# Maths Quiz



## Krazzy Warrior (Dec 1, 2008)

*MATHS QUIZ*
​One friend of mine asked me a question and told me to solve..question is based upon *Simultaneous Linear Equation*..Solve if if u think u r damn good enough..but i think u can't..and don't forget to show the process but first of all u have to solve which is impossible..

Note: We all guys (me and my friend) study in class IX and i m damn sure there are many senior in this forum..so just check this problem...

Here is the question:-

*Find a two digit no. whose sum is equal to 10 and when 18 is subtracted from the no. both the digit of the no. become the same..*..

Look damn easy but do then u will understand...One more question but that I will post afterward..._*

U can even post ur question too...
*_


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## trublu (Dec 1, 2008)

The answer is 73.
7+3=10
73-18=55.

This is what u wanted,right?


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## moverspacker (Dec 1, 2008)

Answers is 73

yes trublu i am agree with you.

thanks for your answer


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## srinivasdevulapally (Dec 1, 2008)

Let the no. be xy

x + y = 10

[(10x + y) - 18 ]/11 = p {Where p is an integer since the two digits are equal it should be divisible by 11}

putting y = 10 - x we have

[(10x + 10 - x) - 18]/11 = p 

x = ( 11p+8 )/9

Substituting p = 5 (no other value of p solves the equation so that x lies b/n 0 and 10)
x = 7; y = 3 hence 73 is the no.


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## Krazzy Warrior (Dec 1, 2008)

U guys rocks....
U can even post ur question too...

Here is another but this is easy but still..

x+1/x = 2 

find value of x^999999919191

damn easy...



srinivasdevulapally said:


> Let the no. be xy
> 
> x + y = 10
> 
> ...



the bold letters proves me that u r excellent in maths...


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## srinivasdevulapally (Dec 1, 2008)

x=1 simply solves the equation and hence answer is (1)^999999919191 = 1

But if u want the solution here it is

x + 1/x = 2

(x^2+1)/x = 2

x^2 - 2x + 1 = 0

(x-1)^2 = 0

Hence x = 1

and (1)^999999919191 = 1

That's all!!!!


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## Krazzy Warrior (Dec 1, 2008)

You are awesome...Give me some question and I will try to solve...

Q) *A lotus from the surface is 2 cm high.whose root is fixed to the surface in a sea.A blow of wind moves the lotus 18 cm away..find the depth of the sea..*


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## srinivasdevulapally (Dec 1, 2008)

Let the depth of the sea be x cm 

(x+2)^2 = x^2 + 18^2

x^2 + 4x + 4 = x^2 + 324

4x = 320

x= 80 cm ---------> depth of the sea

Is it really a sea or a road side pond?

Can u find out what is the last digit(units digit) of the number obtained after this operation?

(1!)^(1!) + (2!)^(2!) + (3!)^(3!) + ...(so on) ... + (100!)^(100!)

Quite simple if u get the logic... try....


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## Krazzy Warrior (Dec 1, 2008)

srinivasdevulapally said:


> (1!)^(1!) + (2!)^(2!) + (3!)^(3!) + ...(so on) ... + (100!)^(100!)
> 
> Quite simple if u get the logic... try....



what is meant by ! sign


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## srinivasdevulapally (Dec 1, 2008)

Factorial
eg: 3! means 3*2*1
     4! means 4*3*2*1


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## Krazzy Warrior (Dec 1, 2008)

huh! what a question...answer plz..

and after this plz post any other question...let me and u rocks in maths..

it is "0" or "1" or I donot know..


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## srinivasdevulapally (Dec 1, 2008)

Its takes time to understand if u don't get the logic but very simple if u are through

1!= 1
2!= 2 
3!= 6
4!= 24
5!= 120

From 5! every no. ends with a zero...
Any no ending with zero raised to any power gives a number which has the last digit as zero.

So all the other no.s from 5! onwards will end in zero so we should calculate only the last digit for the following

(1!)^(1!) + (2!)^(2!) + (3!)^(3!) +(4!)^(4!)

(1!)^(1!) = 1

(2!)^(2!) = 2^2 = 4

(3!)^(3!) = 6^6 =... ends with 6 (since 6 raised to any power gives a no. whose last digit is always 6

(4!)^(4!) = 24^24 = 4^24 * 6^24 =....ends with 6 

{since 6^24 ends with 6 and

4 raised to the odd power  gives no. ending wit 4
4 raised to the even power gives no. ending with 6
eg:
4^1 = 4
4^2 = 16
4^3 = 64
4^4 = 256  and so on...
4^24 ends with 6

6*6 ends with 6}

hence the last digit will be the addition of these last digits
1+4+6+6 = 17  hence ends with *7*

Now a simple question:

There are hens and cows in a farm. When counted they add up to 30 but when their legs are counted they added up to 100. Find the no. of cows and hens in the farm


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## dreamcatcher (Dec 1, 2008)

^^ old trick.. 

Same questions

last digit 1!+2!+3!+4!+5!+6!...100!

last digit 2^12345


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## slugger (Dec 1, 2008)

srinivasdevulapally said:


> There are hens and cows in a farm. When counted they add up to 30 but when their legs are counted they added up to 100. Find the no. of cows and hens in the farm


Cows = 20
Hens = 10


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## dreamcatcher (Dec 1, 2008)

4x+2y=100
x+y=30

x=y-30
4(y-30)+2y=100

120-2y=100
y=10

x+10=30

x=20

So 20 cows and 10 hens


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## The_Devil_Himself (Dec 1, 2008)

Fuall.

:<


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## Faun (Dec 1, 2008)

^^no horses for ya n00b


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## shady_inc (Dec 1, 2008)

Prove that 0.999999....... = 1


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## nvidia (Dec 1, 2008)

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


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## dreamcatcher (Dec 1, 2008)

i^2=i*i cannot be part of the equation.


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## nvidia (Dec 1, 2008)

Why not?


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## Kl@w-24 (Dec 1, 2008)

Hmm... Is this one valid?



> -1 = √-1 * √-1


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## dreamcatcher (Dec 1, 2008)

i^2=i*i

here both sides should cancel out. The same way we can prove 1=2, 1=4, blah balh. Its not feasible.


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## srinivasdevulapally (Dec 2, 2008)

dreamcatcher said:


> 4x+2y=100
> x+y=30
> 
> x=y-30
> ...


 
Dont just go for equations, just use ur brain

For cows or hens, there  will be 2 legs in common. There are 100 legs in total and there are 30 in number. Hence 30 x 2 = 60 legs are common.

But the extra 40 (100 - 60) legs are nothing but cows' legs.

Since each cow has 2 extra legs, 

40/2 = 20 cows

rest (30-20) 10 are hens...That's all !!! 



nvidia said:


> -1 = √1
> -1 = 1
> Why does this happen?


 
 √1 can be (-1)^2 or (1)^2 so u cannot say that  √1 is 1


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## Krazzy Warrior (Dec 2, 2008)

> Its takes time to understand if u don't get the logic but very simple if u are through
> 
> 1!= 1
> 2!= 2
> ...



I more trick(i got that last night):-

(1)^1 + (2)^2.....+(n)^n --->the unit digit will be same as 1+2+.....+n...and to find 1+2+.....+n we apply this formula :- n*(n+1)/2

so for *ur question* 100*(100+1)/2--> 100*101/2 = 505*0* the last digit is zero..therefor the unit digit of (1!)^(1!)....+(100!)^(100!) will be zero....

A question (just u have to explain..you will be shock):-

*Three friends went to a restaurant and had a bill of Rs. 30 ...each decided to pay Rs.10...the owner was impressed by the waiter who served this friend and return Rs. 10 to the waiter telling him to return Rs. 2 to each person (2*3=6) and to keep Rs. 4 with u (waiter).. (6+4=10)...As Rs. 2 is returned so each person paid Rs.8 (10-2) so 8*3=24 (all three together) and Rs. 4 from waiter make 24+4=28 *..*Where is this Rs. 2 gone ?*



shady_inc said:


> Prove that 0.999999....... = 1



totally wrong question...


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## shady_inc (Dec 2, 2008)

Krazzy Warrior said:


> totally wrong question...


Nope..it's true.
0.999..... means 0.9999....forever [till infinity] and this value is exactly equal to 1.I just want someone who can prove it.


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## thewisecrab (Dec 2, 2008)

@nvidia


> -1 = √-1 * √-1
> -1 = √(-1*-1)


This is wrong.



> √-1 * √-1 = -1


This is correct

@krazzy warrior
looks like your concept still needs more shaping up
Regarding that 0.9999999999999 = 1, it is not wrong (simple explanation would mean rounding up of decimal places to 1)
But here is a better method I found on a blog


> First we set:​x=0.999999999……  (infinitely recurring)​Multiplying both sides by 10, we have,10x=9.999999999….. (infinitely recurring)​subtracting the first equation from the second one,10x - x = 9.999999999…… - 0.999999999…….​Therefore,9x = 9​We divide both sides by 9 to get,x = 1​so do we have, from the first statement,1 = .999999999….. ?​



Also see this link which I found on the same blog
*mathforum.org/library/drmath/view/55746.html


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## The_Devil_Himself (Dec 2, 2008)

Krazzy Warrior said:


> *Three friends went to a restaurant and had a bill of Rs. 30 ...each decided to pay Rs.10...the owner was impressed by the waiter who served this friend and return Rs. 10 to the waiter telling him to return Rs. 2 to each person (2*3=6) and to keep Rs. 4 with u (waiter).. (6+4=10)...As Rs. 2 is returned so each person paid Rs.8 (10-2) so 8*3=24 (all three together) and Rs. 4 from waiter make 24+4=28 *..*Where is this Rs. 2 gone ?*


Which 2 rupees you twat?
final:friends paid Rs30 in bills off which Rs20 went to the owner(he returned Rs10),Rs4 as tip to the waiter,and the rest ^ rupees weer distributed amongst friends.I have no Idea what 2 rupees you are blabbing about.

24+4 is wrong as the waiters tip is included in that 24 already.


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## ico (Dec 6, 2008)

shady_inc said:


> Prove that 0.999999....... = 1


Hah easy.....Any 8th standard guy who has learnt to change Recurring decimals to fractions would have solved this.

Let x = 0.999......

10x = 9.999......
9x = 10x - x = 9.999..... - 0.9999 = 9
9x = 9
x = 9/9 = 1

And srinivasdevulapally, Krazzy is still in 9th (if I'm not wrong) and he hasn't learnt about factorials.

Edit: lol thewisecrab solved this already,


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## Pathik (Dec 6, 2008)

nvidia said:


> Check this out -
> i = √-1.
> i² = -1.
> *i² = i * i*
> ...



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

and √1 can be -1 or 1. 

Consider both cases then.


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## QwertyManiac (Dec 6, 2008)

Pathik said:


> 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


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## Pathik (Dec 6, 2008)

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


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## shady_inc (Dec 6, 2008)

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.


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## Pathik (Dec 6, 2008)

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?


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## mrintech (Dec 6, 2008)

*Check out:* *www.contests2win.com/quizzes/49380/BrainMathazz-Quiz

Must Play


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## garfield_56 (Dec 6, 2008)

nvidia said:


> Check this out -
> i = √-1.
> i² = -1.
> i² = i * i
> ...


 


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)*


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## shady_inc (Dec 7, 2008)

Pathik said:


> Hmmm...
> 
> Let there be N people in the room,
> 
> ...


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.


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## slugger (Dec 10, 2008)

*The Birthday Paradox*


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## Edge-of-chaos (Dec 10, 2008)

garfield_56 said:


> This does not happen!!
> 
> Actually, there's a rule dat
> 
> ...




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"


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## Krazzy Warrior (Dec 14, 2008)

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


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## dheeraj_kumar (Dec 14, 2008)

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


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## red_devil (Dec 14, 2008)

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


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## Krazzy Warrior (Dec 15, 2008)

LOL!

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


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## Edge-of-chaos (Dec 17, 2008)

n6300 said:


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


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


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## The Conqueror (Dec 18, 2008)

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


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## MasterMinds (Dec 18, 2008)

ans> E

my ques:-

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


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## jal_desai (Dec 20, 2008)

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?


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## Kl@w-24 (Dec 20, 2008)

^ ^ ^ Hanuman.


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## jal_desai (Dec 20, 2008)

^^ 

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".


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## jal_desai (Dec 29, 2008)

no one??


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## Sykora (Dec 30, 2008)

9 slashes.


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## kanishka (Dec 30, 2008)

jal_desai said:


> ^^
> 
> Another one:
> 
> ...



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 .


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## Sykora (Dec 30, 2008)

(4/4 + 4) * 4


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## kanishka (Dec 30, 2008)

^good one buddy!

Right!


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