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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
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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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84% (00:53) correct 16% (00:50) wrong based on 130 sessions

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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.

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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, 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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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.

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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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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 1761 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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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
Solution:

Given: jk ≠ 0 (j, k) lies in the xy-coordinate system.

To find: In which quadrant does (j, k) lie?

For to find out this we should know that, the polarities of the point (j,k) in different quadrants.

Analysis of statement 1: jk < 0
The product of j and k should give us the value which is less than 0. This means that, one of the values either “j” or “k” must be negative and other must be positive. (As we get product of positive and negative as negative itself. )
That means, the point (j, k) lie either in 2nd quadrant (where “j” is negative and “k” is positive) or (j,k) lie in 4th quadrant (where “j” is positive and “k” is negative).
Hence statement 1 is insufficient to answer. We can eliminate the options A and D.

Analysis of statement 2: j + k > 0
This case is definitely valid for I quadrant. Not at all valid for III quadrant as both j and k will be negative.
Cannot comment about II and IV quadrant. For example: The points (-4, 8) and (8, -4). So the point (j,k) can lie in either of the quadrants.
Hence statement 2 is insufficient to answer. So we can eliminate option B.

Combining the statements 1 and 2: we get:
From statement 1, we know that the point (j,k) can lie in 2nd or 4th quadrant.
From statement 2, we know the point (j,k) can lie in 1st , 2nd or in 4th quadrant.
Even after combining we get that point (j,k) can lie in 2nd or in 4th quadrant.
Hence insufficient.

So the correct answer option is “E”.

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