Is -x < 2y ?

(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.

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

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)

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.

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.

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

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.

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.

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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