Is a positive? (1) a^6

Question

Is a positive?

(1) a^6 < a

(2) a^5 < a

chetan2u · Answer

Is a positive? (1) a^6 < a a-a^6>0.......a(1-a^5)>0 So two cases a) a>0, then 1-a^5>0 or 1>a^5 so 01 or a>1........NOT possible So a>0 as it cannot be 0 or <0 Sufficient (2) a^5 < a a-a^5>0.......a(1-a^4)>0 So two cases a) a>0, then 1-a^4>0 or 1>a^4 so 01...... a^2>1 .....thus either a>1 or a<-1.......so a<-1 So a>0 or a<0 Instead A

PKN · Answer

St1:-  a^6 < a 
Or,  a^6 -a < 0 
Or,  a(a^5-1)<0 
Or, a>0 
Sufficient.

St2:-  a^5 < a 
Or, a^5-a<0 
Or,  a\left(a+1\right)\left(a-1\right)\left(a^2+1\right)<0 
Or, a<-1 or, 0<a<1
Insufficient.

anant327 · Answer

Hello,

I am confused. According to me it should be E. Lets forget the algebra for a while and lets substitute a = 0.5 or a = -0.5

Statement 1

a^6 < a. For either of the values of a (0.5 or -0.5), the answer will be 0.015625 which is lesser than a.

Statement 2

a^5 < a. For 0.5, the answer will be 0.03125 and for -0.5 the answer is -0.03125. The answer is still lesser than a.

If 1 + 2, then also the answer is either +ve or -ve. I am confused as to how the answer is A? Please clarify

chetan2u · Answer

In statement I, when a=-0.5
a^6=0.15... But a is -0.5 so a^6 >a

anant327 · Answer

Ohhhhhh silly mistake. Thank you so much!!!!

vasuca10 · Answer

anant327
Option A alone is sufficient
As per Statement 1
a( a^5 -1 ) <0
Now when you plot on number line marking + and - as alternative signs
The range would come between 0 -1

As per Statement 2
a( a^4-1)<0
a( a^2+1) ( a^2-1) <0
a( a^2+1)(a+1)(a-1)<0
Now when you plot on number line marking + and - as alternative signs
The range would come between 0 -1 and less than -1
So a can be in 0-1 range and can be less than -1 also
Hence B alone is insufficient

Kindly give kudos if my explanation is helpful

fskilnik · Answer

A\,\,\mathop > \limits^? \,\,0

\left( 1 ight)\,\,\underline {{A^6} < A}

\left. \begin{gathered}
 A = 0\,\,\,\, \Rightarrow \,\,\,0 = \underline {{A^6} < A} = 0\,\,\,\, \Rightarrow \,\,\,{	ext{impossible}}\,\,\,\, \hfill \
 A < 0\,\,\,\, \Rightarrow \,\,\,0 < \underline {{A^6} < A} < 0\,\,\,\, \Rightarrow \,\,\,{	ext{impossible}} \hfill \ 
\end{gathered} ight\}\,\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{	ext{YES}}} ightangle

\left( 2 ight)\,{A^5} < A\,\,\,\left\{ \begin{gathered}
 \,Take\,\,A = \frac{1}{2}\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{	ext{YES}}} ightangle \,\,\,\, \hfill \
 Take\,\,A = - 2\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{	ext{NO}}} ightangle \hfill \ 
\end{gathered} ight.\,\,

The above follows the notations and rationale taught in the GMATH method.

Archit3110 · Answer

test the statements with value of a = -1/2 and +1/2
#1
a^6 < a
sufficeint
only when a=1/2 sufficeint
#2
a^5 < a
suffficient when a=+/-1/2
so insufficient
IMO A

CEdward · Answer

chetan2u  for statement 2, when you take the square root of a^2 and 1, shouldn't we only get position 1? -1 is not permissible on the GMAT?

chetan2u · Answer

The square root of a would mean a is positive but square of a will have two values. 
 \sqrt{a} =1 will mean a is positive but a^2=1 can give a as both 1 and -1. 
So if you take square root of a^2... \sqrt{a^2} , a^2 will be positive but a can be negative or positive

TarunKumar1234 · Answer

Is a positive? Stat1: a^6 < a a(a^5-1) <0 So, either a= +ve then, (a^5-1)= -ve, in that case, 00 or a<0 Insufficient So, It is A.

Is a positive? (1) a^6 < a (2) a^5 < a

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