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Bunuel
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scabr
(1) х > 0
if х = 10^(1/9) (approx.1.29) than 10х = х^10 and NO
if 0 < х < 10^(1/9) than YES
If х > 10^(1/9) and x <0 than NO
Unsufficient

(2) -1 < х < 1
Unsufficient, see (1)

(1) + (2) sufficient, see (1)

Answer D

scabr Answer D means that both statements are sufficient to answer the question
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lets examine two stm seperately-
stm 1 x is a +ve number . it measn if x=1 then yes x#1 then its false so stm1 is not suufic.

stm2- it says that x2 is less then 1 it means that x is rational no .

Combine both stm we deduce that x is a positive but rational no hence sufficent ...Its C
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The answer has to be C
The above equality will hold for x=1 but not for higher value of X say 2,3 4 etc so statement 1 insufficient
Statement is also not sufficient as for 0 the the inequality will not hold
So both of them taken together gives x=1
Thus C is the answer
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Bunuel
Is 10x > x^10 ?

(1) x is a positive number.
(2) x^2 – 1 < 0

(1) x is a positive number.

Let x = 1/2.........5 > (1/2)^10.............Answer is Yes

Let x = 2...........20 > (2)^10..............Answer is No

Insufficient

(2) x^2 – 1 < 0

x^2 < 1

-1 < x < 1

Let x = 1/2.........5 > (1/2)^10.............Answer is Yes

Let x =-1/2.........-5 > (-1/2)^10.............Answer is No

Insufficient

Combine 1 & 2

From Statement 1, x > 0

From Statement 2, -1 < x < 1

We can conclude that 0 < x < 1 ...............our example of 1/2 gives clear answer.

Answer: C
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