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YES! the OA is A. Interesting....I never thought to use the xy-plane. I know how to draw the \(y<(-3/2)x + 9\) on the xy-plane. 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

YES! the OA is A. Interesting....I never thought to use the xy-plane. I know how to draw the \(y<(-3/2)x + 9\) on the xy-plane. 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 .. x-y=2 ---> (2,0) and (0,-2) y-x=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 x-y=2 ..... suff statement 2 : area to the right of line y-x=2 ... not suff....

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!!!

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!!!

YES! the OA is A. Interesting....I never thought to use the xy-plane. I know how to draw the \(y<(-3/2)x + 9\) on the xy-plane. 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 .. x-y=2 ---> (2,0) and (0,-2) y-x=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 x-y=2 ..... suff statement 2 : area to the right of line y-x=2 ... not suff....

wow, really nice approach, but I have a question regarding my interpretation of the xy-plane: Now, from statement 1, I can see that there is an intersection between x-2<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 x-2<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 y-x<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

wow, really nice approach, but I have a question regarding my interpretation of the xy-plane: Now, from statement 1, I can see that there is an intersection between x-2<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 x-2<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 y-x<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 )....

y-x < 0, (0,0) is satisfying this 0<2 ... means we are looking at the same side of line y-x=2 which has (0,0).

wow, really nice approach, but I have a question regarding my interpretation of the xy-plane: Now, from statement 1, I can see that there is an intersection between x-2<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 x-2<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 y-x<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 )....

y-x < 0, (0,0) is satisfying this 0<2 ... means we are looking at the same side of line y-x=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?

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 ....

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 x-axis? 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 x-axis in particular? thanks

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 x-axis? 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 x-axis 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, 11:53, edited 1 time in total.

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 x-axis? 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 x-axis 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,

..... 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, 18:35, edited 1 time in total.

again, 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.

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)

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 _________________

cheers Its Now Or Never

gmatclubot

Re: DS: Tricky one
[#permalink]
03 Aug 2008, 08:31