Not clear.
What if a=0 or a=1 => then x and y may be any numbers, positive or negative
0^20/0^(-1)=0^5 => x=-1
(0^10)^4 = 0^12 => y = 10
-1-10 = -11 < 10^2
Why the answer is A?

a cannot be zero because the fraction is = to a^5

So, the result must be an integer.

If a = 0 then also the denominator is = zero which is impossible because the fraction cannot be indefinite.

Regards

Thank you!
Yeh.
It was really stupid to divide by zero.
But in the case of a=1?

no because if a is 1 the two quantities would be 1 as well and the entire question was silly.

Is not what it means to test

I mean this kind of inference, aside from the algebra manipulations, what really the question is asking you is the element that separates the tests takers in the 99 percentile from the rest of the pack...

Regards

Giving a more detailed explantation so posting:
a) a^20/a^x = a^5
Taking denominator to right side by multiplying:
a^20 = a^5*a^x
now to get a^5 common
a^5 * a^15 = a^5 * a^x [removing a to power 5 as it is common]
we get x = 15 (A)

for right side, a^4y = [a^3]^4
removing a^4 on both sides as it is common, we get a^y = a ^3, y = 3

this per equation, x - y = 15 - 9 = 6 and y square is 81. Hence B > A. Option B is correct.

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