A New Beginning..

So people...
Here is the first question..

Its simple but is a very famous one..please do not search for the answers on google..I am sure you'll find it there..so first just try yourself.

So the question is.. You have to arrange eight queens on the chess board in such a way that no queen can kill any other
See.. simple and staright
but there is one more thing..if you can give an ans to above then try to come up with the no of ways in which such a thing is possible.

Sorry Sorry Sorry...

I and the original creators of the blog are really sorry...the regular readers know the reason very well and for the ones who are visiting for the first time,the reason for all the apologies is that we (the creators of this blog) have been underground for a long time now.
You see I came up with the idea of this blog as a tool to help me and my friends to discuss about the day to day problems we faced during the course of preparing for entrance examinations ,but seeing the response we got, i guess it grew to be much more than that..THANX to all the people who have contributed.Well the place has been dead for a while,well actually a very very long time(the last post was in aug 2005) which is an absolute no no in the blogging world.
So I hope that like a pheonix this blog will grow from its ashes.After all the ashes still are there...safe and sound.
So I end this post hoping that it(the blog) lives yet again...and I promise to come up with as many posts as possible.

P.S. The rules still remain the same i.e any questions from the science world just...
NO CHEMISTRY
Also thanks to MANDOLUX for providing such a beautiful background.

Ek rupaiyya!!!

Well, I have question for you... The question is not so costly..... Only of Rs. 1 /-

Once A, B & C went to a restaurant. After having tea and light refreshment they asked for bill, the amount of bill was Rs. 30 /- They decided to pay the bill

amount on equal share. Everybody took out Rs. 10 /- from their pocket and paid the bill. They were yet sitting and gossiping in restaurant, Suddenly the

waiter realized that the amount of their bill was Rs. 25 /- not Rs. 30 /- He took Rs. 5 /- from counter and decided to return them. Then he thought they are 3

how will they share Rs. 5 /- ? So he has decided to keep Rs. 2 /- to him and return only Rs. 3 /- to them.

He came to their table and returned them Rs. 3 /- with apology. Everybody took Rs. 1 /- and put in their pockets.

Now my question is first time everyone paid Rs. 10 /- later they get Rs 1 /- refunded. So everybody paid Rs. 9 /-

Rs. 9 * 3 = Rs. 27 /-

The waiter put Rs 2 /- in his pocket.

Rs. 27 + Rs 2 = 29

So...... Where is the remaining Rs. 1 /-

Find out where is the remaining Rs. 1 /- OR everybody has to pay Rs. 1 /- to me :P

Whats up people??

The original contributors of the blog seem to have vanished away..What's up people?? After the exams it seems no one is interested in Maths. Had too much of it already??
I think this blog also neds a new look. Though black is a good colour but CHANGE is really necessary. Something bright would liven up the blog.
Hope to see some good interesting stuff too.

Sum Word

Following is a problem really interesting..... lets see if it wacks your brain or not...( mite not... if you are too good!!)


N O S I E R
A S T R A L
-------------
7 2 5 6 1 3

there has to be a '+' sign in between that is N O S I E R is being added to A S T R A L
I have removed it as it was causing problems , the letters were not coimng in stright lines i.e. A below N ; S below O ; so on and similarly 7 being the sum of N and A etc..
NOSIER and ASTRAL both denote individual numbers ( ie some lakhs) where each letter stands fro some digit

In the additon above, the sum represents a word.
1) Each letter represents a different digit.
2) No letter represents zero.

Which word is represented by : ' 7 2 5 6 1 3 ' ?

heres another problem... more interesting than the previous

APARTMENTS

A , B , C and D each live in an apartment.

1) Their apartments are arranged like this :
-------------------
| a | b | c | d | ---------> EAST
-------------------
alrite the figure is a bit distorted ... but anyways the idea is tat a, b , c , d are the 4 apartments in which A , B , C , D are residing ( NOT RESPECTIVELY) , the direction too has been shown

2) One of the four is the landlord
3) If C's apartment is not next to B's then A is the landlord and resides in a.
4) If A's apartment is east of C's then D is the landlord and resides in d.
5) If B's apartment is not next to D's then C is the landlord and resides in c.
6) If D's apartment is east of A's then B is landlord and resides in b.

Who is the landlord and what is the arrangemnt of the apartments ( ie who is in which aaprtment) ?????

try this one... its simple though

you have to guess the house number from the following clues:

1) if the house number is a multiple of 3 , then it is a number from 50 through 59.
2) if the house number is not a multiple of 4 , then it is a number from 60 through 69.
3) if the house number is not a multiple of 6 , then it is a number from 70 through 79.

so what is the house number being talked about?

VICTORY!!!!!!!!!!!!!!!!!!!!!!!!

Sweet victory... months of agony and mental torture finally comes to an end... if you people are still wondering well Im just talkin about the puzzle involving 12 coins having 1 defective coin ( for details luk at d q... its below this one only)... well ive cracked it... finally!!!!!

alrite here goes the soln
let the 12 coins be divided into 3 groups of 4 coins each such that
group A - A1, A2, A3, A4 COINS
group B - B1, B2, B3, B4 COINS AND SO ON FOR GROUP C

WEIGHING (1)

MEASURE A AGAINST B
1) A = B
WELL THIS ONE IS SIMPLE...
THE WRONG ONE LIES IN GROUP C...
TAKE C1 , C2..
WEIGH AGAINST ANY 2 COINS OF GROUP A AND B ( SINCE THEY ARE ALL NORMAL) { WEIGHING 2}

3 RD WEIGHIN :
IF UNEQUAL THAT MEANS EITHER C1 OR C2 IS DEFECTIVE..... AGAIN WEIGH OF THEM AGAINST 1 NORMAL COIN AND YOULL GET THE DEFECTIVE COIN.....
IF EQUAL THEN DEFCTIVE COIN LIES IN OTHER 2 C COINS IE C3 OR C4... FOLLOW THE SAME PROCEDURE ... WIEGHING AGANIST 1 RIGHT COIN ETC ETC

2) A > B ...( CAN BE THE OTHER WAY ALSO... BUT JUST FOR INSTANCE)....
WELL THIS MEANS C1- C4 ARE ALL NORMAL COINS

WEIGHING (2)
PAN 1 CONSISTS OF - A1 , C1 , C2 , C3
PAN 2 CONSISTS OF - B1 , B2 , A2 , A3

SO WE 'RE LEFT WITH ( THE OTHER POSSIBLE DEFCTIVE COINS) - A4 , B3 , B4

AFTER THIS THERE ARE 3 CASES
A) PAN 1 = PAN 2
THAT MEANS DEFCTIVE IS AMONG A4 , B3 , B4 ... OUT OF WHICH EITHER A4 IS HEAVY OR ONE OF B3 AND B4 IS LIGHT SINCE WE ALREADY KNOW ( BY 1 WEIGHING ) THAT A > B

SO 3RD WEIGHING WILL BE AS FOLLOWS...
PAN 1 - A4 , B3
PAN 2 - C1, C2 ( IE ANY OF THE NORMAL ONES)
AGAIN 3 POSSIBLITIES,,,,
1) PAN 1 = PAN 2..
SO A4 AND B3 ARE NORMAL... IE B4 IS DEFCTIVE

2) PAN 1 > PAN 2
SINCE WE KNOW THAT OUT OF A4 AND B3 ... A4 CAN POSSIBLY BE THE HEAVY ONE ( AS BELONGS TO A... WHICH WAS HEAVIER)... HENCE A4 IS THE DEFECTIVE COIN

3) PAN 1 < PAN 2...
JUST LIKE THE ABOVE CASE,,,, THIS TIME THE DEFECTIVE COIN WOULD BE B3 ( BELONGS TO B GROUP.. WHICH WAS LIGHTER ETC).......

NOW GOING BACK TO THE INTIAL PAN 1 AND PAN 2
IF PAN 1 NOT EQUAL TO PAN 2
IE DEFECTIVE COIN IS AMONG A1 , B1 , B2 , A2 , A3

B) PAN 1 > PAN 2
SINCE THIS IS THE PREV RECORDING ONLY ( FOR A > B)... THIS MEANS MOVING A2 AND A3 TO THE OTHER SIDE HAD NO EFFECT ... THE DEFECTIVE COIN REMAINS IN ITS ORIGINAL POSITION... SO IT HAS TO BE AMONG A1 , B1 , B2 OUT OF WHICH WE KNOW ( THROUGH WEIGHIN 2 ) THAT EITHER A1 IS HEAVY OR ONE OF B1 AND B2 IS LIGHT AS PAN 1 > PAN 2..
NOW THE DEFTIVE COIN CAN BE OBTAINED BY THE SAME PROCEDURE AS MENTIONED BEFORE ( FOR PAN 1 = PAN 2)

C) PAN 1 < PAN 2
THIS MEANS MOVING A2 AND A3 BROUGHT ABOUT A CHANGE IN THE POSITION OF THE DEFCTIVE COIN SO THE DEFCTIVE IS EITHER A2 OR A3.... WEIGH ANY OF THEM AGAINST A NORMAL ONE ... IF EQUAL THEN THE OTHER ONE IS THE DEFECTIVE COIN ( THE DEFECTIVE COIN WOULD BE HEAVIER PLS NOTE AS PAN 2 > PAN 1) IF NOT EQUAL THEN OBVIOUSLY THAT VERY COIN IS DEFECTIVE

SO I HOPE THE SOLUTION IS FULL PROOF ... ANY QUERIES OR LOOPHOLES PLS DO COMMENT

a prob about coins

alright this peoblem might sound familiar... but somebody please post the solution asap
the q is
there are 12 coins... all are samilar but one..
one has a mass defect ie could be more or less
you have a weighing balance( for convienence take any balance... phy ... tarazu... electronic etc etc)
and got only 3 weighings to find that defective coin....
anybody who comes with the soln please post it.... and if only in certain ciricumstances( like assuming the weight of the coins).. then do mention it
happy solving