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gmatprep - sequence ds

Expert replies
Source: — Data Sufficiency |

by amitansu » Sun Sep 14, 2008 10:37 am
The q is asking a numerical value of x1.

From 1: taking the value of j>1 we can have,

x2=x2-1/2=x1/2; x3=x3-1/2=x2/2...so the sequence is
x4=x3/2,x5=x4/2; x6=x5/2 and so on....but we can't get a value of x1
here since no idea of any value of x2; x3; x4 .
We only know x2=x1/2=>x1=2 * x2

From 2: x5=x4/x4+1 we can't get anything about x1 or x2.

But combining both :
from 1 we know : x5=x4/2and putting that value in x5=x4/x4+1
=>x4/2=x4/x4+1
=>2x4=(x4)^2+x4=>(x4)^2-x4=0=>x4(x4-1)=0 so x4 could be 0 or 1

taking 0 we can't get a sequence like this..hence taking 1 as x4 we know x5=1/1+1=1/2 hence x1 also can be calculated.

Ans should be "C".


Amit
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by amitansu » Sun Sep 14, 2008 10:44 am
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is this expln correct?

by tendays2go » Mon Sep 15, 2008 12:57 pm
one more soln.
IMO: C


as there are no values given and we are asked the value for x1, so it could be only (C) or (E).

therefore,
from(i) => x2 = x1/2
from(ii)=> x2 = x1/(x1 +1)

now equating both of them, we get:
x1/2 = x1/ (x1+1)
or that, x1+1 = 2
=> x1 = 1
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by alescau » Mon Sep 15, 2008 2:40 pm
Using (1):
x5 = x4 / 2

Compare to:
x5 = x4 / (x4+1)

Basically, 2 = x4 + 1 => x4=1

(1) says that previous value is 2 times the current value. Starting with x4, we get x3=2, x2=4, x1=8.

C
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