The "Great Indian" Puzzle Thread!

Puzzles For Month Jan 2

Hi All,

As it was decided earlier to post puzzles in different thread for each mnth,So I m posting my puzzle in different thread for the mnth jan.Here is the puzzle.............

one boy has some money ad he has to go to three temples.He has to pay some money for entering and exiting the money in all temples.(same amount in each temple(same for entering and exiting).one thing to note is when he enters the temple after paying money ,he got the same amount frm temple he is having at that time(u can say his money got doubled in temple).At the end he was not having any money.

the question is.............
1) how much money he was having in starting.
2) how much he paid in each temple for entering and exiting.

temple A- pay money-enter-money doubled-pay money-exit
same for other temples.

 

Solution....

hi Sonia
Here is one trial from me.

The money he has is 21rs at the time of starting.

1st temple
entry------fee 8rs
so he has 21-8=13rs--then double-----26rs
exit-------fee 8rs
then 26-8=18rs
2nd temple
entry------fee 8rs
so he has 18-8=10rs--then double-----20rs
exit-------fee 8rs
then 20-8=12rs
1st temple
entry------fee 8rs
so he has 12-8=4rs--then double-----8rs
exit-------fee 8rs
then 8-8=0rs
So he has nothing when he came out from the third temple.
He has 21rs at the time of starting.
And he paid 8rs to enter and exit from every temple.
Let me know whether this is correct or not.

Gayathri.

 

yes u r right

dear gayathri,

yes , u r right but this is not my answer.
so if u want u can try other answer also.

 

one more........

Using only the numbers 1, 3, 4, and 6, together with the operations +, −, ×, and ÷, and unlimited use of brackets, make the number 24. Each number must be used precisely once. Each operation may be used zero or more times. Decimal points are not allowed, this form 3 × (14 − 6) is also not allowed.

 

Another puzzle

Here is another puzzle for this month.


Four men approach a bridge at night. The bridge is old and rickety and can support, at most, two men going across at one time. They have only one flashlight to share between them. Each man has differing abilities and some take more time to cross than the others. The times it takes for each man to cross are: 1 minute, 2 minutes, 5 minutes, and 10 minutes. When crossing in pairs, they can only cross as fast as the slowest man can go. Calculate the total time taken by the all the four men to cross the bridge, which should be the minumum.

kodi.

 

Hi Surya

Here is mu solution...
Hope this is correct and the minimum time needed to cross:
Time taken for every trip is given in the brackets. Plz note.
First persons with speed 1min, 2min will go. (2min)
Then 1min person will come back.(1 min)
Now persons with speed 10min and 5 min will go.(10 min)
Then 2min person will come back.(2 min)
Finally 1min, 2min persons are left and they will go. (2 min)

Diagrammatically

1,2--------------->1,2 (2 min)
1min person goes back.... (1 min)
10,5--------------->10,5,2 (10 min)
2min person goes back.... (2 min)
1,2---------------->1,2,5,10 (2 min)


So total time is 17 min. (2+1+10+2+2)
This is the minimum of all combinations.
Let me know ur opinion...

Gayathri.

 

Yes Gayatri, your answer is right.

regards,
kodi.

 

Sonia..

Actually this has many solutions.
My view is this. Any wayz, we can get answer for any number mulitiplicative of 21.
Suppose u take 42.
Then we've to set the entry and exit fee as 16rs.
This is another solution.
63,82rs are next combinations with fee as 24, 32 respectively.
For 42rs, the steps are

42----42-16(entry fee)=26-----52(double)-------52-16(exit fee)=36
36-----36-16(entry fee)=20-----40(double)-------40-16(exit fee)=24
24-----24-16(entry fee)=8--------16(double)-------16-16(exit fee)=0

Like this, we can give any solution.

Give ur FB.
Gayathri.

 

u r right Gayathri

Hi Gayathri,

U r right.:clap I did not notice this that it is true for each multiple of 21.
but still my answer is different,Actually when I was asked this puzzle I counted like this .

boy is having o rs in the end. so I suppose he had rs 1 in the 3rd temple as he can not have o rs.Now he got doubled that means 2 rs which he gave for existing last temple(no money in end).This way my answer came 5rs 25 paisa which he was having initially.and 2 rs fee for entering and existing.

Now turn to solve my second puzzle which I already posted yesterday.:tongue

 

Answer

Sonia,

The answer is :

6/(1-(3/4))

Regards,
Preethi