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Is x < 2y ? [#permalink]
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30 Jan 2007, 21:25
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Question Stats:
75% (02:21) correct
25% (01:20) wrong based on 203 sessions
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Is x < 2y ? (1) y > 1 (2) (2^x)  4 > 0
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Director
Joined: 06 Feb 2006
Posts: 897

I choose C...
Is x>2y
1) insufficient, we do not know anything about x
y could be negative: 0,5 or positive 1, 3, 2.5
2) insufficient, we do not know anything about y
x>2
1+2) Sufficient. Take worst case scenario: y=0.99, x=2.1
Plug into
x>2y
2.1>1.98...



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(C) for me
x < 2y <=> y > x/2
In a XY Plan, Y is above the line x/2. I prefer to solve this DS with the XY plan.
Stat1
y > 1. This brings us to y above the line y=1. The area covered could be either above x/2 (when x>0) or below x/2 (wehn x<10). (Fig1)
INSUFF.
Stat2
(2^x)  4 >0
<=> 2^x > 4 = 2^2
<=> x > 2
In the XY plan, it does mean that the concerned area is at right of the line x=2. One more time, their is 2 case, y could be either above the line x/2 (when y > 0) or below the line x/2 (the point (3,100)). (Fig2)
INSUFF.
Both (1) and (2):
x > 2 and y > 1 represents an area with a not attainable vertice at (1;2). By drawing the line x/2 and this point, we observe that this point is on the line. Thus, the whole area of points such that x>2 and y>1 is above the line x/2. (Fig3)
SUFF.
Attachments
Fig1_y sup 1.GIF [ 3.17 KiB  Viewed 3067 times ]
Fig2_x sup 2.GIF [ 3.07 KiB  Viewed 3066 times ]
Fig3_x sup 2 and y sup 1.GIF [ 2.91 KiB  Viewed 3112 times ]



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Is x < 2y
(1) y > 1
(2) (2^x)  4 > 0
is
2y+x>0
from one we ve no info about x ......insuff
from two
2^x > 2^2
ie: x>2........no info about y insuff
both together
x>2 , y>1
thus x+y>1
we need to prove that 2y+x>0 ie +ve
x is always +ve
and the maximum ve value y can be multiplied by two is less than 0
thus C is my answer too



Senior Manager
Joined: 04 Jan 2006
Posts: 279

yezz wrote: Is x < 2y (1) y > 1 (2) (2^x)  4 > 0
is
2y+x>0
from one we ve no info about x ......insuff
from two
2^x > 2^2
ie: x>2........no info about y insuff
both together
x>2 , y>1
a little bit more work from the quote above.
x > 2 (equation 1)
y > 1 or 2y > 2 (equation 2)
(equation 1) + (equation 2)... x + 2y > 2 + (2)
x + 2y > 0
x > 2y
x < 2y (this is the same as the question.)
Thus C. is the answer



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Re: DS: Property of Number [#permalink]
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02 Feb 2007, 00:34
We have to determine whether:x<2y>2y
From (1) y>1  insufficient
From (2)
2^x >2^2
x>2 insufficient
Combining (1) and (2)
We can say x > 2y
(C)
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Re: DS: Property of Number [#permalink]
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03 Dec 2010, 11:06
I go for (C) 1) y>1 NOT SUFF 2) 2^x  4 >0 NOT SUFF
Combine: From 1: y>1 <=> 2y>2 From 2: 2^x 4>0 <=> 2^x>2^2 <=> x>2 <=> x<2 Therefore: x<2<2y <=> x<2y. Here we go.



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Re: DS: Property of Number [#permalink]
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03 Dec 2010, 11:22



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Re: Is x < 2y ? [#permalink]
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01 Jan 2014, 06:42
devilmirror wrote: Is x < 2y ?
(1) y > 1 (2) (2^x)  4 > 0 Statement 1 y>1 but no info about 'x'. Not sufficient Statement 2 x>2, but no info on 'y'. Not sufficient Statements (1) + (2) together y + x/2 > 0? Well, x>2 so x/2>1 And y>1, so yes it will always be more than zero Hence C is sufficient Hope it helps Cheers! J



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Re: DS: Property of Number [#permalink]
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06 Feb 2014, 08:41
Bunuel wrote: devilmirror wrote: Is x < 2y (1) y > 1 (2) (2^x)  4 > 0 Quite an old thread is resurrected. Is x<2y?Is \(x<2y\)? > rearrange: is \(x+2y>0\)? (1) y>1, clearly insufficient as no info about \(x\). (2) (2^x)4>0 > \(2^x>2^2\) > \(x>2\) > also insufficient as no info about \(y\). (1)+(2) \(y>1\), or \(2y>2\) and \(x>2\) > add this inequalities ( remember, you can only add inequalities when their signs are in the same direction and you can only apply subtraction when their signs are in the opposite directions) > \( 2y+x>2+2\) > \(x+2y>0\). Sufficient. Answer: C. I chose C.. thats correct.. bt with different approach bt Bunuel I didnt get this highlighted thing?? How did u do that?
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Re: DS: Property of Number [#permalink]
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07 Feb 2014, 05:12
sanjoo wrote: Bunuel wrote: devilmirror wrote: Is x < 2y (1) y > 1 (2) (2^x)  4 > 0 Quite an old thread is resurrected. Is x<2y?Is \(x<2y\)? > rearrange: is \(x+2y>0\)? (1) y>1, clearly insufficient as no info about \(x\). (2) (2^x)4>0 > \(2^x>2^2\) > \(x>2\) > also insufficient as no info about \(y\). (1)+(2) \(y>1\), or \(2y>2\) and \(x>2\) > add this inequalities ( remember, you can only add inequalities when their signs are in the same direction and you can only apply subtraction when their signs are in the opposite directions) > \( 2y+x>2+2\) > \(x+2y>0\). Sufficient. Answer: C. I chose C.. thats correct.. bt with different approach bt Bunuel I didnt get this highlighted thing?? How did u do that? \(2y+x>2+2\) > rearrange the left hand side as x+2y. As for the right hand side: 2 + 2 = 0. So, we get \(x+2y>0\). Hope it's clear.
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Re: Is x < 2y ? [#permalink]
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14 Jan 2016, 05:06
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Re: Is x < 2y ? [#permalink]
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15 Jan 2016, 19:37
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution. Is x < 2y ? (1) y > 1 (2) (2^x)  4 > 0 In the original condition, there are 2 variables(x,y), which should match with the number of equations. So you need 2 equation. For 1) 1 equation, for 2) 1 equation, which is likely to make C the answer. When 1) & 2), they become y>1, 2y>2 and 2^x>4=2^2 > x>2, x<2, which is x<2<2y > x<2y. So it is yes and sufficient. Therefore, the answer is C. > For cases where we need 2 more equations, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
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