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If (x/y)>2 , is 3x+2y<18? (1) xy is less than [#permalink]
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28 Jul 2008, 21:21
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If \((x/y)>2\), is \(3x+2y<18?\)
(1) \(xy\) is less than \(2\)
(2) \(yx\) is less than \(2\)



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Re: DS: Tricky one [#permalink]
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28 Jul 2008, 21:47
tarek99 wrote: If \((x/y)>2\), is \(3x+2y<18?\)
(1) \(xy\) is less than \(2\)
(2) \(yx\) is less than \(2\) is it A. i used xy plane and i'm not able to upload the file...



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Re: DS: Tricky one [#permalink]
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28 Jul 2008, 23:24
YES! the OA is A. Interesting....I never thought to use the xyplane. I know how to draw the \(y<(3/2)x + 9\) on the xyplane. Would you please show me how you logically approached each statement? Don't worry about the drawing because I can draw it as you explain, so just explain I'll really appreciate it! thanks



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Re: DS: Tricky one [#permalink]
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28 Jul 2008, 23:34
tarek99 wrote: YES! the OA is A. Interesting....I never thought to use the xyplane. I know how to draw the \(y<(3/2)x + 9\) on the xyplane. Would you please show me how you logically approached each statement? Don't worry about the drawing because I can draw it as you explain, so just explain I'll really appreciate it! thanks i'll try it can be a bit confusing first draw following lines on xy plane x=2y  simple just connect (0,0) with (4,2) and extend on both sides 3x+2y=18> x/6+y/9=1  again simple connect (6,0) and (0,9) and extend .. xy=2 > (2,0) and (0,2) yx=2 > (2,0) and (0,2) mark the are x/y > 2. this'll be the area between the line x=2y and x axis on both sides of y axis select target area 3x+2y<18 ... this will be area left of line 3x+2y=18 ..... statemet 1 : area to the left of line xy=2 ..... suff statement 2 : area to the right of line yx=2 ... not suff....



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Re: DS: Tricky one [#permalink]
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29 Jul 2008, 00:49
Thank durgesh79, but when I try to draw xy plane as you discribed above, I am still confused. Sometimes I deal with math questions like this, I don't know how to sovle them. Please explain in more detail. Here are some DS examples 1. If x and y are positive, is 4x > 3y? (1) x>y  x (2) x/y < 1 2. If x and y are positive, is 3x> 7y? (1) x>y+4 (2) 5x<14y Please help me, thanks lot!!!



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Re: DS: Tricky one [#permalink]
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29 Jul 2008, 01:04
fiesta wrote: Thank durgesh79, but when I try to draw xy plane as you discribed above, I am still confused. Sometimes I deal with math questions like this, I don't know how to sovle them. Please explain in more detail. Here are some DS examples 1. If x and y are positive, is 4x > 3y? (1) x>y  x (2) x/y < 1 2. If x and y are positive, is 3x> 7y? (1) x>y+4 (2) 5x<14y Please help me, thanks lot!!! refer below threads 7t669827t675047t669777t671967t671837t67700plotting conditions on xy plane may not be the best approach in all the cases, but with some practise it is one of the easiest....



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Re: DS: Tricky one [#permalink]
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29 Jul 2008, 02:19
durgesh79, Thank you so much, good luck and good study!!!



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Re: DS: Tricky one [#permalink]
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29 Jul 2008, 02:53
durgesh79 wrote: tarek99 wrote: YES! the OA is A. Interesting....I never thought to use the xyplane. I know how to draw the \(y<(3/2)x + 9\) on the xyplane. Would you please show me how you logically approached each statement? Don't worry about the drawing because I can draw it as you explain, so just explain I'll really appreciate it! thanks i'll try it can be a bit confusing first draw following lines on xy plane x=2y  simple just connect (0,0) with (4,2) and extend on both sides 3x+2y=18> x/6+y/9=1  again simple connect (6,0) and (0,9) and extend .. xy=2 > (2,0) and (0,2) yx=2 > (2,0) and (0,2) mark the are x/y > 2. this'll be the area between the line x=2y and x axis on both sides of y axis select target area 3x+2y<18 ... this will be area left of line 3x+2y=18 ..... statemet 1 : area to the left of line xy=2 ..... suff statement 2 : area to the right of line yx=2 ... not suff.... wow, really nice approach, but I have a question regarding my interpretation of the xyplane: Now, from statement 1, I can see that there is an intersection between x2<y and x>2y just before the line 2y=3x+18, if I drew it correctly. so does that I mean that I should consider only the area between x2<y and x>2y until that intersection? if so, then it makes a sense how 2y<3x+18 is sustained. Also, in statement 2: I really don't understand how to interpret it. Does the line yx<2 includes all the area below it including the area below the x=2y line? if so, then I can see that this area will include below and above the line 2y=3x+18 If i'm correct with how I explained it, then this is definitely an amazing approach on how to approach such problems with inequality because it is much safer and a lot easier to visual and see how each equation really works. +1 for your approach. just tell me whether I was correct how I looked at this graph. thanks



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Re: DS: Tricky one [#permalink]
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29 Jul 2008, 03:11
tarek99 wrote: wow, really nice approach, but I have a question regarding my interpretation of the xyplane: Now, from statement 1, I can see that there is an intersection between x2<y and x>2y just before the line 2y=3x+18, if I drew it correctly. so does that I mean that I should consider only the area between x2<y and x>2y until that intersection? if so, then it makes a sense how 2y<3x+18 is sustained.
Also, in statement 2: I really don't understand how to interpret it. Does the line yx<2 includes all the area below it including the area below the x=2y line? if so, then I can see that this area will include below and above the line 2y=3x+18
If i'm correct with how I explained it, then this is definitely an amazing approach on how to approach such problems with inequality because it is much safer and a lot easier to visual and see how each equation really works. +1 for your approach. just tell me whether I was correct how I looked at this graph. thanks yes you are right aboth both st1 and st2 .... one simple way to check whether to take left or right area is to check a simple point (say 0,0 ).... yx < 0, (0,0) is satisfying this 0<2 ... means we are looking at the same side of line yx=2 which has (0,0).



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Re: DS: Tricky one [#permalink]
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29 Jul 2008, 03:44
durgesh79 wrote: tarek99 wrote: wow, really nice approach, but I have a question regarding my interpretation of the xyplane: Now, from statement 1, I can see that there is an intersection between x2<y and x>2y just before the line 2y=3x+18, if I drew it correctly. so does that I mean that I should consider only the area between x2<y and x>2y until that intersection? if so, then it makes a sense how 2y<3x+18 is sustained.
Also, in statement 2: I really don't understand how to interpret it. Does the line yx<2 includes all the area below it including the area below the x=2y line? if so, then I can see that this area will include below and above the line 2y=3x+18
If i'm correct with how I explained it, then this is definitely an amazing approach on how to approach such problems with inequality because it is much safer and a lot easier to visual and see how each equation really works. +1 for your approach. just tell me whether I was correct how I looked at this graph. thanks yes you are right aboth both st1 and st2 .... one simple way to check whether to take left or right area is to check a simple point (say 0,0 ).... yx < 0, (0,0) is satisfying this 0<2 ... means we are looking at the same side of line yx=2 which has (0,0). Actually, I use another method. Whenever I see y<x+2, then I'm actually considering the area BELOW line y=x+2. However, when I see y>x+2, then i'm considering the area ABOVE the line y=x+2. that also works, no?



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Re: DS: Tricky one [#permalink]
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29 Jul 2008, 03:49
tarek99 wrote: Actually, I use another method. Whenever I see y<x+2, then I'm actually considering the area BELOW line y=x+2. However, when I see y>x+2, then i'm considering the area ABOVE the line y=x+2. that also works, no? yes that works as well..... i like to make sure that the area i'm marking is the right area... (0,0) is just a way to double check.. doesnt take much time ....



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Re: DS: Tricky one [#permalink]
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29 Jul 2008, 03:57
Walker's thread for this question, simply great 7t68037



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Re: DS: Tricky one [#permalink]
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30 Jul 2008, 07:08
durgesh79, I just have 1 little question. when drawing the line (x/y)<y, how come we look at only the area between this line and the xaxis? cause i always thought that we should consider the entire space or area above the line (x/y)<y. would you please explain why are you considering only the area between this line and the xaxis in particular? thanks



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Re: DS: Tricky one [#permalink]
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30 Jul 2008, 08:42
tarek99 wrote: durgesh79, I just have 1 little question. when drawing the line (x/y)<y, how come we look at only the area between this line and the xaxis? cause i always thought that we should consider the entire space or area above the line (x/y)<y. would you please explain why are you considering only the area between this line and the xaxis in particular? thanks i think your doubt is for (x/y)>2 not (x/y)<y ..... x/y > 2 can be written as x > 2y when y is +ve x < 2y when y is ve (in equality will change sign if we multiply oth sides with a ve number ) So in the upper half of xy plane (y +ve ) the area will be below the line x=2y and in the lower half of the xy plane ( y ve) the area will be above the line x=2y
Last edited by durgesh79 on 30 Jul 2008, 10:53, edited 1 time in total.



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Re: DS: Tricky one [#permalink]
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30 Jul 2008, 10:07
durgesh79 wrote: tarek99 wrote: durgesh79, I just have 1 little question. when drawing the line (x/y)<y, how come we look at only the area between this line and the xaxis? cause i always thought that we should consider the entire space or area above the line (x/y)<y. would you please explain why are you considering only the area between this line and the xaxis in particular? thanks i think your doubt is for (x/y)<2 not (x/y)<y ..... x/y < 2 can be written as x < 2y when y is +ve x > 2y when y is ve (in equality will change sign if we multiply oth sides with a ve number ) So in the upper half of xy plane (y +ve ) the area will be below the line x=2y and in the lower half of the xy plane ( y ve) the area will be above the line x=2y yeah sorry, i meant x/y<2!! but thanks a lot. that makes it a lot clearer. regards,



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Re: DS: Tricky one [#permalink]
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30 Jul 2008, 17:16
durgesh79 wrote: ..... mark the are x/y > 2. this'll be the area between the line x=2y and x axis on both sides of y axis
I think it will be the region to the right of the line x=2y. Correct me if i am erring... in the first quadrant.. for y=2, x is atleast 4..can be 5,6 or anything greater than 4. so the region to the right of this line. if y=2 then y is atleast 4 and can be 3,2, 2,3 or anything greater than 4. (no??!!) ... this would be the region to the right of the line and below x axis. If i am not mistaken, the region between the line and the x axis would be the case only for mod functions.. Ps. i am not nit picking here.. just trying to brush up my rusty fundamentals.
Last edited by bhushangiri on 30 Jul 2008, 17:35, edited 1 time in total.



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Re: DS: Tricky one [#permalink]
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30 Jul 2008, 17:20
durgesh79 wrote: Walker's thread for this question, simply great 7t68037again, walker just gave me much better way to tackle these problems.. +1 but looking at his diagram, i am convinced that x/y>2 wud be the region to the right of the line. that figure represents at x<2y on the ve y side and x>2y on the +ve y side. but the region we are interested in is x>2y for every y. let me know what u think guys..



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Re: DS: Tricky one [#permalink]
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30 Jul 2008, 19:45
bhushangiri wrote: but looking at his diagram, i am convinced that x/y>2 wud be the region to the right of the line. that figure represents at x<2y on the ve y side and x>2y on the +ve y side. but the region we are interested in is x>2y for every y. let me know what u think guys.. the area we are looking for is (x/y)>2 when y is positve this becomes x>2y when y is negative this becomes x<2y ( if you multiply a ve number (y) to both sides, the sign will change)



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Re: DS: Tricky one [#permalink]
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03 Aug 2008, 07:31
Are these questions from GMAT prep and OG i wonder what am goin to do in Quant hey durgesh superb u really give brilliant analysis in quant.Good luck
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