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# y^x compared with (y^x+1)

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y^x compared with (y^x+1) [#permalink]  30 Aug 2016, 06:12
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Question Stats:

23% (01:03) correct 76% (00:40) wrong based on 30 sessions
$$x>1$$ ; $$y>0$$

 Quantity A Quantity B $$y^x$$ ($$y^x$$)+1

A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
[Reveal] Spoiler: OA

Last edited by Carcass on 28 May 2018, 11:20, edited 5 times in total.
Edited by Carcass
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Re: y^x compared with (y^x+1) [#permalink]  30 Aug 2016, 06:17
1
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I solved it this way but my answer turned out to be wrong,

Since y^x is positive and greater than 0 so we can subtract y^x from both quantities

 Quantity A Quantity B $$y^x$$ ($$y^x$$)+1

Subtracting y^x from both Quantity A and Quantity B we get

 Quantity A Quantity B 0 1

=>

Quantity B is greater than Quantity A, so answer should be (B) but OA is given to be (D)

I need to know why my approach to solving this problem is wrong

Last edited by yasir9909 on 30 Aug 2016, 07:03, edited 1 time in total.
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Re: y^x compared with (y^x+1) [#permalink]  30 Aug 2016, 06:55
1
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@Yasir my friend, hope your well. The answer is a clear D. Take y as 0.5 and x as 2. A is .25 and B is .125. now take y as 3 and maintain x as 2. A is 9 and and B is 27. So it's a D.

Hope this helps. Cheers !!

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Re: y^x compared with (y^x+1) [#permalink]  30 Aug 2016, 07:04
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]  30 Aug 2016, 07:10
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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]  30 Aug 2016, 07:12
1
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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

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]  30 Aug 2016, 07:16
Wasn't that the erroneous version u posted prior to correction ?

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Re: y^x compared with (y^x+1) [#permalink]  30 Aug 2016, 07:20
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]  30 Aug 2016, 10:49
1
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nothing is there to think about it....addition of any positive integer always increases the value irrespective of what it is added to.

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]  31 Aug 2016, 00:24
Expert's post
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Re: y^x compared with (y^x+1) [#permalink]  31 Aug 2016, 03:49
@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]  27 May 2018, 22:05
Carcass wrote:
Please refer to this post http://greprepclub.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]  27 May 2018, 22:51
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Expert's post
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]  29 May 2018, 19:16
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]  09 Jul 2018, 18:11
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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.
Re: y^x compared with (y^x+1)   [#permalink] 09 Jul 2018, 18:11
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