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Manager Joined: 13 Aug 2016
Posts: 77
GRE 1: Q158 V154 Followers: 0

Kudos [?]: 39 , given: 21 00:00

Question Stats: 20% (01:08) correct 79% (00:38) wrong based on 24 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 Manager Joined: 13 Aug 2016
Posts: 77
GRE 1: Q158 V154 Followers: 0

Kudos [?]: 39  , given: 21

Re: y^x compared with (y^x+1) [#permalink]
1
KUDOS
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. Intern Joined: 02 Jul 2016
Posts: 25
Followers: 0

Kudos [?]: 15  , given: 2

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

Posted from my mobile device Manager Joined: 13 Aug 2016
Posts: 77
GRE 1: Q158 V154 Followers: 0

Kudos [?]: 39 , given: 21

Re: y^x compared with (y^x+1) [#permalink]
I have edited my question for the sake of clarity now what do you say? Intern Joined: 02 Jul 2016
Posts: 25
Followers: 0

Kudos [?]: 15  , given: 2

Re: y^x compared with (y^x+1) [#permalink]
1
KUDOS
Trying every combination. Answer is B. Unless the question was misprinted.

Posted from my mobile device  Manager Joined: 13 Aug 2016
Posts: 77
GRE 1: Q158 V154 Followers: 0

Kudos [?]: 39  , given: 21

Re: y^x compared with (y^x+1) [#permalink]
1
KUDOS
@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.
Intern Joined: 02 Jul 2016
Posts: 25
Followers: 0

Kudos [?]: 15 , given: 2

Re: y^x compared with (y^x+1) [#permalink]
Wasn't that the erroneous version u posted prior to correction ?

Posted from my mobile device Manager Joined: 13 Aug 2016
Posts: 77
GRE 1: Q158 V154 Followers: 0

Kudos [?]: 39 , given: 21

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 Manager Joined: 12 Jan 2016
Posts: 144
Followers: 1

Kudos [?]: 73  , given: 17

Re: y^x compared with (y^x+1) [#permalink]
1
KUDOS
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
Founder  Joined: 18 Apr 2015
Posts: 7401
Followers: 125

Kudos [?]: 1449 , given: 6611

Re: y^x compared with (y^x+1) [#permalink]
Expert's post
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
_________________
Manager Joined: 12 Jan 2016
Posts: 144
Followers: 1

Kudos [?]: 73 , given: 17

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.
Intern Joined: 08 Apr 2018
Posts: 44
Followers: 0

Kudos [?]: 15 , given: 16

Re: y^x compared with (y^x+1) [#permalink]
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. GRE Prep Club Legend  Joined: 07 Jun 2014
Posts: 4809
GRE 1: Q167 V156 WE: Business Development (Energy and Utilities)
Followers: 123

Kudos [?]: 1972  , given: 397

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

Sandy
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Intern Joined: 08 Apr 2018
Posts: 44
Followers: 0

Kudos [?]: 15 , given: 16

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. Intern Joined: 09 Jul 2018
Posts: 10
Followers: 0

Kudos [?]: 7  , given: 0

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