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

[vr4indian]

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

(1) x/|x| < x
(2) |x| > x

[khudania]

I think D (Both A & B are sufficient)

[khudania]

I think D (Both A & B are sufficient)

[Morgoth]

IMO C.

[robzoc]

I think D is not right.. In B, "lxl" can be more or less than 1. :

[Morgoth]

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

(1) x/|x| < x
(2) |x| > x

Here is why I think answer should be C.

|x| less than 1

only way this is possible if -1 < x < 1
We have to find the range for the value of x

Statement(1)
x/|x| < x
this shows that x is greater than -1. insufficient

Statement(2)
|x| less than 1
this shows that x is less than 1. insufficient.


Combining (1) & (2) we get -1 <x < 1. sufficient.

Hence C.

[Ian Stewart]

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

(1) x/|x| < x
(2) |x| > x

From 1, we can multiply both sides of the inequality by |x|, since we know |x| must be positive:

x/|x| < x
x < x*|x|

Now, if x is positive, we can divide by x without reversing the inequality to find:

1 < |x|, and since x > 0, this means 1 < x.

On the other hand, if x is negative, we must reverse the inequality when we divide by x:

1 > |x|, and since x < 0, this means -1 < x < 0

So from 1), we know that either -1 < x < 0 or 1 < x. Not sufficient.

2) simply tells us that x is negative, since |x| = x whenever x is positive. Not sufficient.

Together, we know that -1 < x < 0, so C.

[skang357]

Gawd I hate problems like this.

[TryHarder]

I think the answer must be C. Here is why.

x |x| |x| > x x/|x| x/|x| < x
-5 5 Yes -1 No
-4 4 Yes -1 No
-3 3 Yes -1 No
-2 2 Yes -1 No
-1 1 Yes -1 No
-0.75 0.75 Yes -1 Yes
-0.5 0.5 Yes -1 Yes
-0.25 0.25 Yes -1 Yes
-0.1 0.1 Yes -1 Yes
0
0.25 0.25 No 1 No
0.5 0.5 No 1 No
1 1 No 1 No
2 2 No 1 Yes
3 3 No 1 Yes
4 4 No 1 Yes
5 5 No 1 Yes

1 x/|x| < x in two situations
when -1 < x < 0 (where |x| < 1)
OR
when x > 1 (where |x| > 1)
Hence not sufficient
2 |x| > x in one situation
when x < 0 (where |x| > 0 but not always >1)
Sufficient
Together however |x| is always less than 1


The answer is clearly C

[skang357]

I think the answer must be C. Here is why.

x |x| |x| > x x/|x| x/|x| < x
-5 5 Yes -1 No
-4 4 Yes -1 No
-3 3 Yes -1 No
-2 2 Yes -1 No
-1 1 Yes -1 No
-0.75 0.75 Yes -1 Yes
-0.5 0.5 Yes -1 Yes
-0.25 0.25 Yes -1 Yes
-0.1 0.1 Yes -1 Yes
0
0.25 0.25 No 1 No
0.5 0.5 No 1 No
1 1 No 1 No
2 2 No 1 Yes
3 3 No 1 Yes
4 4 No 1 Yes
5 5 No 1 Yes

1 x/|x| < x in two situations
when -1 < x < 0 (where |x| < 1)
OR
when x > 1 (where |x| > 1)
Hence not sufficient
2 |x| > x in one situation
when x < 0 (where |x| > 0 but not always >1)
Sufficient
Together however |x| is always less than 1


The answer is clearly C

Doing it your way, how long do you think it would take to answer the question?

And what is the probability for error. I did it your way at first (plug and chug) and missed the >1 portion. Anyways, I missed one of the possibilities. You have to include negative, positive, negative fraction, positive fraction. It's easy to get something wrong.

I like Ian's way. It tells you right away what the limits are without having to plug in values.

However, seems like you have to plug in value for the second statement but that one is pretty obvious.

[TryHarder]

By no means am I suggesting that this is the optimal way, and frankly I did not solve this within 2:05 mins, but i think these problems have better odds when we use real values for x. Granted you dont have to try it with -4, 5, and -5, -4 etc. as I have done to illustrate. Just one representative value would do. In this case for instance the idea is to use a combination of negatives/positives and fractions/integers. Hope it helps!

[sabeer6870]

(1)
x/|x|<x => x<|x|.x =>x.|x|-x>0 =>x(|x|-1)>0
possible only when x>0 & |x|>1 OR x<0 & |x| <1 clearly NOT SUFF.

(2)x<0 N.S.

combine out of two possibility only one left x<0 & |x| <1 so suff
hence C.

[sachindia]

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

(1) x/|x| < x
(2) |x| > x

From 1, we can multiply both sides of the inequality by |x|, since we know |x| must be positive:

x/|x| < x
x < x*|x|

Now, if x is positive, we can divide by x without reversing the inequality to find:

1 < |x|, and since x > 0, this means 1 < x.

On the other hand, if x is negative, we must reverse the inequality when we divide by x:

1 > |x|, and since x < 0, this means -1 < x < 0

So from 1), we know that either -1 < x < 0 or 1 < x. Not sufficient.

2) simply tells us that x is negative, since |x| = x whenever x is positive. Not sufficient.

Together, we know that -1 < x < 0, so C.

I understand we can multiply |x| on both sides but

my argument is if we do it without multiplying we get ans A

Here's how:

x/|x| < x

above can be true only when x is positive and less than 1.

take x=1/2

in this case it satisfies x/|x| < x and so 0<x<1.

and also answers no to our Q |x|<1 which is nothing but -1<x<1

Please let me know if my reasoning is wrong.

[Ankur87]

Hi Ian,
Can't we say directly that statement 1 is sufficient as:
x/|x| < x
x/x < |x|
1 < |x|
and it is asked whether |x| less than 1
so we can directly say NO as 1 < |x| from statement 1.
Please help.
I would have selected this answer is Main exam.

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

(1) x/|x| < x
(2) |x| > x

From 1, we can multiply both sides of the inequality by |x|, since we know |x| must be positive:

x/|x| < x
x < x*|x|

Now, if x is positive, we can divide by x without reversing the inequality to find:

1 < |x|, and since x > 0, this means 1 < x.

On the other hand, if x is negative, we must reverse the inequality when we divide by x:

1 > |x|, and since x < 0, this means -1 < x < 0

So from 1), we know that either -1 < x < 0 or 1 < x. Not sufficient.

2) simply tells us that x is negative, since |x| = x whenever x is positive. Not sufficient.

Together, we know that -1 < x < 0, so C.