I think the answer to this problem is A. Solving the first equation we get that x=1 or x=-1 in any case the absolute value of X will be 1 which is greater than 1
Second clue doesnt actually help as xNE 3 as in X can take any value.
Just try checking this answer with others I might not be correct though I feel this might be correct.
Roopachet
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PDF800 SET3 Question 5
Source: Beat The GMAT — Data Sufficiency |
roopachet's explanation, as usual, makes no sense and is completely incorrect.
1.) Tells you X = 3 OR 1/3. Insufficient.
2.) Tells you X NE 3. Insufficient.
Together, X = 1/3. Sufficient.
Answer is C.
1.) Tells you X = 3 OR 1/3. Insufficient.
2.) Tells you X NE 3. Insufficient.
Together, X = 1/3. Sufficient.
Answer is C.
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givemeanid
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beny, I have following some other posts of yours too and it seems like you are taking this personally. You need to calm down and stop making personal remarks. Remember that just because a problem/solution/concept is easy for you DOES NOT mean that its easy for everybody else.roopachet's explanation, as usual, makes no sense and is completely incorrect.
So It Goes
i agree .. beny u need to stop targeting a a particular forum user .. dont get me wrong over here .. it is great to see u helping out the members over here .. but u need to understand that there is always going to be someone on the forum who is better than u and also someone who is not as good as u in solving quant .. remember the point behind the whole forum is to learn from each others experience .. So plz stop making personal remarks ..givemeanid wrote:beny, I have following some other posts of yours too and it seems like you are taking this personally. You need to calm down and stop making personal remarks. Remember that just because a problem/solution/concept is easy for you DOES NOT mean that its easy for everybody else.roopachet's explanation, as usual, makes no sense and is completely incorrect.
I answered the question, like only twice in the entire time since I joined beathegmat who are you benny or whatever to take it so personally. I accept that i may be wrong when i say i think so. DO YOU OWN THIS SITE?
Last edited by roopachet on Sat Aug 04, 2007 7:04 am, edited 1 time in total.
Agreed with gabriel--we're all here to learn together! I want Beat The GMAT to be the nicest and most supportive GMAT community on the Internet...
Thanks!
Thanks!
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Could you guys tell me, how do I solve the following absolute value?
|x+1| = 2|x-1|
|x+1| = 2|x-1|
Thank you,
Vj
Vj
Easiest method is Try squaring on both sides,vzzai wrote:Could you guys tell me, how do I solve the following absolute value?
|x+1| = 2|x-1|
so we have :
(x+1)^ 2 = 4 (x-1)^2.
Upon expansion u will get : 3x^2 -10x+3
solve for x . Obviously being a quadriatic equn u will get 2 solns ; namely x = 3 or x = 1/3.
SO insufficient.
St 2 :
we just infer that X can't be 3 so as to satisfy the prompt that |x-3| <>3
Combining both the sts, we get X may be 3 or 1/3 from st 1 & X CAN'T be 3 from ST 2.
SO X is 1/3
|1/3| = 1/3 which is certainly less than 1.
Pick C
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ankur.agrawal
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Hi , Can u pls clear a few doubts:gmatmachoman wrote:Easiest method is Try squaring on both sides,vzzai wrote:Could you guys tell me, how do I solve the following absolute value?
|x+1| = 2|x-1|
so we have :
(x+1)^ 2 = 4 (x-1)^2.
Upon expansion u will get : 3x^2 -10x+3
solve for x . Obviously being a quadriatic equn u will get 2 solns ; namely x = 3 or x = 1/3.
SO insufficient.
St 2 :
we just infer that X can't be 3 so as to satisfy the prompt that |x-3| <>3
Combining both the sts, we get X may be 3 or 1/3 from st 1 & X CAN'T be 3 from ST 2.
SO X is 1/3
|1/3| = 1/3 which is certainly less than 1.
Pick C
1) Does squaring removes the absolute sign.
2) What are the rules for squaring something with a absolute sign.
Thanx in advance.
Ankur,
can u Plz post ur query with some example? I am not able to figure out the wording precisely.
can u Plz post ur query with some example? I am not able to figure out the wording precisely.
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ankur.agrawal
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I mean Is it ok to square both sides of this equation. |x+1| = 2|x-1| . Is it always allowed or does it have some restrictions.?gmatmachoman wrote:Ankur,
can u Plz post ur query with some example? I am not able to figure out the wording precisely.
I hope i make myself clear now.
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kevincanspain
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If |x| = |y| , x^2 = y^2 and vice versa. Thus, by squaring both sides, you always get an equivalent statement
You could also use the fact that if |x| = |y| either x= y or x = - y
Thus either x + 1 = 2(x - 1) i.e. x= 3
or x + 1 = -2(x - 1) i.e x = 1/3
You could also use the fact that if |x| = |y| either x= y or x = - y
Thus either x + 1 = 2(x - 1) i.e. x= 3
or x + 1 = -2(x - 1) i.e x = 1/3
Kevin Armstrong
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gmatmachoman / kevincanspain, Please help to clarify.
Could I use the same logic to |x-2| + |x-3| < 0 and conclude that x < 2.5?
Thank you,
Could I use the same logic to |x-2| + |x-3| < 0 and conclude that x < 2.5?
Thank you,
Thank you,
Vj
Vj
Hi vijay Just answering the post to help you.
Actually this inequality is not possible. so you will not get a value for x. WHY??
absolute value of x-2 + absolute value of x-3
the lowest possible value of both can be zero (since it cant be negative)
which means |x-2|+|x-3| can never be < 0
hence you cannot find a value for x......
hope it helps.. let me know if you need any further clarification.
Actually this inequality is not possible. so you will not get a value for x. WHY??
absolute value of x-2 + absolute value of x-3
the lowest possible value of both can be zero (since it cant be negative)
which means |x-2|+|x-3| can never be < 0
hence you cannot find a value for x......
hope it helps.. let me know if you need any further clarification.
