rahulvsd wrote:Is x greater than 1?

(1) 1/{x} > -1
(2) 1/{x^5} > 1/{x^3}

1.1/x > -1
Let x = 2, 1/x = 0.5 > -1 => Yes
Let x = -2, 1/x = -0.5 > -1 => No
INSUFF

2.1/x^5 > 1/x^3
We have to consider 4 cases.

+ve int => 1/+ve integers will give us fractions and that raised to higher power will yield lower value - does not satisfy the inequality
x = 5, 0.00032 < 0.008

+ve fraction => 1/+ve fractions will give us integer values and that raised to higher power will yield higher values - satisfies the inequality
x = 0.5, 32 > 8

-ve int => 1/-ve integers will give us -ve fractions and that raised to higher power will yield higher value - satisfies the inequality
x = -5, -0.00032 > -0.008

-ve fraction => 1/-ve fractions will give us numerical values and that raised to higher power will yield lower value - does not satisfy the inequality
x = -0.5, -32 < -8

So we can have + ve fractions and -ve integers both are not greater than 1. So ans is No, B is SUFF. We can try with numbers as well. When sign is not given consider both +ve and -ve values, when there is no mention of integers consider fractions in your number cases. That will help us solve it easily. Hope the solution is right.

IMO B.

rahulvsd wrote:Is x greater than 1?

(1) 1/{x} > -1
(2) 1/{x^5} > 1/{x^3}

A. Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient
B. Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient
D. EACH statement ALONE is sufficient
E. Statements (1) and (2) TOGETHER are NOT sufficient

Stat. 1...

Given is 1/x > -1;

Pl. put x = 1/2 => It satisfies 1/x > -1, and x < 1. Ans. No.

Pl. put x = 2 => It satisfies 1/x > -1, but x > 1, but ans. Yes. Not sufficient.

Stat. 2...

Given is 1/x^5 > 1/x^3;

Pl. put x = 1/2 => It satisfies 1/x^5 > 1/x^3, and x < 1. Ans No.

Pl. put x = -2 => It satisfies 1/x^5 > 1/x^3, and x < 1. Ans No.

No value of x greater than 1 will fit into 1/x^5 > 1/x^3. Sufficient.

IMO B.

rahulvsd wrote:Is x greater than 1?

(1) 1/{x} > -1
(2) 1/{x^5} > 1/{x^3}

The CRITICAL POINTS of each inequality are where the left-hand side is equal to the right-hand side or where the inequality is undefined.
To determine the ranges of x that satisfy each inequality, test one value to the left and right of each critical point.

Algebra often is necessary to determine the critical points but is not needed here.

Statement 1: 1/x > -1
The critical points are:
x=-1 (since 1/-1 = -1)
x=0 (since 1/0 is undefined).

x>0:
If x=1, then 1/x > -1.
Thus, x>0 is part of the range.
-1<x<0:
If x=-1/2, then it is not true 1/x > -1.
Thus, -1<x<0 is NOT part of the range.
x<-1:
If x=-2, then 1/x > -1.
Thus, x<-1 is part of the range.

Since it's possible that x>0 or that x<-1, we cannot determine whether x>1.
INSUFFICIENT.

Statement 2: 1/(x^5) > 1/(x^3)
The critical points are:
x=-1 [since 1/(-1^5) = 1/(-1^3)]
x=0 [since 1/(0^5) is undefined]
x=1 [since 1/(1^5) = 1/(1^3)]

x>1:
If x=2, then it is not true that 1/(x^5) > 1/(x^3).
Thus, x>1 is NOT part of the range.

No need to test other values; we know that x is not greater than 1.
SUFFICIENT.

The correct answer is B.

GMATGuruNY wrote:

rahulvsd wrote:Is x greater than 1?

(1) 1/{x} > -1
(2) 1/{x^5} > 1/{x^3}

The CRITICAL POINTS of each inequality are where the left-hand side is equal to the right-hand side or where the inequality is undefined.
To determine the ranges of x that satisfy each inequality, test one value to the left and right of each critical point.

Algebra often is necessary to determine the critical points but is not needed here.

Statement 1: 1/x > -1
The critical points are:
x=-1 (since 1/-1 = -1)
x=0 (since 1/0 is undefined).

x>0:
If x=1, then 1/x > -1.
Thus, x>0 is part of the range.
-1<x<0:
If x=-1/2, then it is not true 1/x > -1.
Thus, -1<x<0 is NOT part of the range.
x<-1:
If x=-2, then 1/x > -1.
Thus, x<-1 is part of the range.

Since it's possible that x>0 or that x<-1, we cannot determine whether x>1.
INSUFFICIENT.

Statement 2: 1/(x^5) > 1/(x^3)
The critical points are:
x=-1 [since 1/(-1^5) = 1/(-1^3)]
x=0 [since 1/(0^5) is undefined]
x=1 [since 1/(1^5) = 1/(1^3)]

x>1:
If x=2, then it is not true that 1/(x^5) > 1/(x^3).
Thus, x>1 is NOT part of the range.

No need to test other values; we know that x is not greater than 1.
SUFFICIENT.

The correct answer is B.

Sir,

May I write the statements like?

1. 1/x > -1 to a. x < -1 as If 1/a > 1/b then a < b if a, b not equal to o.

b. x > 1; simply changing the sign.

2. 1/x^5 > 1/x^3 to a. x^5 < x^3 and can i bring it to x^2 < 1 or 1/x^2 >1?

spider wrote: Sir,

May I write the statements like?

1. 1/x > -1 to a. x < -1 as If 1/a > 1/b then a < b if a, b not equal to o.

b. x > 1; simply changing the sign.

2. 1/x^5 > 1/x^3 to a. x^5 < x^3 and can i bring it to x^2 < 1 or 1/x^2 >1?

Let me be the first user to thank you - I did it accidentally

1/a > 1/b then a < b, 1/x^5 > 1/x^3 to x^5 < x^3 - Both these are right only if you know the signs or else you should consider 2 cases for each inequality. But that will be a waste of time since you have to consider all the cases anyhow if we don't know the signs.

sam2304 wrote:

spider wrote: Sir,

May I write the statements like?

1. 1/x > -1 to a. x < -1 as If 1/a > 1/b then a < b if a, b not equal to o.

b. x > 1; simply changing the sign.

2. 1/x^5 > 1/x^3 to a. x^5 < x^3 and can i bring it to x^2 < 1 or 1/x^2 >1?

Let me be the first user to thank you - I did it accidentally

1/a > 1/b then a < b, 1/x^5 > 1/x^3 to x^5 < x^3 - Both these are right only if you know the signs or else you should consider 2 cases for each inequality. But that will be a waste of time since you have to consider all the cases anyhow if we don't know the signs.

Not with reference to this question, May I write 1. 1/x > -1 to x < -1 or x > 1 ?

2. 1/x^5 > 1/x^3 to x^2 < 1 or 1/x^2 >1?