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# If (x/y)>2 , is 3x+2y<18? (1) x-y is less than

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

(1) $$x-y$$ is less than $$2$$

(2) $$y-x$$ is less than $$2$$
Director
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Re: DS: Tricky one [#permalink]  28 Jul 2008, 21:47
tarek99 wrote:
If $$(x/y)>2$$, is $$3x+2y<18?$$

(1) $$x-y$$ is less than $$2$$

(2) $$y-x$$ 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]  28 Jul 2008, 23:24
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
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Re: DS: Tricky one [#permalink]  28 Jul 2008, 23:34
tarek99 wrote:
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....
Manager
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Re: DS: Tricky one [#permalink]  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
Director
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Re: DS: Tricky one [#permalink]  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

7-t66982
7-t67504
7-t66977
7-t67196
7-t67183
7-t67700

plotting conditions on xy plane may not be the best approach in all the cases, but with some practise it is one of the easiest....
Manager
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Re: DS: Tricky one [#permalink]  29 Jul 2008, 02:19
durgesh79, Thank you so much, good luck and good study!!!
SVP
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Re: DS: Tricky one [#permalink]  29 Jul 2008, 02:53
durgesh79 wrote:
tarek99 wrote:
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
Director
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Re: DS: Tricky one [#permalink]  29 Jul 2008, 03:11
tarek99 wrote:
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).
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Re: DS: Tricky one [#permalink]  29 Jul 2008, 03:44
durgesh79 wrote:
tarek99 wrote:
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?
Director
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Re: DS: Tricky one [#permalink]  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]  29 Jul 2008, 03:57
Walker's thread for this question, simply great

7-t68037
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Re: DS: Tricky one [#permalink]  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 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
Director
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Re: DS: Tricky one [#permalink]  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 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, 10:53, edited 1 time in total.
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Re: DS: Tricky one [#permalink]  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 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,
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Re: DS: Tricky one [#permalink]  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]  30 Jul 2008, 17:20
durgesh79 wrote:
Walker's thread for this question, simply great

7-t68037

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.

let me know what u think guys..
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Re: DS: Tricky one [#permalink]  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]  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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cheers
Its Now Or Never

Re: DS: Tricky one   [#permalink] 03 Aug 2008, 07:31
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