modulo problem

[TOPGMAT]

MGMAT ques

Is |x| < 1 ?

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

(2) |x - 3| > 0


OA: c

I used the following technique...
stmt 1: square both sides and solve.... x=3 or x =1/3
stmt2 : x is not equal to 3.
Hence c
Is this a good method ? Can we just square of both sides when faced
with a module on both sides ?

[wilson4mba]

MGMAT ques

Is |x| < 1 ?

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

(2) |x - 3| > 0


OA: c

I used the following technique...
stmt 1: square both sides and solve.... x=3 or x =1/2
stmt2 : x is not equal to 3.
Hence c
Is this a good method ? Can we just square of both sides when faced
with a module on both sides ?

I think you cannot square both the sides in case of modulus operations. Working with both -ve and +ve signs is a better option

[Night reader]

MGMAT ques

Is |x| < 1 ?

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

(2) |x - 3| > 0


OA: c

I used the following technique...
stmt 1: square both sides and solve.... x=3 or x =1/2
stmt2 : x is not equal to 3.
Hence c
Is this a good method ? Can we just square of both sides when faced
with a module on both sides ?

|x|<1 We need to find that x<1 and x>-1, -1<x<1


statement (1) we could not avoid plethora of plus and negative equations, even if there are absolute values on both sides x+1=2(x-1) <<<undesired.
x+1=2x-2
-x+1=2x-2
x+1=-2x-2
-x+1=-2x-2
algebraically we derive x = {-3;-1;1;3} We don't know yet if -1<x<1 because we got -1 and -3, depending on the sign of x it can be any of the four values

statement (2) |x-3|>0 is equivalent to x-3>0 and -x-3>0 from which x>3 and x<-3 Without statement (1) not much information to decide we still have [less than -3], basically it all depends on the sign of x again

together statements 1&2 We combine them x>3 hence IS NOT x= {1;3} or x={1;3} and NOT x>3

Sorry the answer is actually No interval x<-3 {-3;-1;1;3} x>3

[TOPGMAT]

Hi NightReader,
When you say x <-3 .... How is |x| < 1 ?
x=-4 => |-4| is not less than 1.


Also,
|x-3| > 0 == > x is not equal to 3. For all other values its >0.

And I am not sure about how you removed the mod signs from the first statement.
You seem to substitute -x for x on only one side... I think you should substitute on
both sides.

[Rahul@gurome]

Is |x| < 1 ?

• (1) |x + 1| = 2|x - 1|
  (2) |x - 3| > 0

statement (1) we could not avoid plethora of plus and negative equations, even if there are absolute values on both sides x+1=2(x-1) <<<undesired.
x+1=2x-2
-x+1=2x-2
x+1=-2x-2
-x+1=-2x-2
algebraically we derive x = {-3;-1;1;3} We don't know yet if x<1<-1 because we got -1 and -3, depending on the sign of x it can be any of the four values!

Careful!
Put x = -3 or -1 or 1 in the equation and see!

While treating equations with absolute values you've to be very careful about the initial assumptions you're making! After solving the problem, you've to check whether the solutions satisfies your assumptions. If they don't (which is not rare at all), you've to discard those solutions. Now for this equation,

|x + 1| = 2|x - 1|, there may be three cases x < -1, -1 < x < 1 and x > 1. Observe that x can't be -1 and 1! Now let's analyze the equation for each of these cases...

Case 1: x < -1

• ... |x + 1| = 2|x - 1|
  => -(x + 1) = 2(-(x - 1))
  => -x - 1 = -2x + 2
  => x = 3

Now, our initial assumption was x < -1, but we're getting x = 3! Therefore there is no solution for x < -1!

Case 2: -1 < x < 1

• ... |x + 1| = 2|x - 1|
  => (x + 1) = 2(-(x - 1))
  => x + 1 = -2x + 2
  => 3x = 1
  => x = 1/3

The solution is okay with initial assumption.

Case 3: x > 1

• ... |x + 1| = 2|x - 1|
  => (x + 1) = 2(x - 1)
  => x + 1 = 2x - 2
  => x = 3

The solution is okay with initial assumption.

Therefore, we have to solutions for the equation |x + 1| = 2|x - 1|. And they are x = 1/3 and x = 3.

Now for the question,

Statement 1: |x + 1| = 2|x - 1|
As discussed above x = 1/3 or x = 3.

Not sufficient.

Statement 2: |x - 3| > 0
As absolute value of any number is greater than zero except zero itself, statement 2 simply implies x ≠ 3.

Not sufficient.

1 & 2 Together: x = 1/3

Sufficient.

The correct answer is C.

I used the following technique...
stmt 1: square both sides and solve.... x=3 or x =1/2
stmt2 : x is not equal to 3.
Hence c
Is this a good method ? Can we just square of both sides when faced
with a module on both sides ?

In your method, apparently we're getting correct answer by squaring both the sides... But I'm not sure that it'll always lead to the same! If they do, I'll suggest it's better to cross check the answers by putting them into the equation. In your case though x = 1/2 is wrong. Check that. The solutions to the quadratic are x = 1/3 and x = 3.

[goyalsau]

MGMAT ques

Is |x| < 1 ?

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

(2) |x - 3| > 0


OA: c

I used the following technique...
stmt 1: square both sides and solve.... x=3 or x =1/2
stmt2 : x is not equal to 3.
Hence c
Is this a good method ? Can we just square of both sides when faced
with a module on both sides ?

I not good with inequalities but i have not been anybody squaring the modular....

Is |x| < 1 ?

x < 1
-x<1 or x > -1
If we combine both it will be -1<=x <=1

We Can rephrase the question as -1<=x <=1


(1) |x + 1| = 2|x - 1|
One solution for sure will be x = 3 Then i am not able to understand about the negative Because in generally when its equal to
condition is this.

When mod(something) = some value
|x| = z
then x = z or x=-z


But here Mode is equal to Mode of something so i don't How to solve then.

(2) |x - 3| > 0

When mod(something) > some value
|x| > z
then x<-z OR x>z
Simply to remember: x does not lie between negative and positive values of z

I am not able to open understand How to use it over here as z is 0

Can anybody Help

[TOPGMAT]

Hi Rahul,
Sorry... stmt1 gives value of x = 1/3 and 1/2 as I had put it before...

According to your solution,
stmt1 => x=1/3 or 3
if X =1/3 |x| < 1...
Therefore we need stmt 3 to eliminate option x=3 ?

I have a doubt in your solution...
-1 < x < 1
Why did you take the negative value of only one side or RHS ?

[Rahul@gurome]

Hi Rahul,
Sorry... stmt1 gives value of x = 1/3 and 1/2 as I had put it before...

According to your solution,
stmt1 => x=1/3 or 3
if X =1/3 |x| < 1...
Therefore we need stmt 3 to eliminate option x=3 ?

I have a doubt in your solution...
-1 < x < 1
Why did you take the negative value of only one side or RHS ?

Yes you're right.
Edited the reply.

When -1 < x < 1, then |x + 1| is always positive but |x - 1| is always negative. Thus in LHS, |x + 1| = (x + 1). But in RHS, |x - 1| = -(x - 1).

[TOPGMAT]

Thanks Rahul!!!

[Night reader]

Hi Rahul,
Sorry... stmt1 gives value of x = 1/3 and 1/2 as I had put it before...

According to your solution,
stmt1 => x=1/3 or 3
if X =1/3 |x| < 1...
Therefore we need stmt 3 to eliminate option x=3 ?

I have a doubt in your solution...
-1 < x < 1
Why did you take the negative value of only one side or RHS ?

Yes you're right.
Edited the reply.


total mess in my head now, after 9 hours of quant exercising-since yesterday
ok, it was mistake not to change the sign of arguments above, overlooked

Rahul and others please see if i am correct, I got to clear this problem with absolutes on both sides

Is |x| < 1 ?

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

(2) |x - 3| > 0

stimuli: x<1 and -x<1 or x>-1 [][][][][] Is x>-1 -1<x<1 x<1 ? (I got these only)

st(1) x+1=2x-2, x=3 [][][][] -x-1=-2x+2, x=1/3 Although x=1/3 is within interval -1<x<1 x=3 is not, it can be either depending on the sign of x i.e. x can be 1/3 (satisfies x<1) or 3 (satisfies x>-1) right? Insufficient, because 3 doesn't satisfy -1<x<1, although 1/3 is in the interval -1<x<1

st(2) x-3>0, x>3 [][][][][] -x-3>0, x<-3 depends on the sign of x in the mod x can be greater 3 (satisfies x>-1) or less -3 (satisfies x<1) right? however, both solutions do not satisfy to -1<x<1

By combining st (1) and (2) we can say that x values do not fall within range -1<x<1 except for 1/3. The answer to question |x|<1 is No

[Night reader]

|x + 1| = 2|x - 1|, there may be three cases x < -1, -1 < x < 1 and x > 1. Observe that x can't be -1 and 1! Now let's analyze the equation for each of these cases...

How you get x<-1 and x>1 from |x|<1, by changing the sign in the mod I get only -x<1 and x<1 or -1<x<1


Ok, if this is a rule of four zones and one value of x that is less than -1, one value that is between -1 and 0, one value between 0 and 1, and Â…finally, one value is greater than 1. Then we just execute substitution and find that x in st(1) and st(2) do not satisfy.

[Rahul@gurome]

How you get x<-1 and x>1 from |x|<1, by changing the sign in the mod I get only -x<1 and x<1 or -1<x<1

Ok, if this is a rule of four zones and one value of x that is less than -1, one value that is between -1 and 0, one value between 0 and 1, and Â…finally, one value is greater than 1. Then we just execute substitution and find that x in st(1) and st(2) do not satisfy.

I didn't get x < -1 and x > 1 from |x| < 1.
The equation in the statement 1 has absolute value associated with both the sides. One of the term is |x + 1| and other one is |x - 1|. Thus for various range of values of x their signs becomes different. We have to identify what are those ranges of values. As one of the term is |x + 1|, (x + 1) is positive for x > -1 and negative for x < -1. Similarly (x - 1) is positive for x > 1 and negative for x < 1. Combining these four ranges we have three ranges x < -1, -1 < x < 1 and x > 1.

This is nothing but what you've done by changing the signs. But in this case, we're clearly mentioning the range we're assuming to solve the equation. Which helps us to identify the wrong solutions that occurs with these kind of equations (with absolute value in it) frequently. For example, case 1 in my analysis.

Hope it is clear now.

[davedecibel]

rahul explanations are extremely good as usual.

but what strikes me about this question is... how are you supposed to do it in an acceptable amount of time?

i thought that data sufficiency questions are supposed to test wheter or not you are able to detect if certain data is sufficient to answer a question, without actually wasting time solving the question itself.

in this case it looks to me that to reach a definitive yes or no answer to the question "is |x| < 1"? we have to solve the modular equation (are they called like this in english? sorry, but i'm not a native english speaker) in statement (1), and that takes a relatively long time (in comparison to the infamous average 2'2" that we have for every quant question).

am i right here?

[fskilnik@GMATH]

how are you supposed to do it in an acceptable amount of time?
am i right here?

Using the proper tool; that´s what the GMAT examiners were testing.

i thought that data sufficiency questions are supposed to test wheter or not you are able to detect if certain data is sufficient to answer a question, without actually wasting time solving the question itself.

PERFECT. That´s EXACTLY why I´ve decided to show you the way I would approach and solve the problem in REALLY 20 seconds.

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

From the fact that the distance between any two real numbers A and B in the real line is simply given by |A-B|, please note that sttm (1) simply says that the distance from -1 to x must be twice the distance from 1 to x... draw a real line and it is easy to see (10 seconds) that there are ONLY two possibilities:

> one between -1 and 1, nearer 1 (it is easy to see that the value is 1/3, because you may divide the interval [-1,1] in three equal parts to see that "the point" is one of them, etc... but it is NOT need, for sure)

>> The fact is that this number IS between -1 and 1, therefore its modulus is less than 1 for sure and answers in the POSITIVE.

> one at the right of number 1 (it is in fact number 3, easy to obtain geometrically but, again, no need to know).

>> The fact is that this number IS NOT between -1 and 1, therefore it answers in the NEGATIVE.


(2) It only says that x is different from 3, but this is of course far from enough to decide whether x is between -1 and 1 ...


(1+2) Must I explain??

Explanation: [spoiler]the first choice is NOT 3, the second one you simply test if it is 3, if you did not discover before, at sttm 1, that is was 3. Testing you will see that the second one was the number 3, then you refuse it and leaves just one possibility, therefore (1+2) answers the question asked! [/spoiler]

Regards,
Fabio.


P.S.1: @davedecibel: congrats for you insight in terms of trying to look the problem in the "proper way". That´s what the GMAT is all about, I am sure. Some problems are really time-consuming, others may be not... that´s one of the beauties, because candidates must take decisions during the exam (will I do brutal force, pick numbers, back-solving, "conceptual math", ...?)

P.S.2: have a look at David´s argument at this link... I´m sure you will love it:
https://www.beatthegmat.com/og-12th-ed-d ... 70373.html

P.S.3: I took more than HALF AN HOUR to find a "proper" (iMHO) solution to a goyalsau´s post, called "Escalator" (search it, it is really nice... Rahul´s solution to that problem is also interesting, by the way), because NOW I would take maximum 1 minute to solve it if I had to! That´s what I suggest my students to do: study to build a proper "war armour" at your body. It takes time to be able to build it and to be able to wear it BUT... that´s the proper way of going to battle, isn´t it? (As time passes, problems must get harder and harder to make you suffer... that means you are getting stronger, simple as that.)

P.S.4.: as far as goyalsau is concerned, he puts MANY excellent problems here... have a look at them, because they are really HIGH level. Some people may say they are not "GMAT" stuff, but I disagree. I don´t answer them because I am really short-of-time here, otherwise I would be really happy to contribute to solutions for them! Cheers, goyalsau!