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A gambler bought $3,000 worth of chips at a casino in
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25 Feb 2012, 21:23
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Question Stats:
78% (02:35) correct 22% (02:59) wrong based on 345 sessions
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A gambler bought $3,000 worth of chips at a casino in denominations of $20 and $100. That evening, the gambler lost 16 chips, and then cashed in the remainder. If the number of $20 chips lost was 2 more or 2 less than the number of $100 chips lost, what is the largest amount of money that the gambler could have received back?
Re: A gambler bought $3,000 worth of chips at a casino in
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25 Feb 2012, 22:29
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Chembeti wrote:
A gambler bought $3,000 worth of chips at a casino in denominations of $20 and $100. That evening, the gambler lost 16 chips, and then cashed in the remainder. If the number of $20 chips lost was 2 more or 2 less than the number of $100 chips lost, what is the largest amount of money that the gambler could have received back?
A. $2,040 B. $2,120 C. $1,960 D. $1,920 E. $1,400
In order to maximize the amount of money that the gambler kept, we should maximize # of $20 chips lost and minimize # of $100 chips lost, which means that # of $20 chips lost must be 2 more than # of $100 chips lost.
So, if # of $20 chips lost is x then # of $100 chips lost should be x-2. Now, given that total # of chips lost is 16: x+x-2=16 --> x=9: 9 $20 chips were lost and 9-2=7 $100 chips were lost.
Total worth of chips lost is 9*20+7*100=$880, so the gambler kept $3,000-$880=$2,120.
Re: A gambler bought $3,000 worth of chips at a casino in
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25 Feb 2012, 22:41
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Bunuel wrote:
Chembeti wrote:
A gambler bought $3,000 worth of chips at a casino in denominations of $20 and $100. That evening, the gambler lost 16 chips, and then cashed in the remainder. If the number of $20 chips lost was 2 more or 2 less than the number of $100 chips lost, what is the largest amount of money that the gambler could have received back?
A. $2,040 B. $2,120 C. $1,960 D. $1,920 E. $1,400
In order to maximize the amount of money that the gambler kept, we should maximize # of $20 chips lost and minimize # of $100 chips lost, which means that # of $20 chips lost must be 2 more than # of $100 chips lost. So, if # of $20 chips lost is x then # of $100 chips lost should be x-2. Now, given that total # of chips lost is 16: x+x-2=16 --> x=9: 9 $20 chips were lost and 9-2=7 $100 chips lost.
Total worth of chips lost is 9*20+7*100=$880, so the gambler kept $3,000-$880=$2,120.
Answer: B.
I somehow got confused with the wordings.... _________________
Re: A gambler bought $3,000 worth of chips at a casino in
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13 Nov 2016, 09:24
Answer:B 1- Largest amount of money that the gambler could have received back ⇒ the lost amount is minimum ⇒ $100 amount lost is minimum 2- $20 chips lost was 2 more or 2 less than the number of $100 chips lost ⇒ Lost $100 is minimum and $20 is maximum -1- and -2- ⇒ $20 chips lost is 9 and $100 chips lost is 7 ⇒ total loss=20x9+100x7=$880 ⇒ He received back=3000-880= $2120
Re: A gambler bought $3,000 worth of chips at a casino in
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21 Sep 2018, 02:07
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