RIDDLES -- Dimag ki batti jalao

loginrahul said:
okay let me show you how it gets to EBA

1) engine connect to A
2) they come back and turn from 25m corner and drops A at 10m corner
3) now engine takes the long route from 25m -1000 m corner and connects to B (EB)
4) engine pushes B back and connects to A which is at 10m corner
and you get EBA.
4) now they come back with Engine in front towards 1000 m corner
5)turn from there towards original position of A with A in front
6) leave A there and again turn from 1000 m corner and puts B at 10 m corner
7) E again comes from 1000 m corner and connects to A... Drops A at original position of B

... you guys can do the rest.......
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Think again dude ... when engine coonects to B and pushes B back then it will reach 25m corner where B is on left side and Engine on right. So, the respective postions will be BEA... Engine will be in center not left. Think again and u will find u r wrong :)
 
dkumars said:
Think again dude ... when engine coonects to B and pushes B back then it will reach 25m corner where B is on left side and Engine on right. So, the respective postions will be BEA... Engine will be in center not left. Think again and u will find u r wrong :)
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U guys kidding me ? :)

Let me try and explain why Rahul is right ( assuming that EBA is allowed , which I did not assume at first)

E first deposits A in 10 m space.

Runs back whole way to B and pushes it back to form EBA.

Then takes back EBA and goes to 1000m and then runs back , so basically its A who is at front and goes to the right side of 1000m , puts A there , and then engine comes back , so engine at front , B at back and goes again to 100 m , then backward journey , means B ahead , Engine at back , and pushes B back to 10 m. Then leave B there , then come back and take A and place it at B's place.
and this way it is free to come back alone throught the curve by going thru 1000 m and then goes all thway to 10m where B is there , deposit B safely and done!

See atached sequence of pics for detail!
 
anilsinghd said:
U guys kidding me ? :)

Let me try and explain why Rahul is right ( assuming that EBA is allowed , which I did not assume at first)

E first deposits A in 10 m space.

Runs back whole way to B and pushes it back to form EBA.

Then takes back EBA and goes to 1000m and then runs back , so basically its A who is at front and goes to the right side of 1000m , puts A there , and then engine comes back , so engine at front , B at back and goes again to 100 m , then backward journey , means B ahead , Engine at back , and pushes B back to 10 m. Then leave B there , then come back and take A and place it at B's place.
and this way it is free to come back alone throught the curve by going thru 1000 m and then goes all thway to 10m where B is there , deposit B safely and done!

See atached sequence of pics for detail!
Click to expand...



bhai aur explanation toh baad mein dekhna hai par please ek baar ye EBA ka concept samjha do ... kyukar EBA ho rehya hai ?

See what i am saying and what i am concluding :)
 
dkumars said:
bhai aur explanation toh baad mein dekhna hai par please ek baar ye EBA ka concept samjha do ... kyukar EBA ho rehya hai ?

See what i am saying and what i am concluding :)
Click to expand...

are DK , tumhari 3rd pic mein agar engine ab B ki taraf jaayega and pushes it back till A to EBA hi hoga na ??:eek:
 
anilsinghd said:
are DK , tumhari 3rd pic mein agar engine ab B ki taraf jaayega and pushes it back till A to EBA hi hoga na ??:eek:
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Congrats guys:)
arre u guys were correct ... meri budhhi kharab ho gayi yaar .. although you mentioned 10 m corner but dunno y everytime i kept it at 25m corner :( ... that solution is correct
 
dkumars said:
Congrats guys:)
arre u guys were correct ... meri budhhi kharab ho gayi yaar .. although you mentioned 10 m corner but dunno y everytime i kept it at 25m corner :( ... that solution is correct
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geheeheee budhhi kharab....

bhai budhhi vuddi ka to pata nahi....per solution nikalne me maza aagaya......

yo thi riddles ki baap..... aaissee aur aan do....
 
loginrahul said:
geheeheee budhhi kharab....

bhai budhhi vuddi ka to pata nahi....per solution nikalne me maza aagaya......

yo thi riddles ki baap..... aaissee aur aan do....
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Ya Rahul, that was excellent, this puzzle was very near to that one of crossing the river with Goat grass and the lion. Realy maza aa gaya, keep pouring
 
dkumars said:
haan bahut achhi puzzle thi .. i also loved solving that :)
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though i still believe that without explicit telling EBA does not make sense to me :D i mean at least that is n't obvious!


So whats my score ? 0/1 :( 0 % :eek:


I should be :thappad
 
Hi,

this will prove how strong u are in Mathematics
Here is a curious grid in Attachement; the sums of the shapes are shown at the end the row and at the base of each column. What is the missing sum?
 
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dkumars said:
hi,

this will prove how strong u are in mathematics
here is a curious grid in attachement; the sums of the shapes are shown at the end the row and at the base of each column. What is the missing sum?
Click to expand...


31 ....................
 
loginrahul said:
why???????????????????????????????????
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Rahul, it is like this

A normal year has 365 days whereas a leap year has 366 days. Every year which is divisible by 4 is called a leap year. Also, every 4th century is a leap year but no other century is a leap year.
1 normal year = 365 days = 52 weeks + 1 day
1 leap year = 366 days = 52 weeks + 2 day
Thus, a normal year has 1 odd day whereas a leap year has 2 odd days.
100 years
= 76 normal years + 24 leap years
= 76*[52 weeks + 1 day] + 24*[52 weeks + 2 day]
= (76*52) weeks + 76 days + (24*52) weeks + 48 days
= 5200 weeks + 124 days
= 5217 weeks + 5 days
i.e. 100 years contain 5 odd days
Similarly,
200 years contain 10 odd days i.e. 3 odd days.
300 years contain 15 odd days i.e. 1 odd days.


400 years contain (20+1) odd days i.e. 0 odd days.
Note that 400 years contain one more leap year.
Also, we have Sunday for 0 odd day, Monday for 1 odd day, Tuesday for 2 odd days, and so on...
Thus, last day of first century is Friday. (5 odd days)
Last day of second century is Wednesday. (3 odd days)
Last day of third century is Monday. (1 odd days)
Last day of forth century is Sunday. (0 odd days)
Since the order is repeating in successive cycles, the last day of a century can not be Tuesday, Thursday or Saturday.
 
anilsinghd said:
4 variables , 7 equations , obviously they are not l.i.
one has to find 4 l.i. equations and solve it! :)
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Yes , Anil my point was , if first three cols total also given then difficulty level is two low. there is no need to solve these equations. total of all rows is equal to total of all cols. so solution is available outside the grid only. let it be (the grid squars) totally black.
 
sanjeev_balyan said:
Yes , Anil my point was , if first three cols total also given then difficulty level is two low. there is no need to solve these equations. total of all rows is equal to total of all cols. so solution is available outside the grid only. let it be (the grid squars) totally black.
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Agreed totally! :)
 
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