NEW HERE? LEARN MORE
closeclose
Last visit was: 09 May 2024, 22:57 It is currently 09 May 2024, 22:57
Decision Tracker
My Rewards
New posts
New comers' posts

Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel

M15-05

SORT BY:
Date
Kudos
Tags:
Find Similar Topics 
Show Tags
Hide Tags
User avatar
L
Bunuel
Math Expert
Joined: 02 Sep 2009
Posts: 93124
Own Kudos [?]: 622521 [16]
Given Kudos: 81810
Send PM
CAT Tests Forum Quiz Mode
M15-05 [#permalink] New post   16 Sep 2014, 00:55
1
Kudos
15
Bookmarks
Expert Reply
00:00
A
B
C
D
E

Difficulty:

45% (medium)

Question Stats:

71% (02:04) correct 29% (02:16) wrong based on 140 sessions
Hide Show timer Statistics
For any pair of positive integers \(x\) and \(y\), their arithmetic mean \(A(x, y)\) is defined as \(\frac{x + y}{2}\) while their geometric mean \(G(x, y)\) is defined as \(\sqrt{xy}\). If \(m\) and \(n\) are positive integers, is \(m > n\)?



(1) \(A(G(m, n), m) = m\)

(2) \(A(m, n) - G(m, n) = 0\)
ShowHide Answer
Official Answer
D
User avatar
L
Bunuel
Math Expert
Joined: 02 Sep 2009
Posts: 93124
Own Kudos [?]: 622521 [3]
Given Kudos: 81810
Send PM
CAT Tests Forum Quiz Mode
Re M15-05 [#permalink] New post   16 Sep 2014, 00:55
2
Kudos
1
Bookmarks
Expert Reply
Official Solution:


For any pair of positive integers \(x\) and \(y\), their arithmetic mean \(A(x, y)\) is defined as \(\frac{x + y}{2}\) while their geometric mean \(G(x, y)\) is defined as \(\sqrt{xy}\). If \(m\) and \(n\) are positive integers, is \(m > n\)?

(1) \(A(G(m, n), m) = m\).

\(A(\sqrt{mn}, m) = m\);

\(\frac{\sqrt{mn} + m}{2}= m\);

\(\sqrt{mn} + m= 2m\);

\(\sqrt{mn} = m\);

\(mn = m^2\);

Since \(m\) is positive we can safely reduce the above by it:

\(n = m\)

Therefore, the answer to the question is NO. Sufficient.

(2) \(A(m, n) - G(m, n) = 0\).

\(A(m, n) - G(m, n) =0\);

\(\frac{m + n}{2}-\sqrt{mn}=0\);

\(m + n-2\sqrt{mn}=0\);

\(m -2\sqrt{mn}+ n=0\);

Recognize that the above is the square of the difference:

\((\sqrt{m} - \sqrt{n})^2=0\);

\(\sqrt{m} = \sqrt{n}\);

\(m=n\)

Therefore, the answer to the question is NO. Sufficient.


Answer: D
User avatar
L
Bunuel
Math Expert
Joined: 02 Sep 2009
Posts: 93124
Own Kudos [?]: 622521 [1]
Given Kudos: 81810
Send PM
CAT Tests Forum Quiz Mode
Re: M15-05 [#permalink] New post   11 Mar 2023, 02:43
1
Kudos
Expert Reply
joe123x wrote:
I'm not sure I'm recognizing the square of a difference in statement 2, any hints?


\(m -2\sqrt{mn}+ n\) is very similar to \(a^2 -2ab+ b^2=(a-b)^2\), no? If you consider \(\sqrt{m}=a\) and \(\sqrt{n}=b\), those are basically the same expressions.
User avatar
B
KungFuGmat
Manager
Manager
Joined: 15 Sep 2016
Posts: 61
Own Kudos [?]: 26 [0]
Given Kudos: 65
Location: Pakistan
Concentration: Finance, Technology
Schools: CBS '20
GMAT 1: 640 Q43 V35
Send PM
Reviews Badge
Re: M15-05 [#permalink] New post   24 Nov 2017, 22:44
This is a high quality question. Thanks Bunuel for the amazing solution, it took me quite sometime to see how statement 2 is sufficient.
User avatar
L
Bunuel
Math Expert
Joined: 02 Sep 2009
Posts: 93124
Own Kudos [?]: 622521 [0]
Given Kudos: 81810
Send PM
CAT Tests Forum Quiz Mode
Re: M15-05 [#permalink] New post   10 Mar 2023, 13:08
Expert Reply
I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
User avatar
B
joe123x
Manager
Manager
Joined: 03 Oct 2022
Posts: 89
Own Kudos [?]: 12 [0]
Given Kudos: 53
GMAT 1: 610 Q40 V34
M15-05 [#permalink] New post   11 Mar 2023, 02:38
I'm not sure I'm recognizing the square of a difference in statement 2, any hints?

oh, never mind. just noticed it... gosh.
User avatar
B
joe123x
Manager
Manager
Joined: 03 Oct 2022
Posts: 89
Own Kudos [?]: 12 [0]
Given Kudos: 53
GMAT 1: 610 Q40 V34
Re: M15-05 [#permalink] New post   11 Mar 2023, 02:45
Bunuel wrote:
joe123x wrote:
I'm not sure I'm recognizing the square of a difference in statement 2, any hints?


\(m -2\sqrt{mn}+ n\) is very similar to \(a^2 -2ab+ b^2=(a-b)^2\), no? If you consider \(\sqrt{m}=a\) and \(\sqrt{n}=b\), those are basically the same expressions.


Yep, just noticed that. I wasn't able to see it right away, do you have any recommendations on how to improve this skill? PRACTICE!!!
User avatar
L
Bunuel
Math Expert
Joined: 02 Sep 2009
Posts: 93124
Own Kudos [?]: 622521 [0]
Given Kudos: 81810
Send PM
CAT Tests Forum Quiz Mode
Re: M15-05 [#permalink] New post   11 Mar 2023, 02:50
Expert Reply
joe123x wrote:
Bunuel wrote:
joe123x wrote:
I'm not sure I'm recognizing the square of a difference in statement 2, any hints?


\(m -2\sqrt{mn}+ n\) is very similar to \(a^2 -2ab+ b^2=(a-b)^2\), no? If you consider \(\sqrt{m}=a\) and \(\sqrt{n}=b\), those are basically the same expressions.


Yep, just noticed that. I wasn't able to see it right away, do you have any recommendations on how to improve this skill? PRACTICE!!!


Consistent practice is key. Therefore, keep practicing regularly to improve.
User avatar
S
Aishyk97
Manager
Manager
Joined: 13 Sep 2021
Posts: 92
Own Kudos [?]: 23 [0]
Given Kudos: 77
Send PM
Re: M15-05 [#permalink] New post   04 May 2023, 08:14
Bunuel wrote:
Official Solution:


For any pair of positive integers \(x\) and \(y\), their arithmetic mean \(A(x, y)\) is defined as \(\frac{x + y}{2}\) while their geometric mean \(G(x, y)\) is defined as \(\sqrt{xy}\). If \(m\) and \(n\) are positive integers, is \(m > n\)?

(1) \(A(G(m, n), m) = m\).

\(A(\sqrt{mn}, m) = m\);

\(\frac{\sqrt{mn} + m}{2}= m\);

\(\sqrt{mn} + m= 2m\);

\(\sqrt{mn} = m\);

\(mn = m^2\);

Since \(m\) is positive we can safely reduce the above by it:

\(n = m\)

Therefore, the answer to the question is NO. Sufficient.

(2) \(A(m, n) - G(m, n) = 0\).

\(A(m, n) - G(m, n) =0\);

\(\frac{m + n}{2}-\sqrt{mn}=0\);

\(m + n-2\sqrt{mn}=0\);

\(m -2\sqrt{mn}+ n=0\);

Recognize that the above is the square of the difference:

\((\sqrt{m} - \sqrt{n})^2=0\);

\(\sqrt{m} = \sqrt{n}\);

\(m=n\)

Therefore, the answer to the question is NO. Sufficient.


Answer: D


Hi Bunuel, nice solution.
In statement 1, I did not reduce m.
I took it as msqaure (m-n) = 0
So m = 0 or m-n = 0

How do we know we should cancel m on either side?
Usually with quadratics, we do not reduce a variable as we will lose 1 root. At least according to my understanding.

Please help
User avatar
L
Bunuel
Math Expert
Joined: 02 Sep 2009
Posts: 93124
Own Kudos [?]: 622521 [0]
Given Kudos: 81810
Send PM
CAT Tests Forum Quiz Mode
Re: M15-05 [#permalink] New post   04 May 2023, 08:18
Expert Reply
Aishyk97 wrote:
Bunuel wrote:
Official Solution:


For any pair of positive integers \(x\) and \(y\), their arithmetic mean \(A(x, y)\) is defined as \(\frac{x + y}{2}\) while their geometric mean \(G(x, y)\) is defined as \(\sqrt{xy}\). If \(m\) and \(n\) are positive integers, is \(m > n\)?

(1) \(A(G(m, n), m) = m\).

\(A(\sqrt{mn}, m) = m\);

\(\frac{\sqrt{mn} + m}{2}= m\);

\(\sqrt{mn} + m= 2m\);

\(\sqrt{mn} = m\);

\(mn = m^2\);

Since \(m\) is positive we can safely reduce the above by it:

\(n = m\)

Therefore, the answer to the question is NO. Sufficient.

(2) \(A(m, n) - G(m, n) = 0\).

\(A(m, n) - G(m, n) =0\);

\(\frac{m + n}{2}-\sqrt{mn}=0\);

\(m + n-2\sqrt{mn}=0\);

\(m -2\sqrt{mn}+ n=0\);

Recognize that the above is the square of the difference:

\((\sqrt{m} - \sqrt{n})^2=0\);

\(\sqrt{m} = \sqrt{n}\);

\(m=n\)

Therefore, the answer to the question is NO. Sufficient.


Answer: D


Hi Bunuel, nice solution.
In statement 1, I did not reduce m.
I took it as msqaure (m-n) = 0
So m = 0 or m-n = 0

How do we know we should cancel m on either side?
Usually with quadratics, we do not reduce a variable as we will lose 1 root. At least according to my understanding.

Please help


We are given that m is a positive integer, so m cannot be 0, and thus we can reduce by it.
User avatar
P
BottomJee
DI Forum Moderator
Joined: 05 May 2019
Status:GMAT Club Team member
Affiliations: GMAT Club
Posts: 1030
Own Kudos [?]: 654 [0]
Given Kudos: 1003
Location: India
GMAT Focus 1:
645 Q82 V81 DI82
GMAT 1: 430 Q31 V19
GMAT 2: 570 Q44 V25
GMAT 3: 660 Q48 V33
GPA: 3.26
WE:Engineering (Manufacturing)
Send PM
Reviews Badge CAT Tests Forum Quiz Mode
Re M15-05 [#permalink] New post   12 Aug 2023, 10:05
I think this is a high-quality question and I agree with explanation.
User avatar
P
ChandlerBong
Senior Manager
Senior Manager
Joined: 16 Jan 2022
Posts: 251
Own Kudos [?]: 420 [0]
Given Kudos: 1013
Location: India
GPA: 4
WE:Analyst (Computer Software)
Send PM
Forum Quiz Mode
Re M15-05 [#permalink] New post   27 Oct 2023, 22:13
I think this is a high-quality question.
GMAT Club Bot
gmatclubot
Re M15-05 [#permalink]
27 Oct 2023, 22:13
Moderator:
Bunuel
Math Expert
93124 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne
Main navigation
GMAT Resources
Partners
MBA Resources
www.gmac.com|www.mba.com | | GMAT Club Rules | Terms and Conditions