My experience with the GMAT tells me that the questions on the quant section are NOT of an exceptionally tough variety - just the very SMART variety.
One must, MUST (I cannot emphasis this enough) R E A D the question properly.
Something I need to learn myself (mum's voice going 'haste makes waste' is playing in my head right now!) - Towards this end I would like to start a new thread - problems that may or may not be very tough = but require understanding (and remembering while solving) the question properly.
Lets start the series with :
PS - 1
Is |n| < 1 ?
(1) nx – n < 0
(2) x–1 = –2
(A) Statement (1) ALONE is sufficient to answer the question, but statement (2) alone is not.
(B) Statement (2) ALONE is sufficient to answer the question, but statement (1) alone is not.
(C) Statements (1) and (2) TAKEN TOGETHER are sufficient to answer the question, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient to answer the question.
(E) Statements (1) and (2) TAKEN TOGETHER are NOT sufficient to answer the question.
(picked from MahattanGMAT)
Math!
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- thankont
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I agree with your approach..
I think ans. is --E--
|n|<1 => -1<n<1 (this is what we are looking for
Now 1) nx-n<0 => n(x-1)<0 so either n>0 and x<1 or n<0 and x>1
(A does not hold, take for example. n=3 x=-2)
2) x-1=-2 so x=-1 (this does not say anything about n so B does not hold)
combining both statements we just get n>0 (which does not hold for ex. n=2
so I put --E--
I think ans. is --E--
|n|<1 => -1<n<1 (this is what we are looking for
Now 1) nx-n<0 => n(x-1)<0 so either n>0 and x<1 or n<0 and x>1
(A does not hold, take for example. n=3 x=-2)
2) x-1=-2 so x=-1 (this does not say anything about n so B does not hold)
combining both statements we just get n>0 (which does not hold for ex. n=2
so I put --E--
Last edited by thankont on Mon Jan 15, 2007 12:15 am, edited 1 time in total.
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hiya!
good effort, though thats not the correct ans
would you like to give it another shot before I put up the correct ans?
hint : consider + / - values for everything - including the indice (just re read the question stem - it seems like it didn't copy paste correctly and I lost the original formatting)
It should read :
Question:
Is |n| < 1 ?
(1) n^x – n < 0
(n raised to the power of x) - n < 0
(2) x^–1 = –2
(x raised to the power of -1) = -2
Hope you are able to solve it this time.
good effort, though thats not the correct ans
would you like to give it another shot before I put up the correct ans?
hint : consider + / - values for everything - including the indice (just re read the question stem - it seems like it didn't copy paste correctly and I lost the original formatting)
It should read :
Question:
Is |n| < 1 ?
(1) n^x – n < 0
(n raised to the power of x) - n < 0
(2) x^–1 = –2
(x raised to the power of -1) = -2
Hope you are able to solve it this time.
- aim-wsc
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Great efforts girl
Nice initiative.
OK now that the question is changed altogether
Anybody want to try it one more time?
bytheway with earlier question @ post #1 thanktont's reasonings are correct. the answer should be E for that.
PS:
Dear Pramila,
I am thinking of arranging an offline meet at Mumbai click here to know more
Nice initiative.
OK now that the question is changed altogether
Anybody want to try it one more time?
bytheway with earlier question @ post #1 thanktont's reasonings are correct. the answer should be E for that.
bluestocking wrote:
Question:
Is |n| < 1 ?
(1) n^x – n < 0
(n raised to the power of x) - n < 0
(2) x^–1 = –2
(x raised to the power of -1) = -2
Hope you are able to solve it this time.
PS:
Dear Pramila,
I am thinking of arranging an offline meet at Mumbai click here to know more
Getting started @BTG?
Beginner's Guide to GMAT | Beating GMAT & beyond
Please do not PM me, (not active anymore) contact Eric.
Beginner's Guide to GMAT | Beating GMAT & beyond
Please do not PM me, (not active anymore) contact Eric.
Would you mind explaining this:thankont wrote:I am not sure here but from n^x<n we see that it holds only when
0<n<1 and when x>1
from b we have that x=-1/2 so statement never holds (combining both)
so I would say --c--
"n^x<n we see that it holds only when
0<n<1 and when x>1 "