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two possibilities: a) x+1=2(x-1) or x=3 (consider x+1 and x-1 to be of same sign) b) x+1=2(1-x) or x=1/3 (consider x+1 and x-1 to be of different signs) not sufficient, x may or may not be between -1 and 1

2) |x - 3| ≠ 0 => x≠3 not sufficient, x may or may not be between -1 and 1

together, x≠3 so x is 1/3 which is between -1 and 1 sufficient

I am having trouble understanding how |x| < 1 translates to -1<x<1.Am missing something:

We have two case, a. x>0 eg: x=5, |x|=5 so here |x|>0.Therefore,x < 1 b. when x<0,x=-5,|x| is still 5 how does this become x>-1??

Is |x| < 1?

Is |x| < 1, means is x in the range (-1,1) or is -1<x<1 true?

(1) |x + 1| = 2|x - 1| Two key points: x=-1 and x=1 (key points are the values of x when absolute values equal to zero), thus three ranges to check: ---------{-1}--------{1}---------

A. x<-1 (blue range) --> |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.

(2) |x - 3|\neq{0} Just says that x\neq{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\neq{3} --> means x can have only value \frac{1}{3}, which is in the range (-1,1). Sufficient.

1)|x+1|=2|x-1| There are 2 key points (x=-1 and x=1) where one of the expressions under modules passes 0 and change its sign. x<-1: -(x+1)=-2(x-1) --> -x-1=-2x+2 ---> x = 3. But x=3 does not satisfy x<-1 condition. -1<=x<1: x+1 = -2(x-1) ---> x+1= -2x+2 --> x=1/3. It satisfies the condition. x>=1: x+1=2x-2 ---> x=3. It satisfies the condition. Insufficient

2) |x - 3| ≠ 0 x≠3 Insufficient

1&2) Only x=1/3 satisfies both statements. Sufficient.

For statement 1, you need to actually break it down to 2 separate statements to account for both scenarios:

Option #1 -> x + 1 = 2(x-1) x + 1 = 2x -2 x = 2x -3 3 + x = 2x 3 = x {or also Option #2}

Option #2 -> x + 1 = -2(x-1) x + 1 = -2x +2 x = -2x + 1 3x = 1 x = 1/3

Statement (1) is insufficient because |x| could be 3 or 1/3. So the answer is sometimes yes, sometimes no which means insufficient.

Statement (2) is insufficient because we are told that |x -3| ≠ 0. This means that as long as x ≠ 3, then we're ok. There are far too many options so we certainly have a sometimes yes, and sometimes no.

But together, from Statement (1) we have either 3 or 1/3 and from statement (2) we have everything BUT 3. So the only overlap we have for possible values to make both statements true is 1/3. So the answer is Yes, |x| IS less than 1 and we have enough info to answer the question.

jade3 wrote:

Is |x|< 1? (1) |x + 1| = 2|x - 1| (2) |x - 3| ≠ 0

_________________

------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

However, I used using other methods: a) Squaring both sides of the equation because |x| = (x^2)^(1/2), b) Combining different scenarios between these two possibilities: (x+1)>0, (x+1)<0, (x-1)>0, (x-1)<0. Therefore, I get four possible scenarios.

Bunuel, could you explain the logic behing your method to solve this problem? _________________

"Life’s battle doesn’t always go to stronger or faster men; but sooner or later the man who wins is the one who thinks he can."

However, I used using other methods: a) Squaring both sides of the equation because |x| = (x^2)^(1/2), b) Combining different scenarios between these two possibilities: (x+1)>0, (x+1)<0, (x-1)>0, (x-1)<0. Therefore, I get four possible scenarios.

Bunuel, could you explain the logic behing your method to solve this problem?

If you're more comfortable with "squaring" method (note that it's not always applicable) then you can apply it to the first statement too (and not only to the stem).

Is |x| < 1?

Is |x| < 1 --> is -1<x<1?

(1) |x + 1| = 2|x - 1| --> square both sides: (x+1)^2=4(x-1)^2 --> x^2+2x+1=4x^2-8x+4 --> 3x^2-10x+3=0 --> x=\frac{1}{3} or x=3 --> first is in the range (-1,1) but second is out of the range. Not sufficient.

(2) |x - 3|\neq{0} --> just says that x\neq{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\neq{3} --> means x can have only value \frac{1}{3}, which is in the range (-1,1). Sufficient.

Hi Bunuel , I also used the squaring method.... In what all situations is the squaring method not applicable ??? Can you please explain why ? _________________

The Best Way to Keep me ON is to give Me KUDOS !!! If you Like My posts please Consider giving Kudos

Hi Bunuel , I also used the squaring method.... In what all situations is the squaring method not applicable ??? Can you please explain why ?

Sure, for example it's not always applicable for inequalities.

A. We can raise both parts of an inequality to an even power if we know that both parts of an inequality are non-negative (the same for taking an even root of both sides of an inequality). For example: 2<4 --> we can square both sides and write: 2^2<4^2; 0\leq{x}<{y} --> we can square both sides and write: x^2<y^2;

But if either of side is negative then raising to even power doesn't always work. For example: 1>-2 if we square we'll get 1>4 which is not right. So if given that x>y then we can not square both sides and write x^2>y^2 if we are not certain that both x and y are non-negative.

B. We can always raise both parts of an inequality to an odd power (the same for taking an odd root of both sides of an inequality). For example: -2<-1 --> we can raise both sides to third power and write: -2^3=-8<-1=-1^3 or -5<1 --> -5^2=-125<1=1^3; x<y --> we can raise both sides to third power and write: x^3<y^3.

So for our question we can not square x/|x|< x as we don't know the sign of either of side.

I am having trouble understanding how |x| < 1 translates to -1<x<1.Am missing something:

We have two case, a. x>0 eg: x=5, |x|=5 so here |x|>0.Therefore,x < 1 b. when x<0,x=-5,|x| is still 5 how does this become x>-1??

Is |x| < 1?

Is |x| < 1, means is x in the range (-1,1) or is -1<x<1 true?

(1) |x + 1| = 2|x - 1| Two key points: x=-1 and x=1 (key points are the values of x when absolute values equal to zero), thus three ranges to check: ---------{-1}--------{1}---------

A. x<-1 (blue range) --> |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.

(2) |x - 3|\neq{0} Just says that x\neq{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\neq{3} --> means x can have only value \frac{1}{3}, which is in the range (-1,1). Sufficient.

Answer: C.

Hope it helps.

How is the sign applied for different range?can someone help please X<-1 -x-1=2(-x+1)

I am having trouble understanding how |x| < 1 translates to -1<x<1.Am missing something:

We have two case, a. x>0 eg: x=5, |x|=5 so here |x|>0.Therefore,x < 1 b. when x<0,x=-5,|x| is still 5 how does this become x>-1??

Is |x| < 1?

Is |x| < 1, means is x in the range (-1,1) or is -1<x<1 true?

(1) |x + 1| = 2|x - 1| Two key points: x=-1 and x=1 (key points are the values of x when absolute values equal to zero), thus three ranges to check: ---------{-1}--------{1}---------

A. x<-1 (blue range) --> |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.

(2) |x - 3|\neq{0} Just says that x\neq{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\neq{3} --> means x can have only value \frac{1}{3}, which is in the range (-1,1). Sufficient.

Answer: C.

Hope it helps.

How is the sign applied for different range?can someone help please X<-1 -x-1=2(-x+1)

-1<=x<=1 x+1=2(-x+1)

X>1 x+1=2(x-1)

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;

Applying this to |x + 1| = 2|x - 1|.

If x<-1 then x+1<0 and x-1<0 so |x + 1| =-(x+1) and 2|x - 1|=-2(x-1) which means that |x + 1| = 2|x - 1| becomes: -x-1=2(-x+1) --> x=3.

A. x<-1 (blue range) --> |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).

If understand correctly for (A) you are putting '-' in front of both Left/Right Hand Side Equation. Hence coming to "-x-1=2(-x+1)" Similarly in (C) I guess you are keeping the Left/Right Hand Side Equation both as '+'.

What about (B)?? value of x is between -1 and 1. How did you decide signs' here? How are you deciding the sign and coming here: x+1=2(-x+1) I couldn't understand this

|x+1| if u see will be negative till -1 and after that it will be positive |x-1| will be negative till 1 and will be positive after that.....

Now we have 3 areas where we need to check

x<-1 -1<=x<=1 x>1

If u check the number line till {-1} both the expression's value will be negative......so when we solve x<-1 we take both of them as Negative

but when we solve -1<=x<=1 the expression |x+1| is positive as its sign changed on x=-1 but the other one |x-1| is still Negative as its sign would change on {1}. therefore when we solve of this range we take |x+1|= x+1 and 2|x-1| = -2(x-1) = 2(-x+1)

in the third one both are Positive os both the expressions will be positive....

I hope this help you...as it also took me a lot of time to get this concept in my head.........

A. x<-1 (blue range) --> |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).

If understand correctly for (A) you are putting '-' in front of both Left/Right Hand Side Equation. Hence coming to "-x-1=2(-x+1)" Similarly in (C) I guess you are keeping the Left/Right Hand Side Equation both as '+'.

What about (B)?? value of x is between -1 and 1. How did you decide signs' here? How are you deciding the sign and coming here: x+1=2(-x+1) I couldn't understand this

Absolute value properties: When x\leq{0} then |x|=-x, or more generally when some \ expression\leq{0} then |some \ expression|\leq{-(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|\leq{some \ expression}. For example: |5|=5;

Applying this to |x + 1| = 2|x - 1|.

If -1\leq{x}\leq{1} then x+1>0 and x-1<0 so |x + 1| =x+1 and 2|x - 1|=-2(x-1) which means that |x + 1| = 2|x - 1| becomes: x+1=2(-x+1).