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If x is not equal to 0, is |x| less than 1?
Source: Beat The GMAT — Data Sufficiency |
:Here is why I think answer should be C.vr4indian wrote:If x is not equal to 0, is |x| less than 1?
(1) x/|x| < x
(2) |x| > x
|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.
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From 1, we can multiply both sides of the inequality by |x|, since we know |x| must be positive:vr4indian wrote:If x is not equal to 0, is |x| less than 1?
(1) x/|x| < x
(2) |x| > x
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.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com
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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
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?TryHarder wrote: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
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.
Impossible is nothing
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!
(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.
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.
I understand we can multiply |x| on both sides butIan Stewart wrote:From 1, we can multiply both sides of the inequality by |x|, since we know |x| must be positive:vr4indian wrote:If x is not equal to 0, is |x| less than 1?
(1) x/|x| < x
(2) |x| > x
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.
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.
Regards,
Sach
Sach
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.
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.
Ian Stewart wrote:From 1, we can multiply both sides of the inequality by |x|, since we know |x| must be positive:vr4indian wrote:If x is not equal to 0, is |x| less than 1?
(1) x/|x| < x
(2) |x| > x
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.
