Is |x| 0

Question

Is |x| < 1 ?

(1) |x + 1| = 2|x – 1|

(2) |x – 3| > 0

Bunuel · Accepted Answer

Is |x| < 1 ? Is |x| < 1 , means is x in the range (-1,1) or is -1 |x + 1| = 2|x - 1| becomes: -x-1=2(-x+1) --> x=3 , not OK, as this value is not in the range we are checking ( x<-1 ); B. -1\leq{x}\leq{1} (green range) --> |x + 1| = 2|x - 1| becomes: x+1=2(-x+1) --> x=\frac{1}{3} . OK, as this value is in the range we are checking ( -1\leq{x}\leq{1} ); C. x>1 (red range) --> |x + 1| = 2|x - 1| becomes: x+1=2(x-1) --> x=3 . OK, as this value is in the range we are checking ( x>1 ). So we got TWO values of x (two solutions): \frac{1}{3} and 3 , first is in the range (-1,1) but second is out of the range. Not sufficient. Alternately we could just expand absolute values of RHS and LHS with same sign (for example both LHS and RHS positive: x + 1 = 2(x - 1) ) and then with different sign (for example LHS positive and RHS negative x + 1 = 2(-x + 1) ), solve for x both equations, and finally check whether the solutions satisfy |x + 1| = 2|x - 1| . OR: we can square given equation to get rid of the modulus: (x + 1)^2 = 4(x - 1)^2 --> 3x^2-10x+3=0 --> x=3 or x=\frac{1}{3} . (2) |x - 3|>{0} . Absolute value is always non-negative, more than or equal to zero: |some \ expression|\geq{0} . We are told that absolute value of x-3 is MORE than zero, so just it says that |x-3| eq{0} , which simply means that x eq{3} . But we don't know whether x is in the range (-1,1) or not. (1)+(2) x=\frac{1}{3} or x=3 AND x eq{3} --> means x can have only value \frac{1}{3} , which is in the range (-1,1). Sufficient. Answer: C. Hope it helps.

KarishmaB · Answer

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. Ques6.jpg 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.

Matt1177 · Answer

Hi Bunuel:

Can you please explain why, in B above, we must NOT change the sign in |x+1| leaving it as x+1, and must change the sign in 2|x-1| making it 2(-x+1)?

Thank you.

Bunuel · Answer

Absolute value properties: When x\leq{0} then |x|=-x , or more generally when some \ expression\leq{0} then |some \ expression|={-(some \ expression)} . For example: |-5|=5=-(-5) ; When x\geq{0} then |x|=x , or more generally when some \ expression\geq{0} then |some \ expression|={some \ expression} . For example: |5|=5 ; For B. if -1\leq{x}\leq{1} (green range) --> then x+1\geq{0} (try some value of x from the given range to check: for example x=0 ) so |x+1|=x+1 BUT x-1\leq{0} ] (again try some value of x from the given range to check: for example x=0 ) so |x-1|=-(x-1) thus |x + 1| = 2|x - 1| becomes: x+1=2(-x+1) . For more check: https://gmatclub.com/forum/math-absolute ... 86462.html Hope it's clear.

fluke · Answer

The question is Is |x| < 1 ? Asking whether x falls between 1 and -1, exclusive. if |x|<1 x<1; 0.5, 0.2,0 and x>-1; -0.5,-0.2 or -1 0 Here modulus only on LHS; So, try this with both signs; +(x-3) > 0 x-3>0 x>3 and -(x-3) > 0 -x+3>0 -x>-3 x<3 Here x can be either less than 3 or more than 3. Not 3. However; this doesn't tell us definitively whether x lies between -1 and 1. Not Sufficient. Using both: We know x can be either (1/3 or 3) by 1st statement. Second statement tells us that x can not be 3. The only value thus left is 1/3. Sufficient. Ans: "C"

mikemcgarry · Answer

I'm happy to respond.

I dare say, this problem is a little bit harder than what the GMAT will ask of you. Statement #1 : |x+1| = 2|x-1| If we are given |P| = |Q|, this means: P = Q OR P = -Q. Notice that the word "or" is not a piece of garnish there: rather, it is an essential piece of mathematical equipment. |x + 1| = 2|x - 1| Case I (x + 1) = 2(x - 1) x + 1 = 2x - 2 x = 3 Case II (x + 1) = -2(x - 1) x + 1 = -2x + 2 3x = 1 x = 1/3 This, from statement #1, we have x = 3 or x = 1/3. With this, we do not have sufficient information to answer the prompt question. This statement, by itself, is insufficient . Statement #2 : |x-3| > 0 Forget about everything we did in statement #1. Here, x could equal 10, in which case |x| is not less than 1, or x could equal 0, in which cases |x| is less than 1. We can pick different values that satisfy |x-3| > 0, x = 10 and x = 0, that give two different answers to the prompt question. Therefore, we do not have sufficient information to answer the prompt question. This statement, by itself, is insufficient . Combined : #1 gives us x = 3 or x = 1/3 The value x = 3 does not satisfy the second statement, so we reject that value. The value x = 1/3 is only value that satisfies both statements, and with this, |x| < 1. Combined, the statements are sufficient. Answer = (C) Does all this make sense? Mike

KarishmaB · Answer

Use the number line to solve it quickly. 'Is |x| < 1' implies 'Is distance of x from 0 less than 1?' i.e. does x lie within -1 and 1 (excluding the points -1 and 1)? 1. |x+1| = 2|x-1| This tells you that distance of x from -1 is twice the distance of x from 1. There are two values of x for which this is possible: Ques3.jpg The red line is twice the length of the blue line in both the cases. For the first case, x lies somewhere between 0 and 1 but for the second case, x lies at 3. Hence we can't answer whether x will lie between -1 and 1 from this statement alone. 2. |x-3| > 0 This tells us that x is a point whose distance from 3 is more than 0. That means it is not at 3 but on its left or right. This statement alone doesn't tell us whether x lies between -1 and 1. Both statements together: Stmnt 1 tells us that x lies between -1 and 1 or at 3. Stmnt 2 tells us that x doesn't lie at 3. Then there is only one option left: x must lie between -1 and 1. Answer (C)

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Is |x| < 1 ? (1) |x + 1| = 2|x – 1| (2) |x – 3| > 0

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