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this_time_i_will: Master | Next Rank: 500 Posts; Posts: 268; Joined: Wed Mar 17, 2010 2:32 am; Thanked: 17 times

gambler chips

by this_time_i_will » Wed Apr 28, 2010 7:05 pm

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.

OA is [spoiler]A. But, IMO, we do not need statement 1 or 2 to answer this question.
from question itself: 12p+12 = 14p. so p=6. so number of chips gamblr had before winning = 12p = 72. [p->piles][/spoiler]

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Source: Beat The GMAT — Data Sufficiency | Wed Apr 28, 2010 7:05 pm

rockeyb: Legendary Member; Posts: 537; Joined: Sun Jul 19, 2009 7:15 am; Location: Nagpur , India; Thanked: 41 times; Followed by:1 members

by rockeyb » Wed Apr 28, 2010 9:48 pm

this_time_i_will 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.

OA is [spoiler]A. But, IMO, we do not need statement 1 or 2 to answer this question.
from question itself: 12p+12 = 14p. so p=6. so number of chips gamblr had before winning = 12p = 72. [p->piles][/spoiler]

This is a value question , so any option giving us the single value will be sufficient .

The question stem only tells us that the number of chips the gambler had before winning the hand must be a multiple of 12 and after winning the hand should be a multiple of 14 .

Now we do not know what number of chips the gambler has as there can be any number of common multiple of 12 and 14 .

Ex before winning the hand the number of chips could be 72 (multiple of 12) after winning the hand it could be 84 (a multiple of 14 .)

OR before winning the hand the number of chips could be 828 (multiple of 12 ) and after winning the hand the number of chips could be (840 a multiple of 14 .)

so simply from the the question we can not definitely say how many chips he had .

Lets look at answer option :

1. Before winning the hand of poker, the gambler had fewer than 140 chips.

Now this limits the number of possibilities as we can definitely say the number of chips before winning the hand were 72 and after winning were 84 as there are the only multiple of 12 and 14 that satisfy the condition .

sufficient .

2. Before winning the hand of poker, the gambler had more than 70 chips.

As demonstrated earlier we can not say definitely how many chips he had before winning as more than one number satisfy the condition in the question.

Ans : A .

Last edited by rockeyb on Wed Apr 28, 2010 9:56 pm, edited 1 time in total.

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liferocks: Legendary Member; Posts: 576; Joined: Sat Mar 13, 2010 8:31 pm; Thanked: 97 times; Followed by:1 members

by liferocks » Wed Apr 28, 2010 9:49 pm

before winning the gambler has C=12x chips and after winning he has 12(x+1) chips
now 12(x+1)=2*6*(x+1)
since 12(x+1) is divisible by 14, x+1 has to be multiple of 7
i.e x+1=7k or x=7k-1
so for all k=1,2,3... this relation will hold good
for k=1 we will have x=6 and C=84..this is the one you have considered
for k=2 we will have x=13 and C=156 ..and so on

so multiple numbers are there for C for which the given condition satisfies.

Now from A we can clearly say that C=84..sufficient
but all values of C are greater than 70...B is not sufficient
Ans A

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electrico: Senior | Next Rank: 100 Posts; Posts: 53; Joined: Tue Feb 17, 2009 11:14 pm; Location: India; Thanked: 2 times

by electrico » Thu Apr 29, 2010 12:35 am

rockeyb wrote:
this_time_i_will 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.

OA is [spoiler]A. But, IMO, we do not need statement 1 or 2 to answer this question.
from question itself: 12p+12 = 14p. so p=6. so number of chips gamblr had before winning = 12p = 72. [p->piles][/spoiler]

This is a value question , so any option giving us the single value will be sufficient .

The question stem only tells us that the number of chips the gambler had before winning the hand must be a multiple of 12 and after winning the hand should be a multiple of 14 .

Now we do not know what number of chips the gambler has as there can be any number of common multiple of 12 and 14 .

Ex before winning the hand the number of chips could be 72 (multiple of 12) after winning the hand it could be 84 (a multiple of 14 .)

OR before winning the hand the number of chips could be 828 (multiple of 12 ) and after winning the hand the number of chips could be (840 a multiple of 14 .)

so simply from the the question we can not definitely say how many chips he had .

Lets look at answer option :

1. Before winning the hand of poker, the gambler had fewer than 140 chips.

Now this limits the number of possibilities as we can definitely say the number of chips before winning the hand were 72 and after winning were 84 as there are the only multiple of 12 and 14 that satisfy the condition .

sufficient .

2. Before winning the hand of poker, the gambler had more than 70 chips.

As demonstrated earlier we can not say definitely how many chips he had before winning as more than one number satisfy the condition in the question.

Ans : A .

Here, nothing has been said that the chips are a multiple of 12 or 14.

Let say, no of piles is x. As per question, 12x = 12 +14x

=> x = 6, hence we do not need any of the stment. I don't understand why are we considering multiple of 12 and 14 (like 828, 840 )

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Fiver: Master | Next Rank: 500 Posts; Posts: 114; Joined: Mon Sep 22, 2008 3:51 am; Thanked: 8 times; GMAT Score:680

by Fiver » Thu Apr 29, 2010 1:03 am

this_time_i_will wrote:

OA is A. But, IMO, we do not need statement 1 or 2 to answer this question.
from question itself: 12p+12 = 14p. so p=6. so number of chips gamblr had before winning = 12p = 72. [p->piles]

I made the same assumption when i first attempted this problem.
We are assuming that the number of piles is constant in both cases;

instead what we must have worked out this equation:
12p + 12 = 14q
12 (p + 1) = 14q
12 (6 + 1) = 14 * 6 this fits in when p = q
12 (13 + 1) = 14 * 12 this also fits in when 'p' not equal to 'q'

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rockeyb: Legendary Member; Posts: 537; Joined: Sun Jul 19, 2009 7:15 am; Location: Nagpur , India; Thanked: 41 times; Followed by:1 members

by rockeyb » Thu Apr 29, 2010 1:18 am

electrico wrote:
Here, nothing has been said that the chips are a multiple of 12 or 14.

Let say, no of piles is x. As per question, 12x = 12 +14x

=> x = 6, hence we do not need any of the stment. I don't understand why are we considering multiple of 12 and 14 (like 828, 840 )

Hey ,

If you read the question carefully it says

with 12 chips in each pile and no chips left over

So each pile will have 12 chips and no left over . That means the number of chips will have to completely divisible by 12 . That means the number should be a multiple of 12 .

Adding 12 to such a number will still keep the number a multiple of 12 but now there should be 14 chips in each pile and no left over . So the very same number that is multiple of 12 should be a multiple of 14 after adding 12 to it .

"Know thyself" and "Nothing in excess"

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electrico: Senior | Next Rank: 100 Posts; Posts: 53; Joined: Tue Feb 17, 2009 11:14 pm; Location: India; Thanked: 2 times

by electrico » Thu Apr 29, 2010 1:22 am

rockeyb wrote:
electrico wrote:
Here, nothing has been said that the chips are a multiple of 12 or 14.

Let say, no of piles is x. As per question, 12x = 12 +14x

=> x = 6, hence we do not need any of the stment. I don't understand why are we considering multiple of 12 and 14 (like 828, 840 )
Hey ,

If you read the question carefully it says
with 12 chips in each pile and no chips left over
So each pile will have 12 chips and no left over . That means the number of chips will have to completely divisible by 12 . That means the number should be a multiple of 12 .

Adding 12 to such a number will still keep the number a multiple of 12 but now there should be 14 chips in each pile and no left over . So the very same number that is multiple of 12 should be a multiple of 14 after adding 12 to it .

ok, finally agrred. Thanks.