Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized for You
we will pick new questions that match your level based on your Timer History
Track Your Progress
every week, we’ll send you an estimated GMAT score based on your performance
Practice Pays
we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
Want to score 90 percentile or higher on GMAT CR? Attend this free webinar to learn how to pre-think assumptions and solve the most challenging questions in less than 2 minutes.
Official Answer and Stats are available only to registered users. Register/Login.
_________________
But there’s something in me that just keeps going on. I think it has something to do with tomorrow, that there is always one, and that everything can change when it comes. http://aimingformba.blogspot.com
Re: Is w greater than 1? (1) w + 2 > 0 (2) w^2 > 1
[#permalink]
Show Tags
18 Oct 2015, 07:45
1
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
_________________
Re: Is w greater than 1? (1) w + 2 > 0 (2) w^2 > 1
[#permalink]
Show Tags
12 Dec 2015, 09:48
1
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.?
Re: Is w greater than 1? (1) w + 2 > 0 (2) w^2 > 1
[#permalink]
Show Tags
13 Dec 2015, 02:33
narendran1990 wrote:
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.
_________________
Re: Is w greater than 1? (1) w + 2 > 0 (2) w^2 > 1
[#permalink]
Show Tags
13 Dec 2015, 17:44
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.
_________________
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 \(w>1\)
_________________
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 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.
Re: Is w greater than 1? (1) w + 2 > 0 (2) w^2 > 1
[#permalink]
Show Tags
18 Jul 2016, 04:28
1
aiming4mba wrote:
Is w greater than 1?
(1) w + 2 > 0 (2) w^2 >1
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
_________________
Posting an answer without an explanation is "GOD COMPLEX". The world doesn't need any more gods. Please explain you answers properly. FINAL GOODBYE :- 17th SEPTEMBER 2016. .. 16 March 2017 - I am back but for all purposes please consider me semi-retired.
close
68% (01:03) correct 
