Todays puzzle 11-11-2011

1947 views 7 replies

There is a casino and it has 4 gates (let’s name them as gate A, B, C and D). Every time you come inside the casino, you have to pay $5 and every time you leave the casino, you again have to pay $5. Also know that, whatever amount of money you carry with you inside the casino, it doubles.

For example if you carry $5, it will become $10. You have to go inside the casino through gate A, come out of B, again go inside the casino through gate C and finally come out of gate D. Now, how much money should you carry inside the casino so that when you finally come out of the gate D, you should be left with no (zero amount) money?

 

 

Comment if you know the answer

 

 

Solution of Todays Puzzle 10-11-2011

Replies (7)

its $11.25 which i have to carrry when i enter in casino.

I have to carry $11.25 /- when i enter into the casino 

 

the problem is splved in this way 

[{(x-5)X2}-10] X 2 -5  =0

[{(x-5)X2}-10] = 2.5

{(x-5)X2} = 12.5

(x-5) = 6.25

x = 6.25 + 5

x = 11.25

shortand simple...

he has to pay

entry to gate 1=5 ....double... exit =5

entry to gate 2=5 ....double... exit =5

entry to gate 3=5 ....double... exit =5

entry to gate 4=5 ....double... exit =5

 

the ans is 14.0625...attached is the excel file ..

 

 

7.5$

enter A - 5$

doubles 2.5 $ = 5$

come out of B

Enters C pays 5$

balance Nil

doubles "0"

comeout of D Nil money

Correct ans is $11.25. 

Gate A pays $5 and left with $6.25. Inside doubles and get $12.5. Then exits at Gate B pays $5 and left with $7.5. Then enter from Gate C and pays $5 and left with $2.5. Inside double and get $5 and while finally exits at Gate D pays $5 and left with no money.

Correct answer is 11.5$ for sure,............

sorry my ans is wrong..i counted the gates...wronly

 

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