 Q800 V680 Re: For how many ordered pairs (x, y) that are solutions of the [#permalink] ### Show Tags 05 Jul 2013, 03:54 3 This post received KUDOS $$y=12-2x=2*(6-x).$$ 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. Answer is (d) Kudos [?]: 23 [3], given: 5 Senior Manager Joined: 13 May 2013 Posts: 466 Kudos [?]: 192 [1], given: 134 Re: For how many ordered pairs (x, y) that are solutions of the [#permalink] ### Show Tags 09 Jul 2013, 16:34 1 This post received 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. Kudos [?]: 192 [1], given: 134 GMAT Club Legend Joined: 09 Sep 2013 Posts: 17575 Kudos [?]: 270 [0], given: 0 Re: For how many ordered pairs (x, y) that are solutions of the [#permalink] ### Show Tags 08 Aug 2014, 07:41 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). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________ Kudos [?]: 270 [0], given: 0 GMAT Club Legend Joined: 09 Sep 2013 Posts: 17575 Kudos [?]: 270 [0], given: 0 Re: For how many ordered pairs (x, y) that are solutions of the [#permalink] ### Show Tags 22 Dec 2015, 15:29 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). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________ Kudos [?]: 270 [0], given: 0 Manager Joined: 17 Nov 2013 Posts: 183 Kudos [?]: 33 [0], given: 19 Re: For how many ordered pairs (x, y) that are solutions of the [#permalink] ### Show Tags 09 Apr 2016, 14:15 abs 12 = -12 all the way to 12, which is 25 integers including "0". y= 2(6-x) = even. So y = even. and y is equal the 25 number range. How many possible even numbers are in that range? Answer is 13 possible even y numbers including zero "0". Kudos [?]: 33 [0], given: 19 Director Joined: 22 Jun 2014 Posts: 845 Kudos [?]: 384 [0], given: 108 Location: India Concentration: General Management, Technology GMAT 1: 540 Q45 V20 GPA: 2.49 WE: Information Technology (Computer Software) Re: For how many ordered pairs (x, y) that are solutions of the [#permalink] ### Show Tags 12 Apr 2016, 00:23 tonebeeze wrote: $$2x + y = 12$$ $$|y| \leq 12$$ 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 |y| <= 12 means range of y is -12 <= Y <= +12. which means Y can take any of the value in the set (-12, -11, -10......-1,0,1.....10,11,12). now that we are given 2x + y = 12, y = 12 - 2x we can include all the integer values for X as a solution for y = 12 - 2x as long as y falls in the above range mentioned. Such values of X are (0,1,2....12). 13 is the count for this set. Answer is D. _________________ --------------------------------------------------------------- Target - 720-740 http://gmatclub.com/forum/information-on-new-gmat-esr-report-beta-221111.html http://gmatclub.com/forum/list-of-one-year-full-time-mba-programs-222103.html Kudos [?]: 384 [0], given: 108 Manager Joined: 30 Jun 2015 Posts: 54 Kudos [?]: 2 [0], given: 82 Re: For how many ordered pairs (x, y) that are solutions of the [#permalink] ### Show Tags 23 Oct 2016, 18:43 Thanks Karishma. Great post.. I was under the impression that when we say |X| < a, we get two options x<a OR X >=-a.. How does the equal to sign work? Is the same sign preserved when the modulus sign is opened in both cases? Another eq: |X| <= a X <=a OR X>-a... Will the it be >=-a ?? When does the equal to sign come.. In general I am confused about when we open the modulus, how does the equal to sign come in for both the positive and negative cases.. Any explanation will be really helpful! VeritasPrepKarishma wrote: 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. Kudos [?]: 2 [0], given: 82 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7608 Kudos [?]: 16894 [0], given: 230 Location: Pune, India Re: For how many ordered pairs (x, y) that are solutions of the [#permalink] ### Show Tags 23 Oct 2016, 21:50 cuhmoon wrote: Thanks Karishma. Great post.. I was under the impression that when we say |X| < a, we get two options x<a OR X >=-a.. How does the equal to sign work? Is the same sign preserved when the modulus sign is opened in both cases? Another eq: |X| <= a X <=a OR X>-a... Will the it be >=-a ?? When does the equal to sign come.. In general I am confused about when we open the modulus, how does the equal to sign come in for both the positive and negative cases.. Any explanation will be really helpful! The equal to sign can be used in either case. |x| = x if x >=0 and |x| = -x if x < 0 (it can be x <= 0 too) Note that when x = 0, then |x| = x and |x| = -x too since -0 = 0 only. You can include it in either range. Both work. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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

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24 Oct 2016, 13:42
Thanks Karishma. I do understand that but my doubt is whether the equal to sign comes for both the cases or any one?

If it just comes for once - then how come for |x| <=12, we get both cases -12 and +12

We should consider

X<=12
and
X>-12.. thus -12 should not be included if the equal to sign isn't considered ?? Please help me understand this..

Another example:

Will this open up as

x-4 <=9 and x-4>9

or
x-4<=9 and x-4>=-9

Will the greater/equal to come for both? or just for one case..

In this there is another question on

If x/|x| > x.. here you have solved the question but not using the = sign for any of the cases.. Why is that so?

In summary - i'm confused when does one use the = sign and when doesn't one?

VeritasPrepKarishma wrote:
cuhmoon wrote:
Thanks Karishma. Great post..

I was under the impression that when we say |X| < a, we get two options

x<a OR X >=-a..

How does the equal to sign work? Is the same sign preserved when the modulus sign is opened in both cases?

Another eq:

|X| <= a

X <=a OR X>-a... Will the it be >=-a ?? When does the equal to sign come..

In general I am confused about when we open the modulus, how does the equal to sign come in for both the positive and negative cases.. Any explanation will be really helpful!

The equal to sign can be used in either case.

|x| = x if x >=0 and
|x| = -x if x < 0 (it can be x <= 0 too)

Note that when x = 0, then |x| = x and |x| = -x too since -0 = 0 only.
You can include it in either range. Both work.

Kudos [?]: 2 [0], given: 82

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Re: For how many ordered pairs (x, y) that are solutions of the [#permalink]

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25 Oct 2016, 03:44
1
KUDOS
Expert's post
cuhmoon wrote:
Thanks Karishma. I do understand that but my doubt is whether the equal to sign comes for both the cases or any one?

If it just comes for once - then how come for |x| <=12, we get both cases -12 and +12

We should consider

X<=12
and
X>-12.. thus -12 should not be included if the equal to sign isn't considered ?? Please help me understand this..

Another example:

Will this open up as

x-4 <=9 and x-4>9

or
x-4<=9 and x-4>=-9

Will the greater/equal to come for both? or just for one case..

In this there is another question on

If x/|x| > x.. here you have solved the question but not using the = sign for any of the cases.. Why is that so?

In summary - i'm confused when does one use the = sign and when doesn't one?

VeritasPrepKarishma wrote:
cuhmoon wrote:
Thanks Karishma. Great post..

I was under the impression that when we say |X| < a, we get two options

x<a OR X >=-a..

How does the equal to sign work? Is the same sign preserved when the modulus sign is opened in both cases?

Another eq:

|X| <= a

X <=a OR X>-a... Will the it be >=-a ?? When does the equal to sign come..

In general I am confused about when we open the modulus, how does the equal to sign come in for both the positive and negative cases.. Any explanation will be really helpful!

The equal to sign can be used in either case.

|x| = x if x >=0 and
|x| = -x if x < 0 (it can be x <= 0 too)

Note that when x = 0, then |x| = x and |x| = -x too since -0 = 0 only.
You can include it in either range. Both work.

Include the "=" sign in either range - it won't make a difference. The point is that you need to consider all possible values of x. So you need to consider what is the value of |x| when x is positive, when x is 0 and when x is negative. So you take the cases as x >= 0, x < 0 OR as x > 0, x <= 0 - it won't make any difference. No point including 0 in both ranges. Once you consider it in one range, it is sufficient.

Imagine a case where you are given that x is non-positive. In that case, you need to say
|x| = -x when x <= 0
You need to consider x = 0 as well and you can do that along with x < 0 itself. Because x can take the value 0 too, you cannot ignore it and use only x < 0.
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For how many ordered pairs (x, y) that are solutions of the [#permalink]

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13 May 2017, 09:19
i don't get it

is there a typo in the official guide.

In my book it stated |y|=12
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For how many ordered pairs (x, y) that are solutions of the   [#permalink] 13 May 2017, 09:19

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