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25 Feb 2012, 22:23
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B
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Difficulty:
35%
(medium)
Question Stats:
77% (02:40) correct
23%
(02:51)
wrong
based on 682
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?
$2,040
$2,120
$1,960
$1,920
$1,400
$2,040
$2,120
$1,960
$1,920
$1,400
25 Feb 2012, 23:29
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Chembeti
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
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.
Answer: B.
General Discussion
25 Feb 2012, 23:41
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Bunuel
Chembeti
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
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....
