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# If jk ≠ 0 and (j, k) lies in the xy-coordinate system, in which quadra

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Math Expert
Joined: 02 Sep 2009
Posts: 43830
If jk ≠ 0 and (j, k) lies in the xy-coordinate system, in which quadra [#permalink]

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30 Mar 2015, 04:41
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If jk ≠ 0 and (j, k) lies in the xy-coordinate system, in which quadrant does (j, k) lie?

(1) jk < 0

(2) j + k > 0

Kudos for a correct solution.
[Reveal] Spoiler: OA

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Joined: 17 Feb 2015
Posts: 28
GPA: 3
Re: If jk ≠ 0 and (j, k) lies in the xy-coordinate system, in which quadra [#permalink]

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30 Mar 2015, 05:13
E. Both statements even together are not sufficient.

Statement 1 alone just implies that (j,k) lies in quadrant II or IV.

Statement 2 alone implies that (j,k) is not in III quadrant.

Statement 1 and 2 together are not enough to determine what quadrant (j,k) lie in.
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Joined: 23 May 2013
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Location: United States
Concentration: Technology, Healthcare
Schools: Stanford '19 (M)
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Re: If jk ≠ 0 and (j, k) lies in the xy-coordinate system, in which quadra [#permalink]

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30 Mar 2015, 09:56
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Bunuel wrote:
Attachment:
pointjk_figure.PNG
If jk ≠ 0 and (j, k) lies in the xy-coordinate system, in which quadrant does (j, k) lie?

(1) jk < 0

(2) j + k > 0

Kudos for a correct solution.

1) jk<0. This means that either j,k are both positive, or both negative. Therefore, (j,k) must be in quadrant II or IV. Insufficient.

2) j+k >0. Imagining this as x and y, $$x+y>0.$$, or $$y>-x$$. (j,k) could be in quadrant I,II, or IV. Insufficient.

Taking 1) and 2) together: If (j,k) = (-3,5) then it's in quadrant II. If (j,k) = (3,-5) then it's in quadrant IV. Insufficient.

Math Expert
Joined: 02 Sep 2009
Posts: 43830
Re: If jk ≠ 0 and (j, k) lies in the xy-coordinate system, in which quadra [#permalink]

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06 Apr 2015, 04:34
Bunuel wrote:

If jk ≠ 0 and (j, k) lies in the xy-coordinate system, in which quadrant does (j, k) lie?

(1) jk < 0

(2) j + k > 0

Kudos for a correct solution.

MAGOOSH OFFICIAL SOLUTION:
Attachment:

pointjk_text.PNG [ 16.79 KiB | Viewed 1459 times ]

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Re: If jk ≠ 0 and (j, k) lies in the xy-coordinate system, in which quadra [#permalink]

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25 Sep 2016, 13:12

STMT1) FITS IN QUADRANTS II AND IV BECAUSE FOR THE NUMBER TO BE LESS THAN ZERO THE SIGNS MUST BE OPPOSITE. INSUFF

STMT2) FITS QUADRANTS I, II, AND IV BECAUSE X MUST BE HIGHER THAN Y AND IT CAN BE THE POSSIBLE USING THE QUADRANTS. INSUFF

TOGETHER INSUFF BECAUSE QUADRANTS II AND IV COMBINED STILL IS NOT ENOUGH.
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Joined: 05 Jan 2017
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Location: India
Re: If jk ≠ 0 and (j, k) lies in the xy-coordinate system, in which quadra [#permalink]

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23 Feb 2017, 03:46
PROMPT ANALYSIS

jk ≠ 0 mean (j,k) cannot lie on any axis.

SUPERSET

The answer could be any one of 1st, 2nd,3rd or 4th quadrant.

Translation

In order to get the answer, we need:
Either exact value of j and k
Any condition to reject or accept the quadrant
2 equations to find the exact value of j and k.

STATEMENT TRANSLATION

St 1: jk<0. That means, one of j and k is negative and the other is positive. Therefore 1st quadrant and and 3rd quadrant is rejected. We are stuck with 2nd and 4th quadrant. INSUFFICIENT. Option A and D rejected.

St 2: j + k >0. Definitely valid for 1st quadrant. Not valid for 3rd quadrant. Cannot say anything about 2nd and 4th quadrant.take data like (-2,4), (4,-2). INSUFFICIENT. Hence option B also eliminated.

St 1& St2: INSUFFICIENT as both the statement combined is not helping to make the decision. Option C also eliminated.

Option E
Re: If jk ≠ 0 and (j, k) lies in the xy-coordinate system, in which quadra   [#permalink] 23 Feb 2017, 03:46
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