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

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.