I have edited my question for the sake of clarity now what do you say?

Trying every combination. Answer is B. Unless the question was misprinted.

Posted from my mobile device

@Saurabh03121992

I think answer could be (D) if we could rephrase question as :

x>1;y>0

Quantity A	Quantity B
\(y^x\)	\(y^{(x+1)}\)

As y^x is positive so dividing both Quantity A and Quantity B by y^x we get

Quantity A	Quantity B
1	y

as y>0 so
if y = 1/2 => Quantity A > Quantity B
if y = 1 => Quantity A = Quantity B
if y = 2 => Quantity A < Quantity B

=> answer is (D)

Wasn't that the erroneous version u posted prior to correction ?

Posted from my mobile device

I have gone through answer explanation for this qustion from Quantitative Test01 of Kaplan Test Software,it seems to be a printing mistake to make the question look like what I have posted in my edited first post under this topic

Answer is B
nothing is there to think about it....addition of any positive integer always increases the value irrespective of what it is added to.

Answer will not be D


If col B is (y^x)+1 then answer is B

If col B is y^(x+1) then answer is D

Please refer to this post qq-how-to-post-a-gre-question-the-easy-way-2357.html on how to post properly a question here on the board.

Back to the question: IF it is right in its last version, clearly the answer is D

Regards

@Carcass

If last version is correct , why will it be D, it should be B as subtracting y^x from both sides LHS would be 0 and RHS would be 1 as it is clearly mentioned in the question y>0
if that wouldn't have been mentioned then it would be D.

Your response does not provide clarity. How is the answer to this D? All combinations to the initial question suggest that adding 1 to Qty A will make it larger so confirm why the answer to this is D.
Thank you.

Put y=1 and x=1

Qty A: \(y^x=1^1=1\)

Qty B: \(y^{x+1}y=1^2=1\)

Option C seems correct.

Now put y=2 and x=1

Qty A: \(y^x=2^1=2\)

Qty B: \(y^{x+1}y=2^3=8\)

Here option B seems to be correct.

Now since both cannot be correct thus option D is the best fit.

Thanks for the clarity, @sandy.

But the thing is X can never be equal to 1 based on the restrictions given in the problem (i.e. x > 1 and y > 0).

If we do X = 2 and Y = 1:
Qty A: \(y^x=1^2=1\)
Qty B: \((y^x) + 1 = (1^2) + 1 = 2\)

Here the answer is B, but lets keep trying another scenario.

If we do X = 2 and Y = 0.5:
Qty A: \(y^x=0.5^2=0.25\)
Qty B: \((y^x) + 1 = (0.5^2) + 1 = 1.25\)

Here the answer is still B.

If we do X = 2.5 and Y = 0.5:
Qty A: \(y^x=0.5^2.5=0.17677\)
Qty B: \((y^x) + 1 = (0.5^2.5) + 1 = 1.17677\)

Thus the answer is still B. I am really not sure how D could even be answer when Qty B is always greater.

A = Y^x
B = Y(Y^x)
Divide both by Y^x
A = 1
B = Y
Y could be any value so answer is D

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