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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
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yezz
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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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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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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devilmirror
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.
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devilmirror
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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Bunuel
devilmirror
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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Bunuel
devilmirror
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?

\(2y+x>-2+2\) --> re-arrange 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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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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devilmirror
Is -x < 2y ?

(1) y > -1
(2) (2^x) - 4 > 0

From statement 1 we get that the lowest possible value of 2Y must be greater than -2 .

From statement 1 we get that x > 2 or -x <-2. so even the highest possible value of -x cannot exceed -2

So from the original equation in the qn we can see that even the highest possible value of "-x" is less than the lowest possible value of "2y"
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