Is \(w>1\)?

(1) \(w+2>0\) --> \(w>-2\)
--(-2)-----------------(1)----. Not sufficient.

(2) \(w^2>1\) --> \(w<-1\) or \(w>1\)
-----------(-1)--------(1)----. Not sufficient.

(1)+(2) Intersection of the ranges from (1) and (2) is \(-2<x<-1\) and \(w>1\)
--(-2)----(-1)--------(1)----. Not sufficient to say whether \(w>1\).

Answer: E.
_________________

1. W >-2 which could be greater than 1 INSUFF
2. w > 1/2 INSUFF as it could be less than 1

Combining them you have W >1/2 > -2 which simplifies to W>1/2 which is same as 2 and again INSUFF
so E

Remember when dividing over inequalities if there is negative sign you flip the arrows.

nice explaination

Note that we don't know whether w is an integer.

(1) w+2 > 0 --> w > -2 IS
(2) w² > 1 --> w < -1 or w > 1 IS

Together:
Since w is no integer, -2<w<-1 and w> 1. still IS.

E.

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 w greater than 1?

(1) w + 2 > 0
(2) w^2 >1

In DS questions, if the range of the question includes the range of the conditions, the condition becomes the answer
If we modify the question in this way,
we want to know whether w>1
1) w>-2
2) w<-1, 1<2
The original condition has one variable so we need one equation, so there is high chance (D) will be our answer
1) the question does not include the condition range, so not sufficient
2) the question does not include the condition range as well, so also not sufficient
Even if we combine the 2 conditions, -2<w<-1, 1<w. This condition is also not included in the question, so also not sufficient, so the answer becomes (E).

For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E
_________________

Required: w>1?

Statement 1: w + 2 >0
w > -2
Here w can take integral values which are between -2 and 1 and also values that are greater than 1.
INSUFFICIENT

Statement 2: w^2 > 1
w < -1 or w > 1
w can take values that are greater than 1 and less than 1
INSUFFICIENT

Statement 1 and Statement 2 combined:
On combining both statements,
w take values from (-2,-1) and (1, infinite)

INSUFFICIENT

Correct Answer E

Question:

As per statement 1: w+2>0, in such a case, w=-1,0,1,2. Unlimited possibilities. Hence not sufficient
As per statement 2: w^2>1, in such a case, w=2 satisfies condition whereas w=-2 satisfies condition but is not greater than 1.

Combined: w=2 satisfies both conditions and 2>1. I got option C. Can someone explain my error in reasoning.?

Notice that we are NOT told that w must be an integer and re-read solutions provide above.
_________________

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 w greater than 1?

(1) w + 2 > 0
(2) w^2 >1

In inequality, when the range of que includes the range of con, it is important to know that the con is sufficient. In the original condition, there is 1 variable(w), which should match with the number equation. So, you need 1 equation. For 1) 1 equation, for 2) 1 equation, which is likely to make D the answer.
In 1) w>-2, the range of que doesn’t include the range of con, which is not sufficient.
In 2) w<-1, 1<w, the range of que doesn’t include the range of con, which is not sufficient.
In 1) & 2) which is -2<w<-1, 1<w, the range of que doesn’t include the range of con, which is not sufficient. Therefore, the answer is E.


-> For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.
_________________

Bunuel,

Can you please explain as to how did you get the below step.

\(w^2>1\) --> \(w<-1\) or \(w>1\)

and just for understanding purpose how will you write,
\(W^2<1\) in the same way.

Hi,
\(w^2>1\) can be written as..
w^2-1>0..
(w-1)(w+1)>0...

it gives us two scenarios..
1) both (w-1) and (w+1) are positive..
w>1 for both to be positive
2) both (w-1) and (w+1) are negative..
w<-1 in this scenario..

hence you get
\(w^2>1\) --> \(w<-1\) or \(w>1\)

Hi arshu27,

Remember the following while solving the inequalities questions:

Solving an inequality with a less than sign:
The value of the variable will be greater than the smaller value and smaller than the greater value.
i.e. It will lie between the extremes.

Solving an inequality with a greater than sign:
The value of the variable will be smaller than the smaller value and greater than the greater value.
i.e. It can take all the values except the values in the range.

\(w^2>1\) or \(w^2\) - 1 >0
(w-1)(w+1) > 0

Therefore, w < - 1 or w >1

I think now you can figure out, how \(w^2<1\) would be written.

Is w greater than 1?

(1) w + 2 > 0
w>-2 ; w can be {-1,0,1,2,3,...77.....3456....INFINITY}
w is sometimes -ve (-1) and sometimes +ve(infinite values)
INSUFFICIENT

(2)\(w^2 >1\)
-1>w>1 ; w can be {all possible values except values lying between -1 to 1}
so sometimes -ve (-2,-3,-99,....) and sometimes positive (2,3,99,.....)
INSUFFICIENT


MERGING BOTH

w cannot take these four values -2,-1,0,1
But w can be many other -ve or +ve value
INSUFFICIENT

ANSWER IS E
_________________

\(w>1?\)

\(w>-2\)

\(|w|>1 \implies w<-1 \ \cup \ w>1\)

\(\implies -2<w<-1 \ \cup \ w>1 \implies E\)
_________________

Is w greater than 1?

(1) w + 2 > 0

w > -2

NOT SUFFICIENT.

(2) w^2 > 1

w^2 - 1 > 0
(w+1)(w-1) > 0
w > 1 or w < - 1

NOT SUFFICIENT.

(1&2) -2 < W < -1 or w > 1. INSUFFICIENT.

Answer is E.
_________________

In this inequality: (w+1)(w-1) > 0 ---> w < -1 , w > 1

why do we change the direction of the "greater than" sign into a "less than" sign for the (w+1)? I thought we only do this when multiplying/dividing by negative? Can someone please explain ?