IMO C
ST 1) two ans for x>1 the q is false and for 0<x<1 true
not suff
St 2) diff value for 0<x<1;-1<x<0;x<-1
not suff
together 1 and 2 suff as 0<x<1 any RHS will always be grtr than LHS
OA pls
How to approach this one
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Source: Beat The GMAT — Data Sufficiency |
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xcusemeplz2009
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sanjana
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IMO : A
The given sequence is a GP.
1+X+X^2+X^3+X^4+X^5
Sum to n terms : 1st term * (1-r^n)/(1-r)
1 * (1-x^5)/1-x
Now the question asks
Is (1-x^5)/1-x < 1/1-x ?
or, is 1-x^5 <1
or is x^5 > 0
Statement 1:
------------
If x>0 then x^5 will always be greater than 0
Sufficient.
Statement 2:
------------
x<1
If x=0,then
x^5 = 0
If x is -ve then x^5 < 0
Hence insufficient.
Hence, A
OA please.
The given sequence is a GP.
1+X+X^2+X^3+X^4+X^5
Sum to n terms : 1st term * (1-r^n)/(1-r)
1 * (1-x^5)/1-x
Now the question asks
Is (1-x^5)/1-x < 1/1-x ?
or, is 1-x^5 <1
or is x^5 > 0
Statement 1:
------------
If x>0 then x^5 will always be greater than 0
Sufficient.
Statement 2:
------------
x<1
If x=0,then
x^5 = 0
If x is -ve then x^5 < 0
Hence insufficient.
Hence, A
OA please.
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maihuna
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IMO A
Ok 1+x+x2+x3+x4 = 1.(x4-1)/(x-1) = (1-x4)/(1-x)
so 1/1-x is common in both so we need to decide the sign and value of 1-x4
for x<1 for x=1/2 it will be positive and less than 1 and LHS less
x=0 it will be positive and equal to 1 so both eul
x=-2 it will be negative and LHS less
for x>0 for x=1/2 it will be positive and less than 1
x=2 it will be negative and so LHS will be less
so as we see stmt x>0 is sufficient to say that LHS will be less and so A
Ok 1+x+x2+x3+x4 = 1.(x4-1)/(x-1) = (1-x4)/(1-x)
so 1/1-x is common in both so we need to decide the sign and value of 1-x4
for x<1 for x=1/2 it will be positive and less than 1 and LHS less
x=0 it will be positive and equal to 1 so both eul
x=-2 it will be negative and LHS less
for x>0 for x=1/2 it will be positive and less than 1
x=2 it will be negative and so LHS will be less
so as we see stmt x>0 is sufficient to say that LHS will be less and so A
Charged up again to beat the beast 
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rohan_vus
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Should be C ..
In reducing the expression (1-x^5)/1-x < 1/1-x to 1-x^5 <1 , you are assuming 1-x > 0 ..else if 1 - x < 0 then the inequality changes to (1-x^5) > 1 ...
So stmnt 1 alone cant suffice , but in conjunction with stmnt 2 it works
In reducing the expression (1-x^5)/1-x < 1/1-x to 1-x^5 <1 , you are assuming 1-x > 0 ..else if 1 - x < 0 then the inequality changes to (1-x^5) > 1 ...
So stmnt 1 alone cant suffice , but in conjunction with stmnt 2 it works
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crackgmat007
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Is the above correct? I guess if (1-x) is negative, by dividing the inequality with (1-x), the inequality will be reversed. So without knowing whether (1-x) will be positive, we cannot solve.sanjana wrote:
Is (1-x^5)/1-x < 1/1-x ? ---- 1
or, is 1-x^5 <1 ---- 2
IMO C.
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viju9162
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The answer seems to be A,
From A: X>0
First ex: X =2,
1+2+8 ... < -1
Not true ...
Second ex: X = 1/2
1+1/2+1/8+ .... < 1/1-1/2
=> 1+1/2+1/8+ .... < 1/-1/2
=> 1+1/2+1/8+ .... < - 2
Not true..
A answers the question
From B: x<1
If X = 1/2 - It doesn't answer
If X = -1 - It answers.. The answers are contradicting from B.
From A: X>0
First ex: X =2,
1+2+8 ... < -1
Not true ...
Second ex: X = 1/2
1+1/2+1/8+ .... < 1/1-1/2
=> 1+1/2+1/8+ .... < 1/-1/2
=> 1+1/2+1/8+ .... < - 2
Not true..
A answers the question
From B: x<1
If X = 1/2 - It doesn't answer
If X = -1 - It answers.. The answers are contradicting from B.
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life is a test
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1-1/2 = 1/2 in which case RHS >LHSviju9162 wrote:The answer seems to be A,
=> 1+1/2+1/8+ .... < 1/-1/2
=> 1+1/2+1/8+ .... < - 2
IMO C RHS will always be > LHS providing x>0 and <1