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

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

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New post 03 Nov 2008, 06:49
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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

This one is already discussed.. but I was not able to find it again ...ineqp
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Re: DS : tough inequality [#permalink]

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New post 03 Nov 2008, 08:03
amitdgr wrote:
If (x/y)>2, is 3x+2y<18?

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


stems say x and y have the same sign..

so if x>0, y>0

1) then x-y<2 means y is a fraction, so if x=2, y=1/2; 3x+2y<18
if x<0, y<0
say x=-6, y=-1, then x-y=-6+1=-5<2
3(-6)+2(-1) <18

Suff

2)
y-x<2 ; x>0 y>0 and x/y>2 say if x=1 y=1/4 the 3x+2y<18 is true

if x<0 y<0 also shows that 3x+2y<18
Suff

D it is

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Re: DS : tough inequality [#permalink]

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New post 03 Nov 2008, 11:26
An alternative approach will be to draw inequality graph (sorry for no attachment). You will find that the area enclosed within the X-axis, line x=2y and x=y+2 (as well as y=x+2) satisfies inequality 3x + 2y < 18. Hence, D should be sufficient.

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Re: DS : tough inequality [#permalink]

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New post 03 Nov 2008, 11:35
I found the place from where I got this problem ...

But walker says the answer is A ... http://gmatclub.com/forum/7-t68037
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Re: DS : tough inequality [#permalink]

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New post 06 Nov 2008, 20:19
amitdgr wrote:
If (x/y)>2, is 3x+2y<18?

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

This one is already discussed.. but I was not able to find it again ...ineqp

This is a tough one. Hope I don't get this in my GMAT.

I'm either getting A, but not very sure on my approach.

If I take y>0, then rephrasing the stem, x>2y, and therefore 3x>6y
Substituting into the question, 3x+2y<18, then 6y+2y<3x+2y<18
so 8y<18
y<2.25 for this to be true

(1) x-y<2
x<2+y plus x>2y
2y<x<2+y
y<2 so y<2.25 is true
Sufficient.

(2) y-x<2
x>y-2 plus x>2y
x has many options, eg y=1 x=3 3x+2y<18 OR y=2 x=5 3x+2y>18
Not sufficient.

Taking y<0, then rephrasing the question, then x<2.25 for the statement to be true.
But we know from the stem that when y<0, x is always negative too. So for negative y, any value of x will do (can simply substitute numbers into the equation to prove this, its quite clear it will always be less than 18)

Therefore (A).

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Re: DS : tough inequality [#permalink]

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New post 06 Nov 2008, 22:33
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amitdgr wrote:
If (x/y)>2, is 3x + 2y < 18?

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

This one is already discussed.. but I was not able to find it again ...ineqp


given that: x/y > 2, which implies that x and y have same sign and lxl > lyl.

(1) (x-y) < 2:
only possible if x and y both are -ve. if x and y both are +ve, then x has to be > 2y. in that case, x - y cannot be < 2.
so x and y are -ve and (3x + 2y) < 18. suff.

(2) y-x is less than 2:
2 is possible when x and y both are either -ve or +ve.
if x and y are both +ve, x has to be > 2y.
a. if x = 5 and y = 2, (3x + 2y) > 18.
b. if x = 4.01 and y = 2, (3x + 2y) < 18.

however if x and y both are -ve, then lxl > lyl and in that case (y - x) > 2. so (3x + 2y) > 18.

nsf.




only A.
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Re: DS : tough inequality [#permalink]

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New post 07 Nov 2008, 05:18
GMAT TIGER wrote:
amitdgr wrote:
If (x/y)>2, is 3x + 2y < 18?

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

This one is already discussed.. but I was not able to find it again ...ineqp


given that: x/y > 2, which implies that x and y have same sign and lxl > lyl.

(1) (x-y) < 2:
only possible if x and y both are -ve. if x and y both are +ve, then x has to be > 2y. in that case, x - y cannot be < 2.
so x and y are -ve and (3x + 2y) < 18. suff.

(2) y-x is less than 2:
2 is possible when x and y both are either -ve or +ve.
if x and y are both +ve, x has to be > 2y.
a. if x = 5 and y = 2, (3x + 2y) > 18.
b. if x = 4.01 and y = 2, (3x + 2y) < 18.

however if x and y both are -ve, then lxl > lyl and in that case (y - x) > 2. so (3x + 2y) > 18.

nsf.

only A.


NICE :) +1 !!
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Re: DS : tough inequality   [#permalink] 07 Nov 2008, 05:18
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