Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 24 Aug 2016, 07:01

### 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

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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# In a casino, a gambler stacks a certain number of chips in

Author Message
TAGS:

### Hide Tags

Director
Joined: 25 Aug 2007
Posts: 954
WE 1: 3.5 yrs IT
WE 2: 2.5 yrs Retail chain
Followers: 73

Kudos [?]: 1154 [0], given: 40

In a casino, a gambler stacks a certain number of chips in [#permalink]

### Show Tags

18 Jun 2010, 04:13
00:00

Difficulty:

(N/A)

Question Stats:

60% (05:02) correct 40% (00:00) wrong based on 5 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.
[Reveal] Spoiler: OA

_________________

Tricky Quant problems: http://gmatclub.com/forum/50-tricky-questions-92834.html
Important Grammer Fundamentals: http://gmatclub.com/forum/key-fundamentals-of-grammer-our-crucial-learnings-on-sc-93659.html

Math Expert
Joined: 02 Sep 2009
Posts: 34416
Followers: 6248

Kudos [?]: 79372 [0], given: 10016

Re: Gambling in a casino [#permalink]

### Show Tags

18 Jun 2010, 05:16
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.

Hope it's clear.
_________________
Director
Joined: 25 Aug 2007
Posts: 954
WE 1: 3.5 yrs IT
WE 2: 2.5 yrs Retail chain
Followers: 73

Kudos [?]: 1154 [0], given: 40

Re: Gambling in a casino [#permalink]

### Show Tags

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
Joined: 02 Sep 2009
Posts: 34416
Followers: 6248

Kudos [?]: 79372 [0], given: 10016

Re: Gambling in a casino [#permalink]

### Show Tags

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
Joined: 27 Mar 2010
Posts: 124
Followers: 2

Kudos [?]: 7 [0], given: 17

Re: Gambling in a casino [#permalink]

### Show Tags

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.

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
Joined: 02 Sep 2009
Posts: 34416
Followers: 6248

Kudos [?]: 79372 [0], given: 10016

Re: Gambling in a casino [#permalink]

### Show Tags

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.

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
Joined: 05 Mar 2010
Posts: 221
Followers: 1

Kudos [?]: 28 [0], given: 8

Re: Gambling in a casino [#permalink]

### Show Tags

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 or say both are sufficient)

_________________

Success is my Destiny

Math Expert
Joined: 02 Sep 2009
Posts: 34416
Followers: 6248

Kudos [?]: 79372 [0], given: 10016

Re: Gambling in a casino [#permalink]

### Show Tags

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 or say both are sufficient)

# 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
Joined: 05 Mar 2010
Posts: 221
Followers: 1

Kudos [?]: 28 [0], given: 8

Re: Gambling in a casino [#permalink]

### Show Tags

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 or say both are sufficient)

# 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
Joined: 02 Sep 2009
Posts: 34416
Followers: 6248

Kudos [?]: 79372 [1] , given: 10016

Re: Gambling in a casino [#permalink]

### Show Tags

21 Jun 2010, 02:04
1
KUDOS
Expert's post
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 or say both are sufficient)

# 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.
_________________
Manager
Joined: 05 Mar 2010
Posts: 221
Followers: 1

Kudos [?]: 28 [0], given: 8

Re: Gambling in a casino [#permalink]

### Show Tags

21 Jun 2010, 04:51
Thanks Bunuel for your patience and gr8 explanation +1
I got the point now ........

Just to mention ....... questions say gambler made the piles of 14 chips each. So the number of piles will change not the qty of chips in each pile. you mentioned it other way round.
_________________

Success is my Destiny

Math Expert
Joined: 02 Sep 2009
Posts: 34416
Followers: 6248

Kudos [?]: 79372 [0], given: 10016

Re: Gambling in a casino [#permalink]

### Show Tags

21 Jun 2010, 05:06
hardnstrong wrote:
Thanks Bunuel for your patience and gr8 explanation +1
I got the point now ........

Just to mention ....... questions say gambler made the piles of 14 chips each. So the number of piles will change not the qty of chips in each pile. you mentioned it other way round.

Yes, there was a typo: chips instead of piles, edited it.
_________________
VP
Joined: 15 Jul 2004
Posts: 1473
Schools: Wharton (R2 - submitted); HBS (R2 - submitted); IIMA (admitted for 1 year PGPX)
Followers: 21

Kudos [?]: 155 [0], given: 13

Re: Gambling in a casino [#permalink]

### Show Tags

14 Jul 2010, 12:01
Bunuel - hats off to you boss
Re: Gambling in a casino   [#permalink] 14 Jul 2010, 12:01
Similar topics Replies Last post
Similar
Topics:
18 All boxes in a certain warehouse were arranged in stacks of 12 boxes 10 19 Jun 2016, 09:25
8 A casino pays players using purple and orange chips; each 4 24 Jan 2013, 23:08
10 In a certain game played with red chips and blue chips, each 7 25 Nov 2010, 01:58
4 Are all of the numbers in a certain list of 15 numbers 14 26 Nov 2009, 12:28
5 Are all of the numbers in a certain list of 15 numbers 9 18 Jul 2009, 00:52
Display posts from previous: Sort by