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Source: — Data Sufficiency |
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by rijul007 » Wed Jan 04, 2012 10:33 pm
karthikpandian19 wrote:Is x greater than 1?
(1) 1/x > -1
(2)1/x^5 > 1/x^3
(1) 1/x > -1

1 > -x
x > -1

Insufficient

(2)1/x^5 > 1/x^3

1 > x^2
-1<x<1

Insufficient

Combining the two statements
Still insufficient


Option E
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by neelgandham » Thu Jan 05, 2012 3:01 am
rijul007 wrote:
karthikpandian19 wrote:Is x greater than 1?
(1) 1/x > -1
(2)1/x^5 > 1/x^3
(1) 1/x > -1

1 > -x
x > -1

If x =-2, 1/x = -1/2 > -1. x=-2 doesn't fall under the solution set x >-1 here. It is easy to solve such problems by drawing a graph attached. From the graph(attached) you clearly see that for x <-1 and x > 0 the value of 1/x > -1
Insufficient

(2)1/x^5 > 1/x^3
1 > x^2
-1<x<1
Here again, if x =-1/2(-1<x<1), 1/x^5 = -32 and 1/x^3=-8 and definitely 1/x^5 < 1/x^3 and not 1/x^5 > 1/x^3. You cannot multiply in equations without knowing the sign of the term or value you multiply.
1/x^5 > 1/x^3
1/x > x (Multiplying x^4, which is definitely positive)
From the graph attached,
x<-1 and 0<x<1 is the solution set.
sufficient to say that x <1

Option B
Attachments
Beatthegmat_29.JPG
Statement 1
Beatthegmat_30.JPG
Statement 2
Last edited by neelgandham on Thu Jan 05, 2012 8:42 am, edited 1 time in total.
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by ArunangsuSahu » Thu Jan 05, 2012 8:35 am
This is a "YES" or "NO" Question

We are missing 1 point here about the Statement 2:

Statement 2:

Given,1/x^5 > 1/x^3

or 1/x^3-1/x^5 < 0
or (x^2-1) < 0
or, x^2 <1

This Clearly tells -1<x<1...


Therefore x<1

So, Statement 2 is sufficient

(B)
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by neelgandham » Thu Jan 05, 2012 8:51 am
ArunangsuSahu wrote:This is a "YES" or "NO" Question

We are missing 1 point here about the Statement 2:

Statement 2:

Given,1/x^5 > 1/x^3

or 1/x^3-1/x^5 < 0
or (x^2-1) < 0
or, x^2 <1

This Clearly tells -1<x<1...


Therefore x<1

So, Statement 2 is sufficient

(B)
I agree that B is the solution. I overlooked and read it as Is x greater than 0? instead of Is x greater than 1?. However, there is a flaw in your solution.

1/x^5 > 1/x^3
or 1/x^3-1/x^5 < 0
or (x^2-1) < 0 <- Here you multiplied both sides with x^5 without knowing the sign of x, which shouldn't be done.
or, x^2 <1
This Clearly tells -1<x<1...

If x =-1/2(part of your solution -1<x<1)
1/x^5 = -32
1/x^3 = -8
and definitely 1/x^5 < 1/x^3.

If x = -10(Not part of your solution)
1/x^5 = -1/100000 = -0.00001
1/x^3 = -1/1000 = -0.001
1/x^5 > 1/x^3

The solution should be :
1/x^5 > 1/x^3
1/x > x (Multiplying x^4, which is definitely positive)
From the graph attached,
x<-1 and 0<x<1 is the solution set.

Please correct me if I am wrong !
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by ArunangsuSahu » Thu Jan 05, 2012 9:36 am
@neelgandham

You are right about my flaw

I should have handled in the following way

(x^2-1)/x^5<0

a) x< 0, (doesn't holds good all along. fails for negative fraction) Test with x=-2 and x=-1/2
b)0<x<1 and (holds good)
c)x>1 (the eqn doesn't hold good)
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by karthikpandian19 » Thu Jan 05, 2012 6:39 pm
I have a doubt in Statement I

(1) 1/x > -1

1 > -x
x > -1 (In this step don't we need to change the equality sign as X<-1 as we are multiplying by minus sign from the previous step)


Actually I did like this and found the answer to be D
Insufficient



But the OA is B
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by rijul007 » Thu Jan 05, 2012 11:07 pm
karthikpandian19 wrote:I have a doubt in Statement I

(1) 1/x > -1

1 > -x
x > -1 (In this step don't we need to change the equality sign as X<-1 as we are multiplying by minus sign from the previous step)


Actually I did like this and found the answer to be D
Insufficient



But the OA is B
Statement 1 cant be sufficient

you can check this by plugging no
say x = -2
1/x = -1/2 > -1
x is not greater than 1

now lts say x =2
1/x = 1/2 > -1
x > 1

Insufficient
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by karthikpandian19 » Fri Jan 06, 2012 4:05 am
Thank you for the clarification
rijul007 wrote:
karthikpandian19 wrote:I have a doubt in Statement I

(1) 1/x > -1

1 > -x
x > -1 (In this step don't we need to change the equality sign as X<-1 as we are multiplying by minus sign from the previous step)


Actually I did like this and found the answer to be D
Insufficient



But the OA is B
Statement 1 cant be sufficient

you can check this by plugging no
say x = -2
1/x = -1/2 > -1
x is not greater than 1

now lts say x =2
1/x = 1/2 > -1
x > 1

Insufficient
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