GMAT Question of the Day - Daily to your Mailbox; hard ones only

It is currently 22 Oct 2019, 06:56

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.

Close

Request Expert Reply

Confirm Cancel

M31-53

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Find Similar Topics 
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 58430
M31-53  [#permalink]

Show Tags

New post 20 Jul 2015, 05:49
00:00
A
B
C
D
E

Difficulty:

  65% (hard)

Question Stats:

46% (01:41) correct 54% (01:18) wrong based on 54 sessions

HideShow timer Statistics

Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 58430
Re M31-53  [#permalink]

Show Tags

New post 20 Jul 2015, 05:49
1
2
Official Solution:


If \(a\) and \(b\) are non-negative integers, is \(a > b\)?

(1) \(6^a = 36^b\). Simplify: \(6^a = 6^{2b}\). Bases are equal, hence we can equate the powers: \(a=2b\). If \(a=b=0\), then \(a\) is NOT greater than \(b\) but if \(a=2\) and \(b=1\), then \(a\) IS greater than \(b\). Not sufficient.

(2) \(5^a = 35^b\). If both \(a\) and \(b\) are positive integers, then we'd have that \(5^a\) is equal to some multiple of 7 (because 35=5*7), which is not possible since 5 in any positive integer power has only 5's in it. Therefore, both \(a\) and \(b\) must be 0, giving a NO answer to the question whether \(a\) is greater than \(b\). Sufficient,


Answer: B
_________________
Senior Manager
Senior Manager
User avatar
Joined: 31 Mar 2016
Posts: 375
Location: India
Concentration: Operations, Finance
GMAT 1: 670 Q48 V34
GPA: 3.8
WE: Operations (Commercial Banking)
GMAT ToolKit User
Re M31-53  [#permalink]

Show Tags

New post 17 Jul 2016, 03:20
I think this is a high-quality question and I agree with explanation. Great Trap on choice B! Excellent
Intern
Intern
avatar
B
Joined: 24 Aug 2013
Posts: 39
Re: M31-53  [#permalink]

Show Tags

New post 04 Jun 2017, 08:31
Very good question. I made a mistake assuming that a = 2b => a> b without considering b =0
Intern
Intern
avatar
Joined: 12 Mar 2016
Posts: 4
Re: M31-53  [#permalink]

Show Tags

New post 09 Jul 2017, 03:23
Hi Bunuel,

Since it is given a & b are positive, how can we assume a=0 from statement 1? (0 is neither -ve nor +ve)

Thanks
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 58430
Re: M31-53  [#permalink]

Show Tags

New post 09 Jul 2017, 03:29
Current Student
avatar
G
Joined: 19 Aug 2016
Posts: 146
Location: India
GMAT 1: 640 Q47 V31
GPA: 3.82
Reviews Badge
Re: M31-53  [#permalink]

Show Tags

New post 26 Aug 2017, 04:15
Bunuel wrote:
Official Solution:


If \(a\) and \(b\) are non-negative integers, is \(a > b\)?

(1) \(6^a = 36^b\). Simplify: \(6^a = 6^{2b}\). Bases are equal, hence we can equate the powers: \(a=2b\). If \(a=b=0\), then \(a\) is NOT greater than \(b\) but if \(a=2\) and \(b=1\), then \(a\) IS greater than \(b\). Not sufficient.

(2) \(5^a = 35^b\). If both \(a\) and \(b\) are positive integers, then we'd have that \(5^a\) is equal to some multiple of 7 (because 35=5*7), which is not possible since 5 in any positive integer power has only 5's in it. Therefore, both \(a\) and \(b\) must be 0, giving a NO answer to the question whether \(a\) is greater than \(b\). Sufficient,


Answer: B


Hi bunuel,

Not able to understand the statement 2 explaination. Could you explain in simpler words or with some example?

Thanks
_________________
Consider giving me Kudos if you find my posts useful, challenging and helpful!
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 58430
Re: M31-53  [#permalink]

Show Tags

New post 26 Aug 2017, 05:06
1
ashikaverma13 wrote:
Bunuel wrote:
Official Solution:


If \(a\) and \(b\) are non-negative integers, is \(a > b\)?

(1) \(6^a = 36^b\). Simplify: \(6^a = 6^{2b}\). Bases are equal, hence we can equate the powers: \(a=2b\). If \(a=b=0\), then \(a\) is NOT greater than \(b\) but if \(a=2\) and \(b=1\), then \(a\) IS greater than \(b\). Not sufficient.

(2) \(5^a = 35^b\). If both \(a\) and \(b\) are positive integers, then we'd have that \(5^a\) is equal to some multiple of 7 (because 35=5*7), which is not possible since 5 in any positive integer power has only 5's in it. Therefore, both \(a\) and \(b\) must be 0, giving a NO answer to the question whether \(a\) is greater than \(b\). Sufficient,


Answer: B


Hi bunuel,

Not able to understand the statement 2 explaination. Could you explain in simpler words or with some example?

Thanks


If a and b are not both 0, then we'd have that {not a multiple of 7} = {a multiple of 7}, which cannot be true, thus both \(a\) and \(b\) must be 0.
_________________
Intern
Intern
avatar
B
Joined: 13 Oct 2017
Posts: 38
M31-53  [#permalink]

Show Tags

New post 26 Feb 2018, 05:00
Hi Bunuel,

I worked the second statement out in the following way:

5^a=5^b*7^b

Since exponents with the same bases can be equated...I then got a=b (by equating the exponents of the two 5s)
If a=b, a cannot be greater than b...hence sufficient.

I was still uncomfortable with my working out because I wasn't sure whether ignoring the 7^b was correct...was my approach correct?

Thanks.
Intern
Intern
avatar
B
Joined: 09 Oct 2018
Posts: 6
Location: India
Schools: Tsinghua (S)
GPA: 3.9
Re M31-53  [#permalink]

Show Tags

New post 31 Oct 2018, 15:44
I think this the explanation isn't clear enough, please elaborate.
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 58430
Re: M31-53  [#permalink]

Show Tags

New post 31 Oct 2018, 22:42
Manager
Manager
User avatar
S
Joined: 28 Jun 2018
Posts: 129
Location: Bouvet Island
GMAT 1: 490 Q39 V18
GMAT 2: 640 Q47 V30
GMAT 3: 670 Q50 V31
GMAT 4: 700 Q49 V36
GPA: 4
Re: M31-53  [#permalink]

Show Tags

New post 19 Nov 2018, 09:44
A quick and important tip for such questions!

ALWAYS REMEMBER IN DS questions statement 1 can never contradict statement 2!
So even in case u made mistake of taking a > b from statement 1, while doing statement 2 u realize\(a = b = 0\).
You should immediately stop and check to see why statement 1 and 2 are contradicting.
Probably because u made some mistake.
This simple tip will help u avoid in a lot of DS questions.
Manager
Manager
avatar
S
Joined: 15 Jul 2016
Posts: 101
Location: India
Schools: Oxford "21 (A)
GMAT 1: 690 Q48 V36
CAT Tests
Re: M31-53  [#permalink]

Show Tags

New post 25 Jul 2019, 04:58
1
ashikaverma13 wrote:
Bunuel wrote:
Official Solution:


If \(a\) and \(b\) are non-negative integers, is \(a > b\)?

(1) \(6^a = 36^b\). Simplify: \(6^a = 6^{2b}\). Bases are equal, hence we can equate the powers: \(a=2b\). If \(a=b=0\), then \(a\) is NOT greater than \(b\) but if \(a=2\) and \(b=1\), then \(a\) IS greater than \(b\). Not sufficient.

(2) \(5^a = 35^b\). If both \(a\) and \(b\) are positive integers, then we'd have that \(5^a\) is equal to some multiple of 7 (because 35=5*7), which is not possible since 5 in any positive integer power has only 5's in it. Therefore, both \(a\) and \(b\) must be 0, giving a NO answer to the question whether \(a\) is greater than \(b\). Sufficient,


Answer: B


Hi bunuel,

Not able to understand the statement 2 explaination. Could you explain in simpler words or with some example?

Thanks


In order to understand St 2 better, you can consider the following approach

5^a = 35^b
=> 5^a = (7*5)^b = 7^b*5^b

Bring 5^b to LHS

=> 5^(a-b)=7^b

Now, the rule of cyclicity comes handy

No matter what the power of 7, it will never end in units digit being 5.
No matter what the power of 5, it will always end up in units digit being 5.

Hence, for LHS to be equal to RHS, the powers should be 0 so that both LHS = RHS = 1.

Now, a=b=0 therefore, we get a definite ans that a is not greater than b.
_________________
Please give Kudos if you agree with my approach :)
GMAT Club Bot
Re: M31-53   [#permalink] 25 Jul 2019, 04:58
Display posts from previous: Sort by

M31-53

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  

Moderators: chetan2u, Bunuel






Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne