Maths Quiz

[QwertyManiac]

Originally Posted by Pathik

It can be i*i or even -i * -i

and √1 can be -1 or 1.

Consider both cases then.

That's the reason why

[Pathik]

^ OMG is it or is it not? No idea. Suck in math

[shady_inc]

This question is asked in this month's LFY magazine, and I still haven't found a solution....
You enter a room containing N people.Find probability that atleast someone in the room has his birthday on the same date as yours, assuming no leap years and the probability of any person's birthday being on any one of the 365 days is equally likely.

[Pathik]

Hmmm...

Let there be N people in the room,

The probability of the person 1 having the same birthday as any other is 0/365.

The probability of the person 2 having the same birthday as any other is 1/365.

The probability of the person 3 having the same birthday as any other is 2/365.

Going this way, the probability of any on the N persons having a birthday as any other is
1+2+3...+(N-1)/365?

Is it right?

[mrintech]

Check out: http://www.contests2win.com/quizzes/...inMathazz-Quiz

Must Play

[garfield_56]

Originally Posted by nvidia

Check this out -
i = √-1.
i² = -1.
i² = i * i
-1 = √-1 * √-1
-1 = √(-1*-1)
-1 = √1
-1 = 1
Why does this happen?

This does not happen!!

Actually, there's a rule dat

√(a*b)=√a * √b , if and only if one or both of a & b are +ve (i.e. >0)!!!



So,

√-1 * √-1 is not equal to √(-1*-1)

[shady_inc]

Originally Posted by Pathik

Hmmm...

Let there be N people in the room,

The probability of the person 1 having the same birthday as any other is 0/365.

The probability of the person 2 having the same birthday as any other is 1/365.

The probability of the person 3 having the same birthday as any other is 2/365.

Going this way, the probability of any on the N persons having a birthday as any other is
1+2+3...+(N-1)/365?

Is it right?

Nope...If there are, say 30 people in room, then 1+2+3+4.....+(30-1)/365 = 1.917.... and probability of anything can't exceed 1.

[slugger]

The Birthday Paradox

[Edge-of-chaos]

Originally Posted by garfield_56

This does not happen!!

Actually, there's a rule dat

√(a*b)=√a * √b , if and only if one or both of a & b are +ve (i.e. >0)!!!



So,

√-1 * √-1 is not equal to √(-1*-1)

Thats right! square root of a negative number is undefined except for that of -1 which is "i" hence the basis for the concept of the "imaginary number"

[Krazzy Warrior]

If :- x^2 + y^2 + z^2 = xy+yz+xz
Prove:- x=y=z

[dheeraj_kumar]

^^ dont bring in your homework textbook problems... i remember doing that...

[red_devil]

^^ lol from maths quiz to a maths homework solver, this thread is going places

[Krazzy Warrior]

LOL!

I know the answer..it is a quiz...

[Edge-of-chaos]

Originally Posted by n6300

^^ lol from maths quiz to a maths homework solver, this thread is going places

Lets evolve it to "Dump your assignments here!!!"

[The Conqueror]

okay this was asked today in Maths Olympiad :
ok try to answer this maths question...itz easy :

Each ship that passes through the Panama Canal requires about 52 million gallons of water to move the ship through the canal from the atlantic ocean to the pacific ocean. If 36 ships passed through the canal, moving from the atlantic ocean to the pacific ocean ,which is closes to the number of gallons of water that was required??

A) 1.9 x 10^5 gallons B) 1.9 x 10^9 Gallons C) 8.8 x 10* gallons D) 8.8 x 10^9 gallons E) None of these

[MasterMinds]

ans> E

my ques:-

prove that any number raised to power zero is one..!!!

[jal_desai]

There is an island. Its inmates speak only two sentences at a time. Out of those two sentences, one is right and one is wrong. You go to that island and meet 4 inmates Ram, Sita, Laxman and Hanuman.

You: Whose wife is Sita?
Ram: Sita is the wife of no one. Sita is married to my brother.
Sita: Laxman is my husband. I am married to Hanuman.
Laxman: Sita is unmarried. Ram is not my brother.
Hanuman: I don't talk with strangers. Sita is not unmarried.

pretty simple. Whose wife is Sita?

[Kl@w-24]

^ ^ ^ Hanuman.

[jal_desai]

^^

Another one:

Imagine you have gone to China (from Chandni Chowk, ofcourse) and you have learnt "talwarbaazi" over there. Now you are faced with a dragon and you are supposed to kill it. The dragon has 3 heads and 3 tails. A dragon is considered dead if ALL heads and ALL tails are chopped off. But this dragon is nasty. Here are the conditions:

1) If 1 head is chopped off, 1 extra head grows.
2) If 1 tail is chopped off, 2 extra tails grow.
3) If 2 tails are chopped off together, 1 head grows.
4) If 2 heads are chopped off together, nothing grows.

So what is the minimum number of 'slashes' required to kill the dragon?

this is also preety simple. remember "MINIMUM NUMBER OF SLASHES".

[jal_desai]

no one??

[Sykora]

9 slashes.

[kanishka]

Originally Posted by jal_desai

^^

Another one:

Imagine you have gone to China (from Chandni Chowk, ofcourse) and you have learnt "talwarbaazi" over there. Now you are faced with a dragon and you are supposed to kill it. The dragon has 3 heads and 3 tails. A dragon is considered dead if ALL heads and ALL tails are chopped off. But this dragon is nasty. Here are the conditions:

1) If 1 head is chopped off, 1 extra head grows.
2) If 1 tail is chopped off, 2 extra tails grow.
3) If 2 tails are chopped off together, 1 head grows.
4) If 2 heads are chopped off together, nothing grows.

So what is the minimum number of 'slashes' required to kill the dragon?

this is also preety simple. remember "MINIMUM NUMBER OF SLASHES".

I got the answer.It is damn simple.

1)In First Shot he cut 2 heads together so nothing grows.
Now dragon has 1 head , 3 Tails .
2)next shot he cut 2 tails , so another head grows.
Now 2 heads , 1 tail.
3)He cuts 1 tail so that 2 tails grow.
now 2 heads , 2 tails
4)he cut once again 1 tail so that 2 grows
now 2 heads , 3 tails
5)another 1 tail gone so that 2 tails grow
now 2 heads , 4 Tails
6)He cut 2 tails , 1 head grows
Now 3 heads , 2 tails
7)He cut another 2 tails so 1 head grows
Now 4 head and no Tail
8)He cut 2 heads nothing grows.
Now 2 heads and no tail
9)Again he cut 2 heads and nothing grows.
Now NO HEAD AND NO TAIL!!

Toh mar ke hi mana

Let me shoot you a question.

4_4_4_4=20



using: addition , substraction , Multiplication & Division .Fill in the blanks .

[Sykora]

(4/4 + 4) * 4

[kanishka]

^good one buddy!

Right!