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Re: y^x compared with (y^x+1) [#permalink]
I have edited my question for the sake of clarity now what do you say?
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Re: y^x compared with (y^x+1) [#permalink]
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Trying every combination. Answer is B. Unless the question was misprinted.

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Re: y^x compared with (y^x+1) [#permalink]
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@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)

Originally posted by yasir9909 on 30 Aug 2016, 07:12.
Last edited by yasir9909 on 30 Aug 2016, 07:18, edited 1 time in total.
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Re: y^x compared with (y^x+1) [#permalink]
Wasn't that the erroneous version u posted prior to correction ?

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Re: y^x compared with (y^x+1) [#permalink]
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
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Re: y^x compared with (y^x+1) [#permalink]
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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
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Re: y^x compared with (y^x+1) [#permalink]
Expert Reply
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
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Re: y^x compared with (y^x+1) [#permalink]
@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.
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Re: y^x compared with (y^x+1) [#permalink]
Carcass wrote:
Please refer to this post http://gre.myprepclub.com/forum/qq-how-to- ... -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

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.
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Re: y^x compared with (y^x+1) [#permalink]
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Expert Reply
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.
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Re: y^x compared with (y^x+1) [#permalink]
sandy wrote:
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.
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Re: y^x compared with (y^x+1) [#permalink]
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sandy wrote:
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.


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.
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Re: y^x compared with (y^x+1) [#permalink]
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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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Re: y^x compared with (y^x+1) [#permalink]
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