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 0<a<1
b) a<0, then 1-a^5<0 or a^5>1 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 0<a<1
b) a<0, then 1-a^4<0 or a^4>1......\(a^2>1\).....thus either a>1 or a<-1.......so a<-1
So a>0 or a<0
Instead

A

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.
_________________

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

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

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

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

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

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

\(\left. \begin{gathered}
A = 0\,\,\,\, \Rightarrow \,\,\,0 = \underline {{A^6} < A} = 0\,\,\,\, \Rightarrow \,\,\,{\text{impossible}}\,\,\,\, \hfill \\
A < 0\,\,\,\, \Rightarrow \,\,\,0 < \underline {{A^6} < A} < 0\,\,\,\, \Rightarrow \,\,\,{\text{impossible}} \hfill \\
\end{gathered} \right\}\,\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\text{YES}}} \right\rangle\)


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

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

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

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?
_________________

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

Is a positive?

Stat1: a^6 < a
a(a^5-1) <0 So, either a= +ve then, (a^5-1)= -ve, in that case, 0<a<1
another case, a= -ve then, (a^5-1)= +ve = NOT POSSIBLE. So, we have 0<a<1. Sufficient

Stat2: a^5 < a
a(a^4-1) <0 So, either a= +ve then, (a^4-1)= -ve, in that case, 0<a<1
another case, a= -ve then, (a^4-1)= +ve = POSSIBLE. So, we have a>0 or a<0 Insufficient

So, It is A.