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#### OG - DS - #119

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 Post subject: OG - DS - #119  Posted: Wed Jun 13, 2007 9:21 pm
 Is 2x-3Y2 and Y>0 Can you help me understand how I approach the 2nd statement? Thanks, Mariela :shock:

 Post subject:   Posted: Thu Jun 14, 2007 1:00 am
 Mariela, This is how I would solve the question: (I) Is -2 < x^2 ? Yes, because anything to the power of 2 is going to be either 0 or positive. Therefore, sufficient. (II) I would test out a couple of numbers here. Test 1: X = 2.1 and Y = 0.1 2(2.1) - 0.3 < (2.1)^2 True Test 2: X = 2.1 and Y = 100 2(2.1) - 300 < (2.1)^2 True Test 3: X = 100 and Y = 0.1 200 - 0.3 < 100^2 True Test 4: X = 100 and Y = 100 200 - 300 < 100^2 True I am pretty sure there are better ways to solve, but this is how I would approach it and I would select answer D. It took less than 2 minutes. :) Hope that helps!

 Post subject:   Posted: Thu Jun 14, 2007 1:16 pm
 Is 2x-3Y2 and Y>0 Mariela - To answer whether (2) is sufficient by itself, it helps to rearrange the original inequality (rephase the question): 2x-3y2, then the left side of the inequality is always negative, because the x is positive and the quantity (2-x) will be negative. If y>0 then the quantity 3y is always positive, so it will always be the case that 2x-3y

 Post subject:   Posted: Thu Jun 14, 2007 2:44 pm
 Beautiful, Jeff! Yes, using a plug and play method is risky as you may miss a region (as Jeff stated). Mariela, I would suggest trying to solve something algebraically first, but do not spend a lot of time if you get nowhere or cannot "see" the algebraic solution; if you get stuck, fall back to the plug and play.

 Post subject: Graphing would work, too.  Posted: Fri Jun 15, 2007 4:51 pm
 ManhattanGMAT Staff

Posts: 899
Location: St. Louis, MO
 Good solutions, and excellent point about the potential pitfalls of plug-and-play. Although hard to reproduce here, you may want to try this on your own: graphing. The question rephrases to "Is y > (2/3)x - x^2?" Plot (2/3)x - x^2 and you get an upside-down "U" centered just to the right of the y-axis, with the base of the "U" just above the x-axis. The visual interpretation of the rephrase is "Are all of the points (x, y) in the region above the 'U'?" (1) Rephrases to the equation of a line y = (2/3)x + 2/3, which crosses the y-axis at 2/3 (above the "U") and has a slope of 2/3. The entire line is above the "U." SUFFICIENT. (2) The rectangularly bounded region to the right of x = 2 and above y = 0. This entire region is above the "U," too. SUFFICIENT. Obviously, this method is not required for this problem, and there is a little set-up time involved in drawing the "U." Such a method could be a life-saver on a problems with more complicated constraints, or when you want to plug-and-play but have no idea what values to try. Also, some people are more visually inclined and can follow this approach better than the algebraic one. _________________Emily Sledge Instructor ManhattanGMAT

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