ColumbiaBaby wrote:
could you please help me understand why we are not taking only the overlap of the solutions of x ie. answer to be only x<0 but we take the overlap in this
https://gmatclub.com/forum/is-x-217529.html#p2293754 question for statement I?
Thank you in advance!
I suppose it depends how you solve the question. We have this inequality:
|x + 1| > 2x - 1
which is clearly true for any negative number, because the left side will be positive and the right side will be negative. So we know the answer can only be A or B, and you can just test x = 1 to see that B must be right.
If you do solve these kinds of questions algebraically (something that can almost always be avoided on the GMAT -- there are a few other things you could do) then things can get a bit annoying. I'll explain with a simpler inequality:
|z| < 5
You don't need to use algebra here, but supposing you wanted to, the standard technique is to divide the problem into two cases:
• if the thing inside the absolute value is positive, then the absolute value won't do anything. So here, if z > 0, then |z| = z, and our inequality becomes z < 5. Notice though that we assumed from the outset that z > 0, so our work is only valid when z > 0. So here we must take the 'overlap' of our assumption and our solution, and we find 0 < z < 5 is one set of solutions to the inequality
• if the thing inside the absolute value is negative, then the absolute value will flip its sign. So if z < 0, then |z| = -z, and our inequality becomes -z < 5, or z > -5. Again, we have to combine our solution with our assumption, so a second set of solutions is -5 < z < 0.
You can throw z = 0 into either case, and it won't matter which (z = 0 is also a solution here). So we now have two separate sets of legitimate solutions to our inequality, and since they all work, we now want to combine them: -5 < z < 5 is the complete set of solutions.
So when you analyze each individual case, you must take the overlap of your assumption and your solution, but since each case will, most of the time, give you some valid solutions to the inequality, when you combine the cases you use all of the solutions you found.
I now understand my mistake re the other question that I had tagged in my post.
Statement I is giving the range (i) 10 < x < 50/3 or (ii) 10/3 < x < 10.
So we are essentially taking the overlap in (i) and (ii) and then union of (i) and (ii) hence 10/3 < x < 50/3.
I had earlier missed the (i) 10 < x and (ii) x < 10 part. 😅
thank you for your explanation but the problem was still persisting re Statement I because of my incorrect approach, which I hope I've rectified now as per the above explanation.
Once again thanks folks, I had been going around in circles for the past hour trying to find my error!