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As I have said before, most mod questions are best tackled using a number line. You don't need to do many calculations then. |x| means the distance from 0. |x-3| means the distance from 3. etc. For details of this approach, check out:
Let's go on to this question now. Is |x| < 1 i.e. Is the distance of point x from 0 less than 1?
Statement 1: |x + 1| = 2|x – 1| This means 'distance of x from -1 is twice the distance of x from 1'. Draw the number line now. There will be 2 points where the distance from -1 will be twice the distance from 1.
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For one of these points, distance from 0 is less than 1, for the other it is more than 1. So not sufficient.
Statement 2: |x – 3| > 0 This statement tells us that distance of x from 3 is more than 0 i.e. x does not lie at 3. It can lie anywhere else. You can look at it in another way: Mods are always more than or equal to 0. All this statement tells us is that this mod is not equal to zero i.e. x is not equal to 3. For some of these points, distance from 0 will be less than 1, for the others it will be more than 1. So not sufficient.
Using both statements together, statement 1 says that x is either 3 or a point between 0 and 1 (which I don't really need to calculate). Statement 2 tells us that x is not 3. So together, x must be a point between 0 and 1 and its distance from 0 must be less than 1. Sufficient. Answer C.
This is fine as long as you know why you are doing this:
Am I doing this right:
(1) |x + 1| = 2|x – 1|
(x+1)=2(x-1) (When both the terms are positive, x > 1) x+1=2x-2 3=x (NO) (Valid value for x since 3 >1) & (x+1)=-2(x-1) (When -1 < x < 1, (x+1) is positive but (x-1) is negative so you are put a negative sign here) x+1=-2x+2 3x=1 x=1/3 (YES) (Valid value since -1 < 1/3 < 1)
There would be another case x < -1. In that case both the terms will be negative. -(x+1)=-2(x-1) giving x = 3 (Not a valid value since 3 is not less than -1) I am assuming that you saw the two negatives will get canceled out and give x = 3 which will not be valid so you skipped this step. In some questions, you could get a valid value here. So you have only 2 values for x (3 and 1/3).
(2) |x – 3| > 0
(x-3)>0 x>3 (NO)
-(x-3)>0 -x+3>0 -x>-3 x<3 (MAYBE)
C or E
Statement 1: x=1/3,3 Statement 2: x <> 3
Since x CANNOT equal 3, x = 1/3
Since |1/3| < 1, both statements are sufficient to answer the prompt.
From statement 1, we are able to get two values of x; they are x=3 and x=1/3. Two values of x, hence insufficient. From statement 2, all we know is that the distance from x is more than 0 or it indirectly implies that x is not 0. Not enough information. Hence insufficient.
On combining these two statements, we come to know that x cannot be 3 and x=1/3. Since 1/3 < 1, hence |x|<1. +1C.
Please do add the OA while posting the questions.
This is my approach: Is |x|<1? 1st start from statement 2, cause it is easier, |x – 3| > 0 just tell us x is note equal to 3, so it is insufficient to solve the target question 2nd for statement 2: |x + 1| = 2|x – 1| we have to separate the condition to x<-1, -1<x<1, x>1 that is , |x|<1 and |x|> 1 to do further thinking 1) when |x|<1, we could know we will get the solution in this range after solving equation, thus get the answer "YES" for question |x|<1 2) when |x|>1, we could know we will get the same answer in x<-1 and x>1 condition, and we could assure the answer is "NO" for target question so based on above, statement 2 is insufficient to solve the target question
we only left option C and E now. To test whether statements together will help to solve target question, we could use the denied solution x=3 in statement 2 to statement 1 to see whether it is one of the two solutions of equation.
If it is one of the solutions, then statement 2 will help to reduce the two solutions to one, thus, support the target question. We could feel free to choose option C If it is not one of the solution, then statement 2 will not help to reduce the number of solutions, thus, we could feel free to choose option E.
Let us test now. LS 3+1|=4 RS:2*|3-1|=4, we could know x=3 is one of the two answers. Thus we could choose C