Quadrant of the coordinate plane(Knewton Prep)

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RadiumBall: Master | Next Rank: 500 Posts; Posts: 132; Joined: Thu Dec 02, 2010 2:49 am; Thanked: 5 times

by RadiumBall » Tue Apr 05, 2011 6:42 pm

In which quadrant of the coordinate plane does point (w, z) lie, if wz â‰ 0?

1.(z, w) is in quadrant III.
2.(-w, z) is in quadrant IV

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Source: Beat The GMAT — Data Sufficiency | Tue Apr 05, 2011 6:42 pm

vineeshp: Legendary Member; Posts: 965; Joined: Thu Jan 28, 2010 12:52 amFollowed by: 34 members; Thanked: 156 times; GMAT Score:720

by vineeshp » Tue Apr 05, 2011 6:48 pm

From 1,
we can find the signs of z,w.

From 2 too, we can find the signs of z,w.

Hence D. Either statement is sufficient.

Vineesh,
Just telling you what I know and think. I am not the expert.

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Anurag@Gurome: GMAT Instructor; Posts: 3835; Joined: Fri Apr 02, 2010 10:00 pmFollowed by: 523 members; Location: Milpitas, CA; Thanked: 1854 times; GMAT Score:770

by Anurag@Gurome » Tue Apr 05, 2011 7:05 pm

RadiumBall wrote:In which quadrant of the coordinate plane does point (w, z) lie, if wz â‰ 0?

1.(z, w) is in quadrant III.
2.(-w, z) is in quadrant IV

Solution:
Let us first consider (1) alone.
It says (z,w) is in quadrant III.
So, both z and w are negative, for example (-2,-3), z = -2 and w = -3.
Now, (w,z) is (-3,-2) which again lies in the third quadrant.
Or, statement (1) alone is sufficient to answer the question.
Next, consider statement (2) alone.
It says (-w,z) is in quadrant IV.
So, -w is positive and z is negative.
This also means that w is negative.
Or (w,z) lies in the third quadrant.
Hence, even statement (2) alone is sufficient to answer the question.

The correct answer is (D).

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RadiumBall: Master | Next Rank: 500 Posts; Posts: 132; Joined: Thu Dec 02, 2010 2:49 am; Thanked: 5 times

by RadiumBall » Tue Apr 05, 2011 8:51 pm

Thanks for the explanation Anurag. I get it now...

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Anurag@Gurome: GMAT Instructor; Posts: 3835; Joined: Fri Apr 02, 2010 10:00 pmFollowed by: 523 members; Location: Milpitas, CA; Thanked: 1854 times; GMAT Score:770

by Anurag@Gurome » Tue Apr 05, 2011 9:35 pm

RadiumBall wrote:Thanks for the explanation Anurag. I get it now...

You are welcome.

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