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Positive sequence x1, x2, x3, ....

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Source: — Data Sufficiency |
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by g_beatthegmat » Tue Jun 17, 2008 10:44 am
Oh, I think I understood the problem. I was reading (a) as
Xj = ((Xj) - 1) /2
rather than
Xj = (Xj-1) / 2

as in, if j = 2, I was reading
X2 = (X2 - 1)/2
rather than
X2 = X1/2

:oops:

Thanks.
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by durgesh79 » Tue Jun 17, 2008 10:45 am
statement 1 : we dont know value of J. in suff
statement 2 : we dont know x4. Insuff

Combine : x5 = x4/(x4+1)
from statement 1 x5 = x4/2

x4/2 = x4(x4+1)

let x4 = t

t^2 + t = 2t
t^2 - t = 0
t(t-1)= 0

So x4 = 1 it can not be 0 as its a +ve number

using statement 1 we can find x3,x2 and x1.

Answer C
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by g_beatthegmat » Tue Jun 17, 2008 10:47 am
thanks durgesh79 for the prompt reply. I misread the question statements.

Thanks!
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by atlantic » Tue Jun 17, 2008 10:55 am
I caught this one also during my GMATPrep. And I think I got it right.

(1) we can determine that....

(x2)=(x1)/2
(x3)=(x2)/2
(x4)=(x3)/2
(x5)=(x4)/2
....

No numerical value for x1, so A and D are out.

(2) says nothing about x1 so B is out.

(1+2) we have (x4)/2=(x4)/((x4)-1), x4=1. Knowing X4 we determine x1 from the relations in (1)

OA C
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by durgesh79 » Tue Jun 17, 2008 11:17 am
atlantic wrote:(1+2) we have (x4)/2=(x4)/((x4)-1), x4=1.
Be careful in solving this type of equation, you'll have two values of x4. 0 and 1. if initial the condition was that x is just an integer instead of +ve interger. the answer would have been E.
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by nitya34 » Tue Jun 02, 2009 2:50 am
good Q indeed
afterall its from GMATPrep
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by aj5105 » Tue Jun 02, 2009 6:08 am
(C)

Combined we get x4 = 1
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????

by brick2009 » Tue Jun 02, 2009 6:21 am
1) looks like X_i = X_(i-1)/2

X_2 = X_1/2 ???
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by maaz_gmat » Thu Jul 22, 2010 11:12 pm
stumped!!!
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