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.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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

Show Tags

18 Jun 2010, 04:13

1

This post received KUDOS

00:00

A

B

C

D

E

Difficulty:

55% (hard)

Question Stats:

50% (02:20) correct
50% (01:39) wrong based on 18 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.

Official Answer and Stats are available only to registered users. Register/Login.

_________________

Want to improve your CR: http://gmatclub.com/forum/cr-methods-an-approach-to-find-the-best-answers-93146.html 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

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.

Re: In a casino, a gambler stacks a certain number of chips in piles, with [#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 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 \(12x=156\) --> \(12x+12=156+12=168=14y=14*12\).

Re: In a casino, a gambler stacks a certain number of chips in piles, with [#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.

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 \(12x=156\) --> \(12x+12=156+12=168=14y=14*12\).

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 \(12x=156\) --> \(12x+12=156+12=168=14y=14*12\).
_________________

Re: In a casino, a gambler stacks a certain number of chips in piles, with [#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)

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)

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 \(12x=156\) --> \(12x+12=156+12=168=14y=14*12\). _________________

Re: In a casino, a gambler stacks a certain number of chips in piles, with [#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)

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.
_________________

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)

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.
_________________

Re: In a casino, a gambler stacks a certain number of chips in piles, with [#permalink]

Show Tags

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.
_________________

We’ve given one of our favorite features a boost! You can now manage your profile photo, or avatar , right on WordPress.com. This avatar, powered by a service...

Sometimes it’s the extra touches that make all the difference; on your website, that’s the photos and video that give your content life. You asked for streamlined access...

A lot has been written recently about the big five technology giants (Microsoft, Google, Amazon, Apple, and Facebook) that dominate the technology sector. There are fears about the...

Post today is short and sweet for my MBA batchmates! We survived Foundations term, and tomorrow's the start of our Term 1! I'm sharing my pre-MBA notes...