please help understand this question. thanks
gmatprep - sequence ds
This topic has expert replies
-
- Master | Next Rank: 500 Posts
- Posts: 294
- Joined: Tue Feb 26, 2008 9:05 pm
- Thanked: 13 times
- Followed by:1 members
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
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
- tendays2go
- Senior | Next Rank: 100 Posts
- Posts: 55
- Joined: Sun Sep 14, 2008 6:51 am
- Location: Netherlands
- Thanked: 10 times
- GMAT Score:680
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
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