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# Mark is playing poker at a casino. Mark starts playing with

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Mark is playing poker at a casino. Mark starts playing with [#permalink]

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11 Nov 2004, 04:59
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60% (04:41) correct 40% (03:48) wrong based on 163 sessions

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Mark is playing poker at a casino. Mark starts playing with 140 chips, 20% of which are $100 chips and 80% of which are$20 chips. For his first bet, Mark places chips, 10% of which are $100 chips, in the center of the table. If 70% of Mark's remaining chips are$20 chips, how much money did Mark bet?

A. $1,960 B.$1,740
C. $1,540 D.$3,080
E. $2,640 [Reveal] Spoiler: OA Last edited by Bunuel on 17 Nov 2014, 12:26, edited 1 time in total. Added the OA. Senior Manager Joined: 19 May 2004 Posts: 291 Followers: 1 Kudos [?]: 11 [4] , given: 0 Re: Mark is playing poker at a casino. Mark starts playing with [#permalink] ### Show Tags 11 Nov 2004, 07:23 4 This post received KUDOS A. 1960. Let's say he bet X$100 chips, which means 9X $20 chips. Out of the 140 chips, 112 are$20 chips and 28 are $100 chips. 0.7*(140-10X) = 112-9X. You get X=7. 7*100 + 9*7*20 = 1960. Intern Joined: 03 Aug 2009 Posts: 6 Followers: 0 Kudos [?]: 1 [0], given: 0 Poker question [#permalink] ### Show Tags 24 Sep 2009, 00:14 1 This post was BOOKMARKED Can anyone let me know how you would approach a problem such as this one? Mark is playing poker at a casino. Mark starts playing with 140 chips, 20% of which are$100 chips and 80% of which are
$20 chips. For his first bet, Mark places chips, 10% of which are$100 chips, in the center of the table. If 70% of
Mark's remaining chips are $20 chips, how much money did Mark bet? A.$1,960
B. $1,740 C.$1,540
D. $3,080 E.$2,640

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24 Sep 2009, 01:34
Total chips 140
20% of $100 chips = 28 chips *$100 = $2,800 80% of$20 chips = 112 chips * $20 =$2,240

If x is the number of chips bet and y is the amount of chips remaining:
x + y = 140
y = 140 - x ~ (1)

[First round bet] + [Second round bet] = $2,800 +$2,240
[0.1x * $100 + 0.9x *$20] + [0.3y * $100 + 0.7y *$20] = $5,040 10x + 18x + 30y + 14y = 5040 28x + 44y = 5040 ~ (2) Substitute (1) into (2) to solve for x: 28x + 44(140 - x) = 5040 28x + 6160 - 44x = 5040 16x = 1120 x = 70 Substituting x: [First round bet] = [0.1x *$100 + 0.9x * $20] = 0.1(70)($100)+0.9(70)($20) =$700 + $1,260 =$1,960

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24 Sep 2009, 11:41
Mark is playing poker at a casino. Mark starts playing with
140 chips, 20% of which are $100 chips and 80% of which are$20 chips. For his first bet, Mark places chips, 10% of
which are $100 chips, in the center of the table. If 70% of Mark's remaining chips are$20 chips, how much money did
Mark bet?

A. $1,960 B.$1,740
C. $1,540 D.$3,080
E. $2,640 Thanks is advance. Soln: Total is 140 chips so 100$ chips is 28
and 20$chips is 112 let he bet X chips. of this .1X chips are 100$ chips and .9X chips are 20$chips the left over chips are (140-X) of this .3(140-X) are 100$ chips and .7(140-X) are 20$chips thus equation is .1X + .3(140-X) = 28 solving for X X = 70 thus the betting involved 7 (100$) chips and 63 (20$) chips amount is = 7 * 100 + 63 * 20 = 1960 Manager Joined: 10 Sep 2014 Posts: 99 Followers: 0 Kudos [?]: 35 [0], given: 25 Re: Mark is playing poker at a casino. Mark starts playing with [#permalink] ### Show Tags 17 Nov 2014, 12:22 bump for a good practice question If Mark has 140 chips and 20% are$100 chips, then Mark has 1/5 of 140 or 28 $100 chips. The other 112 are$20 chips.

10% of the chips he bet were $100 chips so for every 10 chips he bet, the total must have been$100 + 9($20) =$280. We can divide each answer choice by this # to see which one works.

A. 1960/280 = 7, meaning that Frank bet 7 $100 chips and 63$20 chips. This would bring his total chip count to 140-7-63=70 chips and his $20 chip count down to 112-63=59 chips. So he has 59/70$20 chips remaining. 70% is what we are looking for which is slighly less than 3/4 and this fraction is much more than that so we need a bigger answer.

B. we need a # larger than answer A so this wont work
C. same reason as above
D. 3080 / 280 = 11 meaning Frank bet 11 $100 chips and 99$20 chips. He now has 30 chips left, 13 of which are $20 chips. 13/20 = 65/100 = 65% this is too small so we know Frank must have bet less money. Answer E! Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 6487 Location: Pune, India Followers: 1761 Kudos [?]: 10507 [6] , given: 206 Mark is playing poker at a casino. Mark starts playing with [#permalink] ### Show Tags 17 Nov 2014, 23:47 6 This post received KUDOS Expert's post 4 This post was BOOKMARKED christoph wrote: Mark is playing poker at a casino. Mark starts playing with 140 chips, 20% of which are$100 chips and 80% of which are $20 chips. For his first bet, Mark places chips, 10% of which are$100 chips, in the center of the table. If 70% of Mark's remaining chips are $20 chips, how much money did Mark bet? A.$1,960
B. $1,740 C.$1,540
D. $3,080 E.$2,640

Initially, he had 80% $20 chips. Then he bet an amount in which he had 90%$20 chips and was left with an amount in which there were 70% $20 chips. Since 80% is the mid of 70% and 90%, it means the number of chips in the two cases were equal. So he bet a total of 140/2 = 70 chips. 7 must be$100 chips with value $700 63 must be$20 chips with value 63*20 = $1260 Total value of the bet =$1960

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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Intern Joined: 25 Jun 2014 Posts: 48 Followers: 0 Kudos [?]: 17 [0], given: 187 Re: Mark is playing poker at a casino. Mark starts playing with [#permalink] ### Show Tags 18 Nov 2014, 01:37 VeritasPrepKarishma wrote: christoph wrote: Mark is playing poker at a casino. Mark starts playing with 140 chips, 20% of which are$100 chips and 80% of which are $20 chips. For his first bet, Mark places chips, 10% of which are$100 chips, in the center of the table. If 70% of Mark's remaining chips are $20 chips, how much money did Mark bet? A.$1,960
B. $1,740 C.$1,540
D. $3,080 E.$2,640

Initially, he had 80% $20 chips. Then he bet an amount in which he had 90%$20 chips and was left with an amount in which there were 70% $20 chips. Since 80% is the mid of 70% and 90%, it means the number of chips in the two cases were equal. So he bet a total of 140/2 = 70 chips. 7 must be$100 chips with value $700 63 must be$20 chips with value 63*20 = $1260 Total value of the bet =$1960

Karishma, I love your weighted method but for this one, i was confused to figure out the reasoning behind it. So sorry but can you elaborate further? Thank you.
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Re: Mark is playing poker at a casino. Mark starts playing with [#permalink]

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18 Nov 2014, 20:48
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vietnammba wrote:
VeritasPrepKarishma wrote:
christoph wrote:
Mark is playing poker at a casino. Mark starts playing with 140 chips, 20% of which are $100 chips and 80% of which are$20 chips. For his first bet, Mark places chips, 10% of which are $100 chips, in the center of the table. If 70% of Mark's remaining chips are$20 chips, how much money did Mark bet?

A. $1,960 B.$1,740
C. $1,540 D.$3,080
E. $2,640 Initially, he had 80%$20 chips. Then he bet an amount in which he had 90% $20 chips and was left with an amount in which there were 70%$20 chips. Since 80% is the mid of 70% and 90%, it means the number of chips in the two cases were equal. So he bet a total of 140/2 = 70 chips.

7 must be $100 chips with value$700
63 must be $20 chips with value 63*20 =$1260
Total value of the bet = $1960 Answer (A) Karishma, I love your weighted method but for this one, i was confused to figure out the reasoning behind it. So sorry but can you elaborate further? Thank you. Look, that's the thing about weighted averages - don't think of them only after you see terms such as average, mixture etc. They are useful whenever you talk about two groups separately and then together or about one group splitting into two groups. Here, he has a bunch of 140 chips first with a concentration of 80%$20 chips. He splits the bunch into two groups:
One he bets with a concentration of 90% $20 chips The other he keeps with a concentration of 70%$20 chips

The mix is the initial bunch. The 70% group combined with 90% group gives the 80% initial bunch of chips.
Now using the usual formula, w1/w2 = (90 - 80)/(80 - 70) = 1:1

The two groups will have equal chips.

Does the use of weighted avgs make sense now?
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Intern Joined: 25 Jan 2013 Posts: 35 Location: Spain Concentration: Finance, Other Schools: Wharton '17 (S) GMAT 1: 740 Q49 V42 Followers: 1 Kudos [?]: 62 [0], given: 12 Re: Mark is playing poker at a casino. Mark starts playing with [#permalink] ### Show Tags 19 Nov 2014, 04:57 Karishma, Thanks for sharing your knowledge. Could you please do the same with the 100$ chips?

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19 Nov 2014, 09:15
Great explaination as usual Karishma, thank you so much
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19 Nov 2014, 20:37
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spla626 wrote:
Karishma,

Thanks for sharing your knowledge. Could you please do the same with the 100 $chips? Thanks in advance Posted from my mobile device Yes, you can do the same with the other "ingredient" as well. The mix has 20%$100 chips, he bet a bunch that has 10% $100 chips and was left with the other bunch that has 30%$100 chips.
w1/w2 = (30 - 20)/(20 - 10) = 1:1

Both bunches are equal.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews GMAT Club Legend Joined: 09 Sep 2013 Posts: 9287 Followers: 455 Kudos [?]: 115 [0], given: 0 Re: Poker question [#permalink] ### Show Tags 17 Jul 2015, 09:50 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. _________________ Re: Poker question [#permalink] 17 Jul 2015, 09:50 Similar topics Replies Last post Similar Topics: 8 The playing area for a certain outdoor game requires markings accordin 18 07 Sep 2015, 23:05 7 Harold plays a game in which he starts with$2. 10 04 Sep 2012, 18:18
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