If x is not equal to 0, is |x| less than 1?

[punzo]

If x is not equal to 0, is |x| less than 1?

(1) x/|x| < x

(2) |x| > x

My explaination for Statement 1 is

x / lxl < x
x< x. lxl

If x is +ve
x < x.x
x > 1

IF x is -ve
-x < -x.lxl
Div both sides by-1
x > x. lxl
x < 1

Therefore, insufficient

Is there something wrong with my explaination??

[sandeepgupta176]

May be we can go through in this way

in statement 1

x / [x] < x

now reciprocating

[x]/x > 1/x

cancelling x in denominator

[x] > 1

which shows its not less than 1.


Tell me if i am incorrect

[htchanit]

Hello,

In my opinion, reciprocating is somehow dangerous and eliminating 1/x without knowing 1/x negative or positive is not right.

since x < x.[x] and x not equal 0 :

1. if x >0 then 1 < [x] (eliminate x both sides)
2. if x <0 then 1 > [x] (eliminate x both sides)

=> cant give final result of [x] => (insufficient)

On the other hand, if "[x] > x" => "x < 0" => (insufficient)

So I will pick "C" (both together makes anything) :D !

May be we can go through in this way
in statement 1
x / [x] < x
now reciprocating
[x]/x > 1/x
cancelling x in denominator
[x] > 1
which shows its not less than 1.

Tell me if i am incorrect

[shailesh244]

statement : 1
x/[x]< x

x=2 ,so 2/[2]< 2

we cannot take any -ve value as doing so will result as below

x= -3 ,so -3/[-3] = -3/3 = -1 < -3 (which is not true)

so we conclude that x should be positive and greater equal to 2

if x=1, 1/[1]=1 < 1 (which is also not true)

so x >=2

So statement 1 is Sufficient as answer to [x]< 1 is NO.


Statement 2 :

[x]> x

if x= -1 , [-1]>1 1>1 (which cannot be possible) NO

x=-2, [-2]> -2 --->2>-2 Yes

1 Yes and 1 No so insufficient.


My Answer would be A. Please suggest flaws if any. Thanks.

[shailesh244]

Also we should abstain from dividing / multiplying as we do not know the signs of the variable.

[htchanit]

Hello,

I suggest you to plug-in with x = -1/2 to see
let's try:
x/[x] < x <=> (-1/2)/(1/2) < -1/2

clearly : [-1/2] < 1 , in this case [x] < 1

statement : 1
x/[x]< x
x=2 ,so 2/[2]< 2
we cannot take any -ve value as doing so will result as below
x= -3 ,so -3/[-3] = -3/3 = -1 < -3 (which is not true)
so we conclude that x should be positive and greater equal to 2
if x=1, 1/[1]=1 < 1 (which is also not true)
so x >=2
So statement 1 is Sufficient as answer to [x]< 1 is NO.
Statement 2 :
[x]> x
if x= -1 , [-1]>1 1>1 (which cannot be possible) NO
x=-2, [-2]> -2 --->2>-2 Yes
1 Yes and 1 No so insufficient.
My Answer would be A. Please suggest flaws if any. Thanks.

[Ben Ku]

If x is not equal to 0, is |x| less than 1?

(1) x/|x| < x

(2) |x| > x

My explaination for Statement 1 is

x / lxl < x
x< x. lxl

If x is +ve
x < x.x
x > 1

IF x is -ve
-x < -x.lxl
Div both sides by-1
x > x. lxl
x < 1

Therefore, insufficient

Is there something wrong with my explaination??

This response is fine. The key for this response is if x < x^2 and x > 0, then x > 1. If x < x^2, then 0 < x < 1.

[Ben Ku]

May be we can go through in this way

in statement 1

x / [x] < x

now reciprocating

[x]/x > 1/x

cancelling x in denominator

[x] > 1

which shows its not less than 1.


Tell me if i am incorrect

The process of "cancelling x in denominator" is the same as multiplying both sides by x. Since we don't know whether x is positive or negative, we cannot do this.

[Ben Ku]

Hello,

In my opinion, reciprocating is somehow dangerous and eliminating 1/x without knowing 1/x negative or positive is not right.

since x < x.[x] and x not equal 0 :

1. if x >0 then 1 < [x] (eliminate x both sides)
2. if x <0 then 1 > [x] (eliminate x both sides)

=> cant give final result of [x] => (insufficient)

On the other hand, if "[x] > x" => "x < 0" => (insufficient)

So I will pick "C" (both together makes anything) :D !

May be we can go through in this way
in statement 1
x / [x] < x
now reciprocating
[x]/x > 1/x
cancelling x in denominator
[x] > 1
which shows its not less than 1.

Tell me if i am incorrect

This solution is good!

[Ben Ku]

statement : 1
x/[x]< x

x=2 ,so 2/[2]< 2

we cannot take any -ve value as doing so will result as below

x= -3 ,so -3/[-3] = -3/3 = -1 < -3 (which is not true)

so we conclude that x should be positive and greater equal to 2

if x=1, 1/[1]=1 < 1 (which is also not true)

so x >=2

So statement 1 is Sufficient as answer to [x]< 1 is NO.


Statement 2 :

[x]> x

if x= -1 , [-1]>1 1>1 (which cannot be possible) NO

x=-2, [-2]> -2 --->2>-2 Yes

1 Yes and 1 No so insufficient.


My Answer would be A. Please suggest flaws if any. Thanks.

The problem with trying different numbers is choosing the number to choose. We tried x = 2, 1, and -3. However, this doesn't represent all possible values for x. Working out the algebra and using theory will help getting to the answer. We cannot conclude that x >= 2.

[sprparvathy]

May be I am asking this too late? Is the answer C?

Combining st 1 and 2 we know that X<1 and X<0. However [X] need not be less than 1 right? Then how can C be the answer?

[vijaykumar.kondepudi]

Yes the Answer is C.
Check out htchanit solution.
Combiniing St1 and St2, we know that,

If x< 0, then |x| < 1. (NOT x < 1).

That gives you the required result.

[RonPurewal]

statement (2) means that x is negative.
this is not enough information to tell whether |x| is less than 1.
insufficient.

--

to interpret statement (1), note that the fraction x/|x| is equal to 1 for any positive value of x, and equal to -1 for any negative value of x.
therefore, to solve this equation, and just consider the positive and negative cases separately.
if x is a positive number, then this inequality can be rewritten as 1 < x.
if x is a negative number, then this inequality can be rewritten as -1 < x. since this only applies to negative values, we can amend this to give -1 < x < 0.

therefore, statement (1) means that EITHER x > 1 OR -1 < x < 0.
for the first possibility, |x| is greater than 1; for the second, |x| is less than 1. insufficient.

--

together:
the only interval that satisfies both statements is -1 < x < 0, in which all numbers satisfy |x| < 1.
sufficient.

[monira.linda]

Hi,

After reading all explanations, I am still confused :-(

For statement 1, when X<0

We derived x/-x < x but if we are trying for x<0, then are not we suppose to change all Xs to -X like below:

-x/-x < -x => X<- 1

Pls help!!!

[jnelson0612]

Hi,

After reading all explanations, I am still confused :-(

For statement 1, when X<0

We derived x/-x < x but if we are trying for x<0, then are not we suppose to change all Xs to -X like below:

-x/-x < -x => X<- 1

Pls help!!!

Okay, here's statement 1:
(1) x/|x| < x

Rather than do all the manipulation you suggest, which is scary considering that we have both an inequality and an absolute value sign, let's think about what is means when x is negative by plugging in a real number. Let's use x=-2. Thus:
-2/|-2| < -2
which is:
-2/2 < -2, or -1<-2. This is NOT true, so x cannot be a negative integer. You can test it out with other negative integers such as -1 and -5, for example, and see that this statement, which we have to regard as true, does not allow us to use negative integers, since they do not work out in the statement.

Now test a fraction between -1 and 0 such as -1/2. Does -1/2 work in this statement? You will see that it does, so thus x cannot be a negative integer but CAN be between -1 and 0.

This illustrates how testing numbers is often very valuable on data sufficiency.