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

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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 Thanks is advance. Intern Joined: 23 Sep 2009 Posts: 9 Own Kudos [?]: 1 [0] Given Kudos: 0 Re: Poker question [#permalink] 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 Answer is A Manager Joined: 27 Oct 2008 Posts: 97 Own Kudos [?]: 299 [3] Given Kudos: 3 Re: Poker question [#permalink] 3 Kudos 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

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

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Re: Mark is playing poker at a casino. Mark starts playing with [#permalink]
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. Tutor Joined: 16 Oct 2010 Posts: 15125 Own Kudos [?]: 66741 [1] Given Kudos: 436 Location: Pune, India Re: Mark is playing poker at a casino. Mark starts playing with [#permalink] 1 Bookmarks Expert Reply 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

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% $20 chips The mix is the initial bunch. The 70% group combined with 90% group gives the 80% initial bunch of chips. Now using the usual formula, w1/w2 = (90 - 80)/(80 - 70) = 1:1 The two groups will have equal chips. Does the use of weighted avgs make sense now? Intern Joined: 25 Jan 2013 Posts: 33 Own Kudos [?]: 81 [0] Given Kudos: 12 Location: Spain Concentration: Finance, Other GMAT 1: 740 Q49 V42 Re: Mark is playing poker at a casino. Mark starts playing with [#permalink] Karishma, Thanks for sharing your knowledge. Could you please do the same with the 100$ chips?

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Re: Mark is playing poker at a casino. Mark starts playing with [#permalink]
Great explaination as usual Karishma, thank you so much
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Re: Mark is playing poker at a casino. Mark starts playing with [#permalink]
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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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Mark is playing poker at a casino. Mark starts playing with [#permalink]
I prefer creating a list to solve such before/after work problems:

$-----------original#--------- 1st bet#----------reminding# 100--------0.2n---------------0.1m--------------0.2n-0.1m 20---------0.8n----------------0.9m--------------0.8n-0.9m total--------n-------------------m------------------n-m we have known that (n-m)70%=0.8n-0.9m, and n=140 So m=70 Consequently, the money mark bet is 0.1*70*100+0.9*70*20=1960 Thanks, SVP Joined: 24 Nov 2016 Posts: 1712 Own Kudos [?]: 1360 [0] Given Kudos: 607 Location: United States Re: Mark is playing poker at a casino. Mark starts playing with [#permalink] 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

Total=140; $100=20%140=28,$20=80%140=112;
$$x=20.chips.bet…y=100.chips.bet$$
$$\frac{x}{x+y}=9/10…10x=9x+9y…x=9y$$
$$\frac{112-x}{28-y}=7/3…\frac{112-9y}{28-y}=7/3…336-27y=196-7y…y=7…x=63$$
$$7*100=700…63*20=1260…bet=1260+700=1960$$

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Re: Mark is playing poker at a casino. Mark starts playing with [#permalink]
Think of the “chips bet” as “Solution A”

Think of the “chips remaining” as “Solution B”

(# of chips bet) + (# of chips remaining at end) = Total Beginning Chips

It would be akin to mixing:

(Solution A) + (Solution B) = New Mix

in which the “New Mix” is the Number of Total Beginning Chips

Solution A (“# of chips bet”): has a 90% “concentration” of $20 Chips Solution B (“# of chips remaining at end”): has a 70% “concentration” of$20 Chips

If we mix all of solution A with all of solution B we will get a NEW Mix (#of BEGINNING Chips) that has ———-> 80% “concentration” of $20 Chips Since 80% is the Arithmetic Mean of 70% and 90%, the 2 solutions must have EQUAL Weight ———-> in other words, each Solution must make up (1/2) of the Final Mix = Total Number of Beginning Chips Since there 140 Chips: He must have bet 70 Chips And There must be 70 chips remaining. Out of 70 chips bet, 10% are$100 chips and 90% are $20 chips (7) * ($100) + (63) * ($20) =$1,960 was bet

(A)

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

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