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Ten Rs.5.00 bills, ten Rs.10.00 bills and ten Rs.20.00 bills are in a box. A blind-folded personmust remove the bills from the box. The contest ends when a person removes three samebills(example: three Rs.5.00 bills). What is the most amount of money a person can removefrom the box?
Read Solution (Total 4)
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- 90rs maximum a person remove from a box.
First he remove 2 Rs.5 bills = 2* 5 = 10rs
Then he remove 2 Rs.10 bills = 2* 10 = 20rs
Then he remove 3 Rs.20 bills = 3* 20 = 60rs
So, totally he remove = 10+20+60 = 90rs - 12 years agoHelpfull: Yes(10) No(3)
- The answer is 90
The person takes two bills from each Rs.5, Rs.10, Rs.20 bills.
so the total amount of money so far is
2*5=10 2*10=20 2*20=40
10+20+40=70
Next he might take Rs.20 bill. (This will give the most amount of money)
70+20=90
The most amount of money will be Rs.90 - 12 years agoHelpfull: Yes(5) No(3)
- 2 Rs.20 bills
2 Rs.10 bills
2 Rs.5 bills
1 Rs.20 bills
So the max amount he can draw is Rs.90 - 12 years agoHelpfull: Yes(1) No(2)
- answer 70
the person can take 6 ills it may be of 2 bills of each type .taking seventh bill the contest ends. therefore maximum amount =2*5+2*10+2*20=70 - 12 years agoHelpfull: Yes(0) No(12)
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