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If x and y are negative numbers, is x<y? [#permalink]
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If x and y are negative numbers, is x<y? (1) 3x+4<2y+3 (2) 2x−3<3y−4 Bunuel, can you please explain the graphical method to solve this question? Thanks
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nitin6305 wrote: If x and y are negative numbers, is x<y?
(1) 3x+4<2y+3
(2) 2x−3<3y−4
Bunuel, can you please explain the graphical method to solve this question?
Thanks Dear nitin6305, I'm happy to help with this. Idea #1: to graph an inequality, solve for slopeintercept form (i.e. y = mx + b form). When we do that Graph (1) becomes: y > 3x/2 + 1/2 Graph (2) becomes: y > 2x/3 + 1/3 We plot the exact lines (y = 3x/2 + 1/2 and y = 2x/3 + 1/3) to determine the boundaries of the regions. If the inequality is y > mx + b, then the region representing the inequality is above the line; if the inequality is y < mx + b, then the region representing the inequality is below the line. Idea #2: any question about whether x is > or = or < y is question about the line y = x. Read this post about the magical properties of the line y = x. http://magoosh.com/gmat/2012/gmatmath ... lineyx/Here's the image: Attachment:
inequalities with y = x.JPG [ 82.84 KiB  Viewed 8065 times ]
The darker green region, above the line y > 3x/2 + 1/23, is the region representing the first inequality. The yellow region, above the line y = 2x/3 + 1/3, is the region representing the second inequality. The bright spring green region is their overlap, the region satisfied by both inequalities. Notice that everything in that bright spring green region is above the solid green line, which is y = x. Points above the line y = x always have y > x. Does all this make sense? Mike
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Re: If x and y are negative numbers, is x<y? [#permalink]
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13 Aug 2013, 11:18
mikemcgarry wrote: nitin6305 wrote: If x and y are negative numbers, is x<y?
(1) 3x+4<2y+3
(2) 2x−3<3y−4
Bunuel, can you please explain the graphical method to solve this question?
Thanks Dear nitin6305, I'm happy to help with this. Idea #1: to graph an inequality, solve for slopeintercept form (i.e. y = mx + b form). When we do that Graph (1) becomes: y > 3x/2 + 1/2 Graph (2) becomes: y > 2x/3 + 1/3 We plot the exact lines (y = 3x/2 + 1/2 and y = 2x/3 + 1/3) to determine the boundaries of the regions. If the inequality is y > mx + b, then the region representing the inequality is above the line; if the inequality is y < mx + b, then the region representing the inequality is below the line. Idea #2: any question about whether x is > or = or < y is question about the line y = x. Read this post about the magical properties of the line y = x. http://magoosh.com/gmat/2012/gmatmath ... lineyx/Here's the image: Attachment: inequalities with y = x.JPG The darker green region, above the line y > 3x/2 + 1/23, is the region representing the first inequality. The yellow region, above the line y = 2x/3 + 1/3, is the region representing the second inequality. The bright spring green region is their overlap, the region satisfied by both inequalities. Notice that everything in that bright spring green region is above the solid green line, which is y = x. Points above the line y = x always have y > x. Does all this make sense? Mike Dear MikeThanks a lot for the prompt response. All of the above makes sense. Nitin



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Re: If x and y are negative numbers, is x<y? [#permalink]
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31 Aug 2013, 11:20
Is there a quicker way to solve?



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Re: If x and y are negative numbers, is x<y? [#permalink]
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Re: If x and y are negative numbers, is x<y? [#permalink]
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07 Oct 2015, 06:03
mikemcgarry wrote: nitin6305 wrote: If x and y are negative numbers, is x<y?
(1) 3x+4<2y+3
(2) 2x−3<3y−4
Bunuel, can you please explain the graphical method to solve this question?
Thanks Dear nitin6305, I'm happy to help with this. Idea #1: to graph an inequality, solve for slopeintercept form (i.e. y = mx + b form). When we do that Graph (1) becomes: y > 3x/2 + 1/2 Graph (2) becomes: y > 2x/3 + 1/3 We plot the exact lines (y = 3x/2 + 1/2 and y = 2x/3 + 1/3) to determine the boundaries of the regions. If the inequality is y > mx + b, then the region representing the inequality is above the line; if the inequality is y < mx + b, then the region representing the inequality is below the line. Idea #2: any question about whether x is > or = or < y is question about the line y = x. Read this post about the magical properties of the line y = x. http://magoosh.com/gmat/2012/gmatmath ... lineyx/Here's the image: Attachment: inequalities with y = x.JPG The darker green region, above the line y > 3x/2 + 1/23, is the region representing the first inequality. The yellow region, above the line y = 2x/3 + 1/3, is the region representing the second inequality. The bright spring green region is their overlap, the region satisfied by both inequalities. Notice that everything in that bright spring green region is above the solid green line, which is y = x. Points above the line y = x always have y > x. Does all this make sense? Mike Hi mikemcgarryThanks for sharing the succinct graphical approach. I am having trouble understanding how the graph 2 gives a definite answer. per the graph2 y > 2x/3 + 1/3 > the yellow region in the graph I can see in the yellow region we have one small region where x>y and one bigger region where y>x. If that's true then we can't have a definite answer. Please help, I think I have confused things here since the OA is B. Thanks in advance ! Regards, SR



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Re: If x and y are negative numbers, is x<y? [#permalink]
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07 Oct 2015, 10:21
solitaryreaper wrote: Hi mikemcgarry
Thanks for sharing the succinct graphical approach. I am having trouble understanding how the graph 2 gives a definite answer.
per the graph2 y > 2x/3 + 1/3 > the yellow region in the graph
I can see in the yellow region we have one small region where x>y and one bigger region where y>x. If that's true then we can't have a definite answer.
Please help, I think I have confused things here since the OA is B.
Thanks in advance !
Regards, SR Dear solitaryreaper, I'm happy to help. Statement #2 is not sufficient by itself. For most of the region 2x−3<3y−4, it's true that x < y. For example, if (x, y) = (0, 5), then that is consistent with the statement #2 inequality, and it gives a "yes" answer to the prompt question, "is x < y?" By contrast, if (x, y) = (9, 7), then this is also consistent with the statement #2 inequality, but it gives a "no" answer to the prompt question. We can pick different values consistent with the statement #2 inequality that give different answers to the prompt question, so Statement #2 by itself is most certainly not sufficient. Part of the problem is that whoever posted the question posted the WRONG OA! The correct OA for this question is (C), as the graphs I posted above show. I don't have the ability to made that edit, but perhaps Bunuel can make that change. Does this answer your questions? Mike
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Re: If x and y are negative numbers, is x<y? [#permalink]
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07 Oct 2015, 10:33
mikemcgarry wrote: solitaryreaper wrote: Hi mikemcgarry
Thanks for sharing the succinct graphical approach. I am having trouble understanding how the graph 2 gives a definite answer.
per the graph2 y > 2x/3 + 1/3 > the yellow region in the graph
I can see in the yellow region we have one small region where x>y and one bigger region where y>x. If that's true then we can't have a definite answer.
Please help, I think I have confused things here since the OA is B.
Thanks in advance !
Regards, SR Dear solitaryreaper, I'm happy to help. Statement #2 is not sufficient by itself. For most of the region 2x−3<3y−4, it's true that x < y. For example, if (x, y) = (0, 5), then that is consistent with the statement #2 inequality, and it gives a "yes" answer to the prompt question, "is x < y?" By contrast, if (x, y) = (9, 7), then this is also consistent with the statement #2 inequality, but it gives a "no" answer to the prompt question. We can pick different values consistent with the statement #2 inequality that give different answers to the prompt question, so Statement #2 by itself is most certainly not sufficient. Part of the problem is that whoever posted the question posted the WRONG OA! The correct OA for this question is (C), as the graphs I posted above show. I don't have the ability to made that edit, but perhaps Bunuel can make that change. Does this answer your questions? Mike Hi Mike, The stem says that x and y are negative numbers, so the OA is correct.
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Re: If x and y are negative numbers, is x<y? [#permalink]
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07 Oct 2015, 21:07
statement 1. 3x+1<2y,now both no's are negative ,x<y/(3/2)1/3=y/1.51/3 (y/1.5 is increased as negative no,e.g, 1.5<1(1.5/1.5) and then 1/3 added,we can't say x<y .
Statement 2. 2x<3y1 or x<(y x 1.5) 1/2, negative number multiplied by a positive will give a lesser number then 1 added ..again decreased
when x<z where z is definitely lesser than y so x<y ,2 is enough



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Re: If x and y are negative numbers, is x<y? [#permalink]
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10 Oct 2015, 05:16
Bunuel wrote: mikemcgarry wrote: solitaryreaper wrote: Hi mikemcgarry
Thanks for sharing the succinct graphical approach. I am having trouble understanding how the graph 2 gives a definite answer.
per the graph2 y > 2x/3 + 1/3 > the yellow region in the graph
I can see in the yellow region we have one small region where x>y and one bigger region where y>x. If that's true then we can't have a definite answer.
Please help, I think I have confused things here since the OA is B.
Thanks in advance !
Regards, SR Dear solitaryreaper, I'm happy to help. Statement #2 is not sufficient by itself. For most of the region 2x−3<3y−4, it's true that x < y. For example, if (x, y) = (0, 5), then that is consistent with the statement #2 inequality, and it gives a "yes" answer to the prompt question, "is x < y?" By contrast, if (x, y) = (9, 7), then this is also consistent with the statement #2 inequality, but it gives a "no" answer to the prompt question. We can pick different values consistent with the statement #2 inequality that give different answers to the prompt question, so Statement #2 by itself is most certainly not sufficient. Part of the problem is that whoever posted the question posted the WRONG OA! The correct OA for this question is (C), as the graphs I posted above show. I don't have the ability to made that edit, but perhaps Bunuel can make that change. Does this answer your questions? Mike Hi Mike, The stem says that x and y are negative numbers, so the OA is correct. Thanks a lot mikemcgarry for explaining the whole thing. Kudos to Bunuel for pointing out our mistake. In wake of x<0 (given in question stem) , inequality y>x holds true. Hence statement 2 is sufficient. OA should be B. Regards, SR



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If x and y are negative numbers, is x<y? [#permalink]
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10 Oct 2015, 07:37
mikemcgarry wrote: nitin6305 wrote: If x and y are negative numbers, is x<y?
(1) 3x+4<2y+3
(2) 2x−3<3y−4
Bunuel, can you please explain the graphical method to solve this question?
Thanks Dear nitin6305, I'm happy to help with this. Idea #1: to graph an inequality, solve for slopeintercept form (i.e. y = mx + b form). When we do that Graph (1) becomes: y > 3x/2 + 1/2 Graph (2) becomes: y > 2x/3 + 1/3 We plot the exact lines (y = 3x/2 + 1/2 and y = 2x/3 + 1/3) to determine the boundaries of the regions. If the inequality is y > mx + b, then the region representing the inequality is above the line; if the inequality is y < mx + b, then the region representing the inequality is below the line. Idea #2: any question about whether x is > or = or < y is question about the line y = x. Read this post about the magical properties of the line y = x. http://magoosh.com/gmat/2012/gmatmath ... lineyx/Here's the image: Attachment: inequalities with y = x.JPG The darker green region, above the line y > 3x/2 + 1/23, is the region representing the first inequality. The yellow region, above the line y = 2x/3 + 1/3, is the region representing the second inequality. The bright spring green region is their overlap, the region satisfied by both inequalities. Notice that everything in that bright spring green region is above the solid green line, which is y = x. Points above the line y = x always have y > x. Does all this make sense? Mike Hi Mike, Can you please explain this concept about evaluating the relevant region of an inequality? I tried reading your previous links but could not grasp it fully! ;( Would really appreciate an explanation with graphs!! Thanks!



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If x and y are negative numbers, is x<y? [#permalink]
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10 Oct 2015, 08:22
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longfellow wrote:
Hi Mike,
Can you please explain this concept about evaluating the relevant region of an inequality? I tried reading your previous links but could not grasp it fully! ;(
Would really appreciate an explanation with graphs!! Thanks! I am no expert but let me give it a try. As per mikemcgarry 's graph above, y=x is a line passing through (0,0). All points on this line will be such that their ycoordinates will be = xccordinates. Thus examples points on this line will be (0,0),(1,1),(10,10) etc. Remember that for this question, we need to only look in the 3rd quadrant (as both x and y are <0). ....(a) Now take statement 1, 3x+4<2y+3. You can treat this as the linear equation, 3x+4=2y+3, giving you y=1.5x+0.5. You now plot y=1.5x+0.5 and points following the inequality 3x+4<2y+3 or y>1.5x+0.5 will lie ABOVE the line y=1.5x+0.5 (additionally, had the inequality been y<1.5x+0.5, then you would have looked at all the points BELOW y=1.5x+0.5). Additionally, see that y=1.5x+0.5 intersects y=x at (1,1) and this point is valid as per (a) above. Since there is a distinct (and valid) point of intersection between y=x and y=1.5x+0.5, you will have 2 scenarios with 1 scenario giving you a "yes" for y>x while for the other you will get a "no" for "y>x", making statement 1 not sufficient.You can adopt the method above for statement 2 and see that as there are no points of intersection for y=x and y=0.67x+0.33 satisfying (a) above. Thus statement 2 will give a definite answer for "is y>x". Thus this statement is sufficient. Hence B is the correct answer. Hope this helps.



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Re: If x and y are negative numbers, is x<y? [#permalink]
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03 Nov 2017, 06:12
nitin6305 wrote: If x and y are negative numbers, is x<y?
(1) 3x+4<2y+3
(2) 2x−3<3y−4
Bunuel, can you please explain the graphical method to solve this question?
Thanks For such question, I intend to use a personalised method (Valid region & Confusion region). Please correct me if my method is wrong. Statement 1: 3x+4<2y+3 => 3x+1<2y My target is to compare whether x<y? And, I have 2y in the right side. If 3x+1 is equal to 2x (or 2x<3x+1), i can say that 2x<2y. Thus, x<y. So my mission is two compare when 2x < 3x+1? 1< x [So this is the valid region, but any value of x less than 1 & any value of y in consistency with the statement 1 will confuse the x & y relationship i.e. in which (i) x=y, or (ii) x<y or (iii) x>y] For example, if x=5, the value of y can be 6 or 5 or 4. Thus confuses x & y relationship. I think detecting confusion region is highly enough to discard statement 1. Statement 2: 2x−3<3y−4 => 2x<3y1 => 2x<3y1<2y => 3y1<2y => y<1 [So this is the valid region. Thus, for all negative value & any value of y in consistency with the statement 2 will ensure that 2x<2y. (i am not concern about positive numbers because x and y are negative numbers). Thus, x<y.] So, Ans: B.



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Re: If x and y are negative numbers, is x<y? [#permalink]
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06 Nov 2017, 22:36
nitin6305 wrote: If x and y are negative numbers, is x<y?
(1) 3x+4<2y+3
(2) 2x−3<3y−4
Bunuel, can you please explain the graphical method to solve this question?
Thanks Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution. Since we have 2 variables and 0 equation, C is most likely to be the answer. Condition 1) & 2) \(3x + 4 < 2y + 3\) \(2x  3 < 3y  4\) If we add both inequalities, we have \(5x + 1 < 5y  1\). \(5x + 1 < 5y  1 < 5y + 1 ⇔ 5x < 5y or x < y\). Both conditions together are sufficient. Since this question is one of key questions, we need to check A or B by CMT(Common Mistake Type) 4 (A). Condition 1) \(3x + 4 < 2y + 3\) \(x = 2\), \(y = 1\) : Yes \(x = 20\), \(y = 21\) : No. This is not sufficient. Condition 2) \(2x  3 < 3y  4 ⇒ 3x  4 < 3x  3 < 2x  3 < 3y  4\) since \(x < 0\). We have \(3x < 3y\) or\(x < y\). This is sufficient. Therefore, the answer is B. Normally for cases where we need 2 more equations, such as original conditions with 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, we have 1 equation each in both 1) and 2). Therefore C has a high chance of being the answer, which is why we attempt to solve the question using 1) and 2) together. Here, there is 70% chance that C is the answer, while E has 25% chance. These two are the key questions. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer according to DS definition, we solve the question assuming C would be our answer hence using 1) and 2) together. (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
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