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

# 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.

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

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

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.

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.

Bunuel - hats off to you boss