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

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

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18 Jun 2010, 04:13
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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

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Re: Gambling in a casino [#permalink]

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18 Jun 2010, 05:16
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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.
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Re: Gambling in a casino [#permalink]

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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.
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Re: Gambling in a casino [#permalink]

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18 Jun 2010, 06:58
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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.
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Re: Gambling in a casino [#permalink]

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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.
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Re: Gambling in a casino [#permalink]

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18 Jun 2010, 23:56
Expert's post
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 $$12x=156$$ --> $$12x+12=156+12=168=14y=14*12$$.
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Re: Gambling in a casino [#permalink]

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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)

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Re: Gambling in a casino [#permalink]

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19 Jun 2010, 04:13
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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 $$12x=156$$ --> $$12x+12=156+12=168=14y=14*12$$.

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Re: Gambling in a casino [#permalink]

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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.
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Re: Gambling in a casino [#permalink]

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21 Jun 2010, 02:04
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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.
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Re: Gambling in a casino [#permalink]

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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.
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Re: Gambling in a casino [#permalink]

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21 Jun 2010, 05:06
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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.
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Re: Gambling in a casino [#permalink]

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14 Jul 2010, 12:01
Bunuel - hats off to you boss
Re: Gambling in a casino   [#permalink] 14 Jul 2010, 12:01
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