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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)

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
11 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
11 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
11 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
11 years agoHelpfull: Yes(0) No(12)

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