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

It is currently 24 May 2019, 12:18

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

In a casino, a gambler stacks a certain number of chips in piles, with

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

Hide Tags

 
Director
Director
User avatar
Joined: 25 Aug 2007
Posts: 713
WE 1: 3.5 yrs IT
WE 2: 2.5 yrs Retail chain
In a casino, a gambler stacks a certain number of chips in piles, with  [#permalink]

Show Tags

New post 18 Jun 2010, 04:13
1
1
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

68% (02:20) correct 32% (02:26) wrong based on 33 sessions

HideShow timer Statistics

In a casino, a gambler stacks a certain number of chips in piles, with 12 chips in each pile and no chips left over. After winning 12 extra chips in a hand of poker, the gambler again stacks his chips in piles, with 14 chips in each pile and no chips left over. How many chips did the gambler have before winning the hand of poker?

(1) Before winning the hand of poker, the gambler had fewer than 140 chips.
(2) Before winning the hand of poker, the gambler had more than 70 chips.

_________________
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 55271
Re: In a casino, a gambler stacks a certain number of chips in piles, with  [#permalink]

Show Tags

New post 18 Jun 2010, 05:16
1
ykaiim wrote:
In a casino, a gambler stacks a certain number of chips in piles, with 12 chips in each pile and no chips left over. After winning 12 extra chips in a hand of poker, the gambler again stacks his chips in piles, with 14 chips in each pile and no chips left over. How many chips did the gambler have before winning the hand of poker?

1) Before winning the hand of poker, the gambler had fewer than 140 chips.
2) Before winning the hand of poker, the gambler had more than 70 chips.


Let # of piles before the winning be \(x\), so the question is \(12x=?\) Also given that \(12x+12=14y\) --> \(6(x+1)=7y\) --> \(x+1\) must be multiple of 7.

(1) \(12x<140\) --> \(x<11\frac{2}{3}\) --> the only integer value of \(x\), satisfying \(x<11\frac{2}{3}\), for which \(x+1\) is a multiple of 7 is when \(x=6\) --> \(12x=72\). Sufficient.

(2) \(12x>70\) --> \(x>5\frac{5}{6}\) --> multiple values are possible for \(12x\), for instance if \(x=6\), then \(12x=72\) but if \(x=13\), then \(12x=156\). Not sufficient.

Answer: A.

Hope it's clear.
_________________
Director
Director
User avatar
Joined: 25 Aug 2007
Posts: 713
WE 1: 3.5 yrs IT
WE 2: 2.5 yrs Retail chain
Re: In a casino, a gambler stacks a certain number of chips in piles, with  [#permalink]

Show Tags

New post 18 Jun 2010, 05:23
Thanks Bunuel, but S2 is still not clear.

I think the number of piles remains same before winning 12 more chips. So, if the earlier chips = 156, then there would be 13 piles of 12 chips each. Now, adding 12 more chips, which is 168, but it (168) is not divisible by 13. So, S2 is insufficient.

I dont know where I am missing.
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 55271
Re: In a casino, a gambler stacks a certain number of chips in piles, with  [#permalink]

Show Tags

New post 18 Jun 2010, 06:58
ykaiim wrote:
Thanks Bunuel, but S2 is still not clear.

I think the number of piles remains same before winning 12 more chips. So, if the earlier chips = 156, then there would be 13 piles of 12 chips each. Now, adding 12 more chips, which is 168, but it (168) is not divisible by 13. So, S2 is insufficient.

I dont know where I am missing.


# of piles before and after the winning 12 extra chips may or may not be the same. If we knew from the beginning that they are equal than \(12x+12=14x\) --> \(x=6\) --> \(12x=72\). So in this case statements are not needed to answer the question.

Next: you are right statement (2) is not sufficient, that's why answer is A and not D.

As stated in my previous post from (2) there are multiple values of x possible:
If \(x=6\), then \(12x=72\) --> \(12x+12=72+12=84=14y=14*6\);
If \(x=13\), then \(12x=156\) --> \(12x+12=156+12=168=14y=14*12\).

Hope it's clear.
_________________
Manager
Manager
avatar
Joined: 27 Mar 2010
Posts: 88
Re: In a casino, a gambler stacks a certain number of chips in piles, with  [#permalink]

Show Tags

New post 18 Jun 2010, 22:45
Hi Brunel,

I am not able to understand as for how can the number of piles be different in the two cases.As new no of chips to be added =12(given) and the new piles will have 14 chips each then it is mandatory that to accomodate 12 chips fully to make new set of 14 chips each can be done only when initiall no of piles would be 6 only.

Can you please explain with an example that contradicts that, for my understanding.

Thanks in advance!!!

utin


Bunuel wrote:
ykaiim wrote:
Thanks Bunuel, but S2 is still not clear.

I think the number of piles remains same before winning 12 more chips. So, if the earlier chips = 156, then there would be 13 piles of 12 chips each. Now, adding 12 more chips, which is 168, but it (168) is not divisible by 13. So, S2 is insufficient.

I dont know where I am missing.


# of piles before and after the winning 12 extra chips may or may not be the same. If we knew from the beginning that they are equal than \(12x+12=14x\) --> \(x=6\) --> \(12x=72\). So in this case statements are not needed to answer the question.

Next: you are right statement (2) is not sufficient, that's why answer is A and not D.

As stated in my previous post from (2) there are multiple values of x possible:
If \(x=6\), then \(12x=72\) --> \(12x+12=72+12=84=14y=14*6\);
If \(x=13\), then \(12x=156\) --> \(12x+12=156+12=168=14y=14*12\).

Hope it's clear.
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 55271
Re: In a casino, a gambler stacks a certain number of chips in piles, with  [#permalink]

Show Tags

New post 18 Jun 2010, 23:56
utin wrote:
Hi Brunel,

I am not able to understand as for how can the number of piles be different in the two cases.As new no of chips to be added =12(given) and the new piles will have 14 chips each then it is mandatory that to accomodate 12 chips fully to make new set of 14 chips each can be done only when initiall no of piles would be 6 only.

Can you please explain with an example that contradicts that, for my understanding.

Thanks in advance!!!

utin


Bunuel wrote:
ykaiim wrote:
Thanks Bunuel, but S2 is still not clear.

I think the number of piles remains same before winning 12 more chips. So, if the earlier chips = 156, then there would be 13 piles of 12 chips each. Now, adding 12 more chips, which is 168, but it (168) is not divisible by 13. So, S2 is insufficient.

I dont know where I am missing.


# of piles before and after the winning 12 extra chips may or may not be the same. If we knew from the beginning that they are equal than \(12x+12=14x\) --> \(x=6\) --> \(12x=72\). So in this case statements are not needed to answer the question.

Next: you are right statement (2) is not sufficient, that's why answer is A and not D.

As stated in my previous post from (2) there are multiple values of x possible:
If \(x=6\), then \(12x=72\) --> \(12x+12=72+12=84=14y=14*6\);
If \(x=13\), then \(12x=156\) --> \(12x+12=156+12=168=14y=14*12\).

Hope it's clear.


Please read the solution carefully. Examples are given in the text you quote.

# of piles before the winning \(x\), after the winning \(y\), so \(12x+12=14y\). Two possible scenarios:

If \(x=6\), then \(12x=72\) --> \(12x+12=72+12=84=14y=14*6\);
If \(x=13\), then \(12x=156\) --> \(12x+12=156+12=168=14y=14*12\).
_________________
Manager
Manager
avatar
Joined: 05 Mar 2010
Posts: 165
Re: In a casino, a gambler stacks a certain number of chips in piles, with  [#permalink]

Show Tags

New post 19 Jun 2010, 04:03
It is really a GMAT question

I dont think we need additional information from any of the two statements to solve it
Question says "a gambler stacks a certain number of chips in piles, with 12 chips in each pile and no chips left over. After winning 12 extra chips in a hand of poker, the gambler again stacks his chips in piles, with 14 chips in each pile and no chips left over. How many chips did the gambler have before winning the hand of poker?"

This means gambler added 2 chips (out of 12) in every pile to make it a pile of 14 from a pile of 12. That means number of piles were 12/2 = 6, because there is no left over. There cannot be any other scenario then 6 piles.

so whatever information we have in both statement is basically useless (new option "F" if GMAT want to add :lol: or say both are sufficient)

any comments ?
_________________
Success is my Destiny
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 55271
Re: In a casino, a gambler stacks a certain number of chips in piles, with  [#permalink]

Show Tags

New post 19 Jun 2010, 04:13
hardnstrong wrote:
It is really a GMAT question

I dont think we need additional information from any of the two statements to solve it
Question says "a gambler stacks a certain number of chips in piles, with 12 chips in each pile and no chips left over. After winning 12 extra chips in a hand of poker, the gambler again stacks his chips in piles, with 14 chips in each pile and no chips left over. How many chips did the gambler have before winning the hand of poker?"

This means gambler added 2 chips (out of 12) in every pile to make it a pile of 14 from a pile of 12. That means number of piles were 12/2 = 6, because there is no left over. There cannot be any other scenario then 6 piles.

so whatever information we have in both statement is basically useless (new option "F" if GMAT want to add :lol: or say both are sufficient)

any comments ?


# of piles before and after the winning 12 extra chips may or may not be the same. If we knew from the beginning that they are equal than \(12x+12=14x\) --> \(x=6\) --> \(12x=72\). So in this case statements are not needed to answer the question.

# of piles before the winning \(x\), after the winning \(y\), so \(12x+12=14y\). Two possible scenarios (out of many):

If \(x=6\), then \(12x=72\) --> \(12x+12=72+12=84=14y=14*6\);
If \(x=13\), then \(12x=156\) --> \(12x+12=156+12=168=14y=14*12\).

_________________
Manager
Manager
avatar
Joined: 05 Mar 2010
Posts: 165
Re: In a casino, a gambler stacks a certain number of chips in piles, with  [#permalink]

Show Tags

New post 21 Jun 2010, 00:07
Bunuel wrote:
hardnstrong wrote:
It is really a GMAT question

I dont think we need additional information from any of the two statements to solve it
Question says "a gambler stacks a certain number of chips in piles, with 12 chips in each pile and no chips left over. After winning 12 extra chips in a hand of poker, the gambler again stacks his chips in piles, with 14 chips in each pile and no chips left over. How many chips did the gambler have before winning the hand of poker?"

This means gambler added 2 chips (out of 12) in every pile to make it a pile of 14 from a pile of 12. That means number of piles were 12/2 = 6, because there is no left over. There cannot be any other scenario then 6 piles.

so whatever information we have in both statement is basically useless (new option "F" if GMAT want to add :lol: or say both are sufficient)

any comments ?


# of piles before and after the winning 12 extra chips may or may not be the same. If we knew from the beginning that they are equal than \(12x+12=14x\) --> \(x=6\) --> \(12x=72\). So in this case statements are not needed to answer the question.

# of piles before the winning \(x\), after the winning \(y\), so \(12x+12=14y\). Two possible scenarios (out of many):

If \(x=6\), then \(12x=72\) --> \(12x+12=72+12=84=14y=14*6\);
If \(x=13\), then \(12x=156\) --> \(12x+12=156+12=168=14y=14*12\).



We cannot take x=13 (please see the highlighted part above) because if we take x=13 then its not possible to make piles of 14 chips each (Note - each pile of 14 chips). In that case gambler need 26 more chips to make it a pile of 14 chips from 12 chips. But he won only 12 chips more.
_________________
Success is my Destiny
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 55271
Re: In a casino, a gambler stacks a certain number of chips in piles, with  [#permalink]

Show Tags

New post 21 Jun 2010, 02:04
2
hardnstrong wrote:
Bunuel wrote:
hardnstrong wrote:
It is really a GMAT question

I dont think we need additional information from any of the two statements to solve it
Question says "a gambler stacks a certain number of chips in piles, with 12 chips in each pile and no chips left over. After winning 12 extra chips in a hand of poker, the gambler again stacks his chips in piles, with 14 chips in each pile and no chips left over. How many chips did the gambler have before winning the hand of poker?"

This means gambler added 2 chips (out of 12) in every pile to make it a pile of 14 from a pile of 12. That means number of piles were 12/2 = 6, because there is no left over. There cannot be any other scenario then 6 piles.

so whatever information we have in both statement is basically useless (new option "F" if GMAT want to add :lol: or say both are sufficient)

any comments ?


# of piles before and after the winning 12 extra chips may or may not be the same. If we knew from the beginning that they are equal than \(12x+12=14x\) --> \(x=6\) --> \(12x=72\). So in this case statements are not needed to answer the question.

# of piles before the winning \(x\), after the winning \(y\), so \(12x+12=14y\). Two possible scenarios (out of many):

If \(x=6\), then \(12x=72\) --> \(12x+12=72+12=84=14y=14*6\);
If \(x=13\), then \(12x=156\) --> \(12x+12=156+12=168=14y=14*12\).



We cannot take x=13 (please see the highlighted part above) because if we take x=13 then its not possible to make piles of 14 chips each (Note - each pile of 14 chips). In that case gambler need 26 more chips to make it a pile of 14 chips from 12 chips. But he won only 12 chips more.


OK. Let me clear this once more:

If # of piles BEFORE winning was \(13\), then TOTAL # of chips would be \(13(piles)*12(chips \ in \ each)=156\).

AFTER winning \(12\) chips TOTAL # of chips would become \(156+12=168\), which IS a multiple of \(14\). So we can redistribute \(168\) chips in \(12\) piles \(14\) chips in EACH, \(12(piles)*14(chips \ in \ each)=168\).

Another scenario possible:
If # of piles BEFORE winning was \(20\), then TOTAL # of chips would be \(20(piles)*12(chips \ in \ each)=240\).

AFTER winning \(12\) chips TOTAL # of chips would become \(240+12=252\), which IS a multiple of \(14\). So we can redistribute \(252\) chips in \(18\) piles \(14\) chips in EACH, \(18(piles)*14(chips \ in \ each)=252\).

OR:
If # of piles BEFORE winning was \(27\), then TOTAL # of chips would be \(27(piles)*12(chips \ in \ each)=324\).

AFTER winning \(12\) chips TOTAL # of chips would become \(324+12=336\), which IS a multiple of \(14\). So we can redistribute \(336\) chips in \(24\) piles \(14\) chips in EACH, \(24(piles)*14(chips \ in \ each)=336\).

...

OR:
If # of piles BEFORE winning was \(699\), then TOTAL # of chips would be \(699(piles)*12(chips \ in \ each)=8388\).

AFTER winning \(12\) chips TOTAL # of chips would become \(8388+12=8400\), which IS a multiple of \(14\). So we can redistribute \(8400\) chips in \(600\) piles \(14\) chips in EACH, \(600(piles)*14(chips \ in \ each)=8400\).

...

Basically if the number of piles BEFORE winning was 1 less than multiple of 7 (see the solution in my first post), gambler would be able to redistribute chips AFTER winning in 14 chips.
_________________
VP
VP
avatar
Joined: 15 Jul 2004
Posts: 1206
Schools: Wharton (R2 - submitted); HBS (R2 - submitted); IIMA (admitted for 1 year PGPX)
Re: In a casino, a gambler stacks a certain number of chips in piles, with  [#permalink]

Show Tags

New post 14 Jul 2010, 12:01
Bunuel - hats off to you boss
Non-Human User
User avatar
Joined: 09 Sep 2013
Posts: 11011
Re: In a casino, a gambler stacks a certain number of chips in piles, with  [#permalink]

Show Tags

New post 11 Aug 2017, 03:27
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
GMAT Club Bot
Re: In a casino, a gambler stacks a certain number of chips in piles, with   [#permalink] 11 Aug 2017, 03:27
Display posts from previous: Sort by

In a casino, a gambler stacks a certain number of chips in piles, with

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


Copyright

GMAT Club MBA Forum Home| About| Terms and Conditions and Privacy Policy| GMAT Club Rules| Contact| Sitemap

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

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.