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alphonsa
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Is x^(k+1) > x^k? Updated on: 18 Aug 2014, 09:47
Originally posted by alphonsa on 18 Aug 2014, 08:57.
Last edited by Bunuel on 18 Aug 2014, 09:47, edited 1 time in total.
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

75% (hard)

Question Stats:

54% (02:11) correct 46% (01:48) wrong based on 173 sessions

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Is x^(k+1) > x^k ?

(1) |x| > 1

(2) k+1 is divisible by 6

source: 4gmat
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C
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TheRunawayFound
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Is x^(k+1) > x^k? 18 Aug 2014, 09:10
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alphonsa
Is \(x^(k+1) > x^k\) ?

1) |x| >1

2) k+1 is divisible by 6



source: 4gmat
Show more


C)


i) NS,

since X could be positive or negative, and since (k+1) could be odd or even.

ii) NS

Since X could be a fraction



i) + ii)

Sufficient since i) makes sure X it is not a fraction <1, and, ii) makes sure (k+1) is even.
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Is x^(k+1) > x^k? Updated on: 17 Jul 2021, 09:40
Originally posted by BrushMyQuant on 18 Aug 2014, 09:27.
Last edited by BrushMyQuant on 17 Jul 2021, 09:40, edited 2 times in total.
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STAT1
means that x < -1 or x > 1
If x is +ve then x^(k+1) will be > x^k for all values of k ...(1)
If x is -ve then x^(k+1) will be > x^k for all odd values of k ....(2)
If x is -ve then x^(k+1) will be < x^k for all even values of k
So, INSUFFICIENT

STAT2
k+1 is divisible by 6 means k+1 is either 0 or even, so k will be odd
For x=0 x^(k+1) = x^k =0
So, other values of x we can get x^(k+1) > x^k and also, x^(k+1) < x^k
So, INSUFFICIENT

STAT1 and STAT2 together we have only ...(1) and ...(2) possible
so, x^(k+1) will be > x^k
so, SUFFICIENT

So, Answer will be C
Hope it helps!

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Is x^(k+1) > x^k? 19 Aug 2014, 00:03
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alphonsa
Is x^(k+1) > x^k ?

(1) |x| > 1

(2) k+1 is divisible by 6

source: 4gmat
Show more

The question can be reduced \(x^{k}(x-1)>0\)

St 1 says |x|>1 or x>1 or x<-1....Now if x<-1 and k is odd then the expression will be greater than zero but x<-1, K is even then the expression is less than zero. Since 2 ans are possible st1 is not sufficient

St 2 says \(\frac{K+1}{6}=I\) , Where I is an Integer. So K=6I-1 or K =even-odd=odd

So we know K is odd but we don't know anything about x not sufficient.

Combined we have, x>1 or x<-1 and k is odd
Case 1: x>1 and k is odd, the expression is postive
Case: x<-1 and k is odd so we have \(x^{k}\) negative and \((x-1)\) also negative..so the expression will be positive.

Ans is C
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Is x^(k+1) > x^k? 20 Sep 2018, 19:45
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alphonsa
Is x^(k+1) > x^k ?

(1) |x| > 1

(2) k+1 is divisible by 6

source: 4gmat
Show more

Bunuel, I am not understanding why the answer is C.
Here is my reasoning:
1.
X can be positive or negative, x is not equal to zero. X can be any number except 0 and greater than 1 and less than -1
Insufficient

2. K+1 is even so K is odd.
If X is 0 , x^(k+1) = x^k=0
If X is +ve,x^(k+1) > x^k
If X is negative, x^(k+1) > x^k
Also, there is condition for fractions
Insufficient

1+2
If X is +ve,x^(k+1) > x^k
If X is negative, x^(k+1) > x^k
and for fractions the statement changes.

Ans: E
I do not know why we are not considering fractions
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Is x^(k+1) > x^k? 20 Sep 2018, 20:34
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alphonsa
Is x^(k+1) > x^k ?

(1) |x| > 1

(2) k+1 is divisible by 6

source: 4gmat

Bunuel, I am not understanding why the answer is C.
Here is my reasoning:
1.
X can be positive or negative, x is not equal to zero. X can be any number except 0 and greater than 1 and less than -1
Insufficient

2. K+1 is even so K is odd.
If X is 0 , x^(k+1) = x^k=0
If X is +ve,x^(k+1) > x^k
If X is negative, x^(k+1) > x^k
Also, there is condition for fractions
Insufficient

1+2
If X is +ve,x^(k+1) > x^k
If X is negative, x^(k+1) > x^k
and for fractions the statement changes.

Ans: E
I do not know why we are not considering fractions
Show more

Not clear how you concluded the red part there.

This following post shows why the answer is C: answerhttps://gmatclub.com/forum/is-x-k ... l#p1394406
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Is x^(k+1) > x^k? 20 Sep 2018, 23:10
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alphonsa
Is x^(k+1) > x^k ?

(1) |x| > 1

(2) k+1 is divisible by 6

source: 4gmat
Show more

Simplifying the prompt we get,

\(x^k*(x - 1) > 0\)

Hence , we have 3 cases
i) x > 0, k - odd/even, then the given Expression is > 0
ii) x < 0, k - odd, then the given Expression is > 0
iii)x < 0, k - even, then the given Expression is < 0


Statement 1 - |x| > 1, hence x < -1 or x > 1

However, No information about k

Statement 1 alone is Not Sufficient.


Statement 2: (k + 1) = 6p, hence k = 6p - 1 = (even) - 1.
k = odd

However, No information about x

Statement 2 alone is Not Sufficient.


Combining both, we get x > 1 or x < -1 & k = odd

We can see from the 3 cases above, the given conditions satisfy cases i) & ii).

Hence the given expression > 0

Combining is Sufficient.


Hence Answer is C.



Thanks,
GyM
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