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Absolute Value!!!!

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by john1234 » Sat Jul 19, 2008 7:18 pm
If y is greater than or equal to 0,what is the value of x

1) lx-3l is greater than or equal to y

2) lx-3l is greater than or equal to -y


Thanx for any help
Even impossible says I M possible
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Source: — Data Sufficiency |
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Re: Absolute Value!!!!

by gabriel » Sun Jul 20, 2008 5:18 am
john1234 wrote:If y is greater than or equal to 0,what is the value of x

1) lx-3l is greater than or equal to y

2) lx-3l is greater than or equal to -y


Thanx for any help
The answer is B. The second statement says that "lx-3l is greater than or equal to -y". But since a modulus can never be negative so y has to be 0 and hence lx-3l is equal to 0 and x =3
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by john1234 » Sun Jul 20, 2008 8:08 pm
Thanks. OA is B.
Even impossible says I M possible
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Re: Absolute Value!!!!

by ravibits » Wed Jul 23, 2008 10:08 pm
gabriel wrote:
john1234 wrote:If y is greater than or equal to 0,what is the value of x

1) lx-3l is greater than or equal to y

2) lx-3l is greater than or equal to -y


Thanx for any help
The answer is B. The second statement says that "lx-3l is greater than or equal to -y". But since a modulus can never be negative so y has to be 0 and hence lx-3l is equal to 0 and x =3
I don't get it!

Since the modulus of a value can never be negative, |x-3| will always be greater than -y(which is always negative), not just for y=0.

But, the explanation above looks good for me when the condition is |x-3|<=-y and this is the case where |x-3| can take the value of 0 ONLY since -y is always zero or negative.

Please let me know if I am missing anything here...
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Re: Absolute Value!!!!

by enniguy » Thu Jul 24, 2008 3:00 am
ravibits wrote:
gabriel wrote:
john1234 wrote:If y is greater than or equal to 0,what is the value of x

1) lx-3l is greater than or equal to y

2) lx-3l is greater than or equal to -y


Thanx for any help
The answer is B. The second statement says that "lx-3l is greater than or equal to -y". But since a modulus can never be negative so y has to be 0 and hence lx-3l is equal to 0 and x =3
I don't get it!

Since the modulus of a value can never be negative, |x-3| will always be greater than -y(which is always negative), not just for y=0.

But, the explanation above looks good for me when the condition is |x-3|<=-y and this is the case where |x-3| can take the value of 0 ONLY since -y is always zero or negative.

Please let me know if I am missing anything here...
Look at it this way,
consider y = 2, as y >=0 from question.

2nd statement:
|x - 3| {LHS} >= -y {RHS}

Then,
|x - 3| {LHS} >= -2 {RHS}.

This means, LHS can be -1, 0, 1, 2 , ... etc.
But since LHS should always be positive or 0, RHS should also be positive or 0.

Now, RHS is -y. RHS can be 0 when y is 0.
RHS can be positive when y is negative. e.g: -(-3)
But y>=0 from question. so RHS cant be positive and y cannot be greater than 0.

So |x-3| has to be zero and hence x =3.
Hope I answered it right!
Last edited by enniguy on Thu Jul 24, 2008 7:11 am, edited 1 time in total.
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by enniguy » Thu Jul 24, 2008 6:05 am
Ans: E

I did not conclude in my previous explanation.

Look at it this way,
consider y = 2, as y >=0 from question.

2nd statement:
|x - 3| {LHS} >= -y {RHS}

Then,
|x - 3| {LHS} >= -2 {RHS}.

This means, LHS can be -1, 0, 1, 2 , ... etc.
But since LHS should always be positive or 0, RHS should also be positive or 0.

Now, RHS is -y. RHS can be 0 when y is 0.
RHS can be positive when y is negative. e.g: -(-3)
But y>=0 from question and y is not greater than 0 from above (RHS cant be positive and y cannot be greater than 0).

That leaves us y = 0.

This brings us to the equation,
|x-3| >= 0

Now,
|x-3| >= 0 leaves 2 cases.

Case 1:- -(x-3) >= 0 and
Case 2:- +(x-3) >= 0.

Consider, Case 1.
Step 2 : x-3 <= 0
Step 3 : x <= 3

Consider Case 2.
Step 2 : x-3 >= 0
Step 3: x >= 3

From both Case 1 and case 2,
x can be any integer.

So, Statement 2 is insufficient.
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by reachac » Thu Jul 24, 2008 10:21 am
I am not convinced with the explanation provided for B being the answer...can anyone explain better please.
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by Ian Stewart » Thu Jul 24, 2008 5:29 pm
reachac wrote:I am not convinced with the explanation provided for B being the answer...can anyone explain better please.
And you should be unconvinced- B is not the correct answer. The solutions above would be correct if the question said 'less than', but they are certainly wrong for the question as presented, which says 'greater than'.

If y is greater than or equal to 0,what is the value of x

1) lx-3l is greater than or equal to y

2) lx-3l is greater than or equal to -y

Clearly 1) does not give you the value of x. 2) does not either; if y is positive, then -y is negative. It is obvious then that |x-3| is greater than -y, a negative number; indeed, statement 2) does not tell us anything at all. The answer must be E, since 2) is entirely useless, and 1) is insufficient.
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by reachac » Fri Jul 25, 2008 5:08 am
Thanks Ian :)
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by maihuna » Mon Apr 13, 2009 10:32 am
what is the oa?
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by ketkoag » Mon Apr 13, 2009 11:19 am
Ian, Please correct me if i am wrong.
I feel that if we take statement 2 alone then |x-3|>=-y and in the ques it is given that y>=0. So, that means that y is positive or 0 also -y is negative or 0. Now in the equation |x-3|>=-y |x-3| will always be positive or 0, right. and we know that -y will always be negative or 0. By this we get that |x-3| =0 . therefore x=3.
By this, i feel that statement 2 alone is sufficient.

But if i do it the other way in which i take a value for y and then do it, i am not able to get the answer by B alone. let say y = 3, then by statement 2 |x-3|>=-3 we get a range of x that is not a unique value hence not sufficient.

In this case i am getting answer as E.

Please help me to understand ambiguity as explained above.
thanks.
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by Ian Stewart » Mon Apr 13, 2009 2:40 pm
ketkoag wrote:Ian, Please correct me if i am wrong.
I feel that if we take statement 2 alone then |x-3|>=-y and in the ques it is given that y>=0. So, that means that y is positive or 0 also -y is negative or 0. Now in the equation |x-3|>=-y |x-3| will always be positive or 0, right. and we know that -y will always be negative or 0. By this we get that |x-3| =0 . therefore x=3.
By this, i feel that statement 2 alone is sufficient.
When you picked a number, your analysis was correct. I'm not sure I understand your conclusion to the quoted text above, however. What you've called an 'equation' is not an equation; it's an inequality. If there were an '=' sign instead of '>=', your analysis would be correct; y would need to be zero, and x would need to be 3. However, since we have an inequality, statement 2 is not at all useful. |x-3| must be greater than or equal to zero, and -y must be less than or equal to zero, so it is clearly true that |x-3| >= -y. So Statement 2 doesn't provide us with any new information; we could have deduced it from information in the question stem.

Whenever that happens (the second statement is clearly useless), the only possible answers are A or E, and since Statement 1 is insufficient alone, the answer is E, as you correctly concluded when you picked a value for y.
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by todor.atev » Mon Oct 19, 2009 1:58 am
the official answer on this one is actually B, the question is exactly as it is posted here. So, what's the logic? could we accept that x-3=0 or not?
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by girish3131 » Tue Mar 09, 2010 3:05 am
need more comments...

IMO B
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by kevincanspain » Tue Mar 09, 2010 3:15 am
This thread shows how important it is to be VERY careful when posting questions. Statement (2) should read
lx-3l is LESS than or equal to -y.
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