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

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.

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

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

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!

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?

Karishma,

Thanks for sharing your knowledge. Could you please do the same with the 100 $ chips?

Thanks in advance

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Great explaination as usual Karishma, thank you so much

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.

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,

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

Answer (A)

Think of the “chips bet” as “Solution A”


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


If we add the following:


(# 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)









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