Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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