# Post mathematic related questions here



## rst (Apr 16, 2013)

If you have any mathematic related question 
post here
we can try to solve them

I know yahoo answer is good platform for such things.
But there is problem of uploading images in yahoo answer
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maths symbols
 £ ω ∴ ∂ Φ γ δ μ σ Є Ø Ω ∩ ≈ ≡ ≈ ≅ ≠ ≤ ≥ • ~ ± ∓ ∤ ◅∈ ∉ ⊆ ⊂ ∪ ⊥ ô ∫ Σ → ∞ Π Δ Ψ Γ ∮ ∇∂ √ °α β γ δ ε ζ η θ ι κ λ ν ξ ο π ρ σ τ υ φ χ ψ 
x⁰ ¹ ² ³ ⁴ ⁵ ⁶ ⁷ ⁸ ⁹ ⁺ ⁻ ⁿ 
 x₀ ₁ ₂ ₃ ₄ ₅ ₆ ₇ ₈ ₉ ₊ ₋

⅓ ⅔ ⅕ ⅖ ⅗ ⅘ ⅙ ⅚ ⅛ ⅜ ⅝ ⅞


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## thetechfreak (Apr 16, 2013)

*Re: mathematic related questions*

Yeah. I also want to help(should be able to do school level sums  )


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## rst (Apr 16, 2013)

*Re: mathematic related questions*

thanks

Can you solve this question
find missing letter ???




(Also you can ask your mathematics related questions or queries
 we will try to solve them)


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## thetechfreak (Apr 16, 2013)

*Re: mathematic related questions*

Is the answer "L" ?

I'm bad at this


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## rst (Apr 16, 2013)

*Re: mathematic related questions*

Answer is "Z"

--------------------------------------------
Its really hard question

in upper and lower box 
there are 3 continuous letters E,F,G (then 2 letter gap) J,K (then 3 letters gap) O

similarly in left and right box
there are 3 continuous letters P,Q,R (then 2 letter gap) U,V (then 3 letters gap) "Z"


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## digit.sh (Apr 16, 2013)

*Re: mathematic related questions*

Solve this one 

3 kids plucked a total of 10 flowers. In how many ways they can distribute the flowers among themselves?

Like:
way 1) kid no. 1 takes 0 flower,  kid 2 takes 0 flower  and kid 3 takes 10 flowers.
way 2) kid no. 1 takes 0 flower,  kid 2 takes 1 flower  and kid 3 takes 9 flowers.
.
.
.
and so on.

How many ways in total?


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## rst (Apr 16, 2013)

*Re: mathematic related questions*

I think using permutation P(10,3) ways
P(10,3)= 10 ! / 7!
= 10 X 9 X8
=720 ways

IS it correct ??


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## digit.sh (Apr 16, 2013)

*Re: mathematic related questions*

No.
Think like this:
You are given 10 identical coins and three boxes. Your task is to distribute the coins in the three boxes in the following manner:
left box: x coins,  middle box: y coins,  right box: z coins,   such that x+y+z=10

This is definitely not P(10,3).


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## rst (Apr 16, 2013)

*Re: mathematic related questions*

Ok
Lets try this way
possibilities are
1) 0,0,10 
2) 0,1,9
3)0,2,8
.
.
.
11) 0,10,0

here I fixed 0 in first place 

similarly I can fix 1 in first place
1) 1,0,9
2) 1,1,8
3)1,2,7
.
.
.
10) 1,9,0

similarly I can fix 2 in first place
1) 2,0,8
2) 2,1,7
3)2,2,6
.
.
.
9) 2,8,0

so in this way 
we get = 11+10+9...+1
=66 ways


Is it correct ??

If not,plz give correct answer


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## digit.sh (Apr 16, 2013)

*Re: mathematic related questions*



rst said:


> Ok
> Lets try this way
> possibilities are
> 1) 0,0,10
> ...




Correct. 
But there is a better and easy way to solve problems like these. I shall write about that if you are interested. Also, this particular problem is the base problem of the family of problems derived from it. While you could solve the base problem using the brute force method, its much harder to solve the other problems of this family using this method.


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## Vignesh B (Apr 17, 2013)

*Re: mathematic related questions*



digit.sh said:


> Correct.
> But there is a better and easy way to solve problems like these. I shall write about that if you are interested. Also, this particular problem is the base problem of the family of problems derived from it. While you could solve the base problem using the brute force method, its much harder to solve the other problems of this family using this method.


(n+r-1)C(r-1) where n is 10 and r is 3. Is it correct?


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## rst (Apr 17, 2013)

*Re: mathematic related questions*



digit.sh said:


> Correct.
> But there is a better and easy way to solve problems like these. I shall write about that if you are interested. Also, this particular problem is the base problem of the family of problems derived from it. While you could solve the base problem using the brute force method, its much harder to solve the other problems of this family using this method.



yeah, 
everyone prefer short method



Vignesh B said:


> (n+r-1)C(r-1) where n is 10 and r is 3. Is it correct?


It is also giving the same answer


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## NoasArcAngel (Apr 17, 2013)

*Re: mathematic related questions*

1. if i divide a square into 8 boxes how many squares will i get? (max possible)

2. if i divide a square into 8 equal squares how many squares will i get? 

EDIT :

not counting original square in both


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## digit.sh (Apr 17, 2013)

*Re: mathematic related questions*



Vignesh B said:


> (n+r-1)C(r-1) where n is 10 and r is 3. Is it correct?



Yes this one. But why is it correct? The explanation is more interesting.  

@rst, vignesh, can you explain it?


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## rst (Apr 17, 2013)

*Re: mathematic related questions*



NoasArcAngel said:


> 1. if i divide a square into 8 boxes how many squares will i get? (max possible)
> 
> 2. if i divide a square into 8 equal squares how many squares will i get?
> 
> ...


I don't understand the first question
By boxes , do you mean rectangle ??


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## Niilesh (Apr 17, 2013)

*Re: mathematic related questions*



NoasArcAngel said:


> 1. if i divide a square into 8 boxes how many squares will i get? (max possible)
> 
> 2. if i divide a square into 8 equal squares how many squares will i get?
> 
> ...


2 --> 2^2 + ....... + 8^2 = 203



digit.sh said:


> Yes this one. But why is it correct? The explanation is more interesting.
> 
> @rst, vignesh, can you explain it?



I Can explain it


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## Vignesh B (Apr 17, 2013)

*Re: mathematic related questions*



rst said:


> I don't understand the first question
> By boxes , do you mean rectangle ??


A box is a 4 sided close figure! 



Niilesh said:


> 2 --> 2^2 + ....... + 8^2 = 203


Nope, wrong answer.


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## rst (Apr 17, 2013)

*Re: mathematic related questions*



NoasArcAngel said:


> 1. if i divide a square into 8 boxes how many squares will i get? (max possible)
> 
> 2. *if i divide a square into 8 equal squares how many squares will i get?
> *
> ...



Is it possible to divide a square into 8 equal square?


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## Niilesh (Apr 17, 2013)

*Re: mathematic related questions*



Vignesh B said:


> Nope, wrong answer.


Oh sorry i thought it was 8*8
New ans 4+9+16=29



rst said:


> Is it possible to divide a square into 8 equal square?


Think of it as a chess board


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## rst (Apr 17, 2013)

*Re: mathematic related questions*



Niilesh said:


> Think of it as a chess board



But chess board is not square

In other words, we cannot divide a square into 8 equal squares
so question is meaningless


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## Niilesh (Apr 17, 2013)

*Re: mathematic related questions*

A question for you guys

Take 5 points in a plane so that no two of the straight lines joining them are parallel, perpendicular or coincident. From each point perpendeculars are drawn to all the lines joining the other 4 pts. Then find total no. of intersection of these perpendiculars.


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## Vignesh B (Apr 17, 2013)

*Re: mathematic related questions*



Niilesh said:


> Oh sorry i thought it was 8*8
> New ans 4+9+16=29


Correct.



rst said:


> But chess board is not square
> 
> In other words, we cannot divide a square into 8 equal squares
> so question is meaningless


Exactly!


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## Niilesh (Apr 17, 2013)

*Re: mathematic related questions*

I think chess board is a square
yup you are right it is not possible to divide it in 8 squares,  I thought he ment 8*8=64 squares


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## Vignesh B (Apr 17, 2013)

*Re: mathematic related questions*



Niilesh said:


> A question for you guys
> 
> Take 5 points in a plane so that no two of the straight lines joining them are parallel, perpendicular or coincident. From each point perpendeculars are drawn to all the lines joining the other 4 pts. Then find total no. of intersection of these perpendiculars.


We can use nC2 right? So, is 10 the answer?


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## rst (Apr 17, 2013)

*Re: mathematic related questions*



Niilesh said:


> A question for you guys
> 
> Take 5 points in a plane so that no two of the straight lines joining them are parallel, perpendicular or coincident. From each point perpendeculars are drawn to all the lines joining the other 4 pts. Then find total no. of intersection of these perpendiculars.



315 point of intersection


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## Vignesh B (Apr 17, 2013)

*Re: mathematic related questions*



rst said:


> 315 point of intersection


How? . . .


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## rst (Apr 17, 2013)

*Re: mathematic related questions*



Vignesh B said:


> We can use nC2 right? So, is 10 the answer?




there are 5 points
for drawing a line we require 2 points
so using  nC2 , we can draw 10 lines

but it is not related with point of intersection



Vignesh B said:


> How? . . .



it requires lot of explanation

its really a tough question


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## Niilesh (Apr 17, 2013)

*Re: mathematic related questions*



rst said:


> 315 point of intersection


correct,  how much time it took you to get the anwser


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## Vignesh B (Apr 17, 2013)

*Re: mathematic related questions*



rst said:


> there are 5 points
> for drawing a line we require 2 points
> so using  nC2 , we can draw 10 lines
> 
> ...


Ok, my bad.
Care to explain the answer  in brief? 

One from my side too - 
How many pairs of integers are there such that a^2 + 6b^2 and b^2+6a^2 are perfect squares?
Source : I found it in one of my books.


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## rst (Apr 17, 2013)

*Re: mathematic related questions*



Vignesh B said:


> One from my side too -
> How many pairs of integers are there such that a^2 + 6b^2 and b^2+6a^2 are perfect squares?
> Source : I found it in one of my books.



you all are asking tough and time consuming question
Is there any way, other than hit and trial method for this question

I think (0,0) is the only pair (by hit and trial method)


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## Vignesh B (Apr 17, 2013)

*Re: mathematic related questions*



rst said:


> you all are asking tough and time consuming question
> Is there any way, other than hit and trial method for this question
> 
> I think (0,0) is the only pair (by hit and trial method)


Correct.
Only the trivial solution (0,0) works.
Suppose there is a nontrivial solution (a, b). Without loss of generality, we may assume it is a minimal solution, say in the sense that |a|+|b| is as small as possible.
The only squares mod 4 are 0 and 1.
If a is even and b is odd, then
a^2 + 6b^2 = 2 (mod 4),
so a^2 + 6b^2 is not a square.  Similarly, if a is odd and b is even we arrive at a contradiction by using the assumption that b^2 + 6a^2 is a square.
If instead a is odd and b is odd, then 
a^2 + 6b^2 +3 ( mod 4),
so again we obtain a contradiction.
The only possibility then is that both a & b are even.  But if you look at a/2, b/2 then one easily sees that both
(a/2)^2 + 6(b/2)^2
(b/2)^2 + 6(a/2)^2
are squares (since the squares a^2 + 6b^2 and b^2 + 6a^2 are both even, hence divisible by 4).  This contradicts the minimality of our original solution.


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## rst (Apr 17, 2013)

*Re: mathematic related questions*

objective question



Also plz show some work

(Also you can ask your mathematics related questions or queries
 we will try to solve them)


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## Niilesh (Apr 17, 2013)

*Re: mathematic related questions*



rst said:


> objective question
> View attachment 10057
> 
> Also plz show some work



I am getting ans as 5  

can someone point out mistake in the following solution? 
PQC is concurrent to PQA by SSS
ABQ is concurrent to CDP by AAS

this means that PC=AP=AQ=QC and PD=QB, now I equated the area and got PD=0=PD.  so PQ will be the diagonal so it will be equal to five


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## rst (Apr 17, 2013)

*Re: mathematic related questions*



Niilesh said:


> I am getting ans as 5
> 
> can someone point out mistake in the following solution?
> PQC is concurrent to PQA by SSS
> ...



< B= <D (each 90)
AB=CD (opposite sides of rectangle)
what is another angle


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## Niilesh (Apr 17, 2013)

*Re: mathematic related questions*



rst said:


> < B= <D (each 90)
> AB=CD (opposite sides of rectangle)
> what is another angle


since <pcq = <qap
<qab=<pcd


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## rst (Apr 17, 2013)

*Re: mathematic related questions*



Niilesh said:


> since <pcq = <qap
> <qab=<pcd



right (then it should be ASA)

I also don't understand  * I equated the area and got PD=0=PD
*


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## Niilesh (Apr 17, 2013)

*Re: mathematic related questions*



rst said:


> right (then it should be ASA)
> 
> I also don't understand  * I equated the area and got PD=0=PD
> *


I found the area of the 4 triangles and equated its sum to 4*3=12


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## rst (Apr 17, 2013)

*Re: mathematic related questions*



Niilesh said:


> I found the area of the 4 triangles and equated it's sum to 4*3=12



OK
how did you find area of ΔPCQ and ΔPAQ


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## Niilesh (Apr 17, 2013)

*Re: mathematic related questions*



rst said:


> ok
> how did you find area of Δpcq and Δpaq


1/2 * pa * 3


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## rst (Apr 17, 2013)

*Re: mathematic related questions*

Then by solving 
you will get , 24=24
so by this method you will not be able to find anything
It doesnot mean PD=0


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## Niilesh (Apr 17, 2013)

*Re: mathematic related questions*



rst said:


> Then by solving
> you will get , 24=24
> so by this method you will not be able to find anything
> It doesnot mean PD=0


3*PD + PA*3 = 12
PD + PA = 4
EDIT : You are right, actually i had put PD=PA-4 instead of 4-PA so i got the wrong result


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## rst (Apr 18, 2013)

*Re: mathematic related questions*



rst said:


> objective question
> View attachment 10057
> 
> Also plz show some work



ans is 15/4

Can anyone explain ??


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## Niilesh (Apr 18, 2013)

*Re: mathematic related questions*



rst said:


> ans is 15/4
> 
> Can anyone explain ??



I can  

I am on mobile right now so I will just give you a hint. take PD as 'x' then find the length of hypotenuse and then find the value of X by qc+QB =4,  I think you will be able to figure out the rest of the solution


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## rst (Apr 18, 2013)

*Re: mathematic related questions*

NO

It can't be solved in this way


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## Niilesh (Apr 18, 2013)

*Re: mathematic related questions*

*i.imgur.com/9BcPIWp.jpg
In AQB, AQ= √9+x² = QC
now, x+√9+x² = 4
9+x² =16+x²-8x
x=7/8
now, 
PQ² =  3² + (√9+x² - x)²
PQ= 3*5/4=15/4


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## rst (Apr 18, 2013)

*Re: mathematic related questions*



Niilesh said:


> *i.imgur.com/9BcPIWp.jpg
> In AQB, AQ= √9+x² = QC
> now, x+√9+x² = 4
> 9+x² =16+x²-8x
> ...



Absolutely right

But bolded above step is complicated
instead of it you can use:-
PQ² = PR² + QR² (here R is used where dotted line meet PA)
PQ² = (4-2x)² + 3²
PQ² = (4-7/4)² + 9
PQ² = (81/16) + 9
pQ=15/4


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## Niilesh (Apr 18, 2013)

*Re: mathematic related questions*



Vignesh B said:


> Correct.
> Only the trivial solution (0,0) works.
> Suppose there is a nontrivial solution (a, b). Without loss of generality, we may assume it is a minimal solution, say in the sense that |a|+|b| is as small as possible.
> *The only squares mod 4 are 0 and 1.*
> ...


umm.. What? Can someone explain that?


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## rst (Apr 18, 2013)

*Re: mathematic related questions*



Niilesh said:


> umm.. What? Can someone explain that?


Here square means numbers like 1,4,9,16,25,36,49,................
when you divide above mentioned numbers by 4 then remainder will be either 0 or 1
thats why, *The only squares mod 4 are 0 and 1.*


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## Niilesh (Apr 18, 2013)

*Re: mathematic related questions*



rst said:


> Here square means numbers like 1,4,9,16,25,36,49,................
> when you divide above mentioned numbers by 4 then remainder will be either 0 or 1
> thats why, *The only squares mod 4 are 0 and 1.*


I can't understand the proof,  can you prove that only pair of a and b that are even are 0,0?(in that question)


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## rst (Apr 18, 2013)

*Re: mathematic related questions*



Niilesh said:


> I can't understand the proof,  can you prove that only pair of a and b that are even are 0,0?(in that question)


this proof is quite good
lets see this proof in simple language

 every square number will give remainder 0 and 1 if it is divisible by 4 
this means *If a number gives  remainder other than 0 and 1 then it is not square number *
we will this result
but it doesnot means that if remainder is o or 1 then it is perfect square
for example 40 gives remainder 0 when divisible by 4 (but it is not perfect square)

 Now in a^2 + 6b^2 
a ,b can be even or odd

case 1
a is even and b is odd
lets take a=2 and b=3
then a^2 + 6b^2 =58
58 will give remainder 2 when divided by 4
so a^2 + 6b^2  is not perfect square (as remainder is other than 0 and 1)



case 2
a is odd and b is even
similar to case one

case 3
both a and b are odd
let a=1 ,b=3
then a^2 + 6b^2 =1+18=19
19 will give remainder 3 when divided by 4
so a^2 + 6b^2  is not perfect square (as remainder is other than 0 and 1)


case 4
both a and b are even
let a=2,b=4
then a^2 + 6b^2 =100
so a^2 + 6b^2  is  perfect square (as 100 is perfect square)

let a=2,b=4
then (b)^2 + 6(a)^2 =16+24=40
 (b)^2 + 6(a)^2 is not perfect square

so a^2 + 6b^2  and (b)^2 + 6(a)^2 is not perfect square for a and b both even [ (0,0) is exception]

so answer is (0,0)


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## Niilesh (Apr 18, 2013)

*Re: mathematic related questions*

I only needed solution for case 4, you just took some random values of a and b :-\
can I get a proof in general form ? (like take a and b as 2k and 2p where k and p are integers)


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## rst (Apr 18, 2013)

*Re: mathematic related questions*

If we want to proof that a theorem or result is not correct
Then in such situation even examples are enough

But If we want to proof a theorem or result is correct
Then  we have to prove it with general terms
---------------------------------------------------------
Anyway I didn't try to prove any thing 
I just tried to explain the proof with examples


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## Niilesh (Apr 18, 2013)

*Re: mathematic related questions*

I know but you proved that it is not valid for all even no. but you didn't prove that it will not be possible for some no. s


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## rst (Apr 19, 2013)

*Re: mathematic related questions*

I didn't try to proof the problem
proof is already given by Vignesh B
I just tried to explain the proof with examples

----------------------------------------------------------------------------------------------
*NEW OBJECTIVE QUESTION*



Also plz explain your answer

(Also you can ask your mathematics related questions or queries
 we will try to solve them)


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## Niilesh (Apr 19, 2013)

*Re: mathematic related questions*

It seems too easy :/
My ans - 50Kg
Solution- 
Note: Assuming 99/98% is given weight by weight
Original condition - 1 Kg real matter + 99 Kg water
New condition - 1 Kg real matter + 'x' Kg water
Now it is given that
x/(x+1) = 98/100
2x=98
x=49
So ans = x+1= 50Kg


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## rst (Apr 19, 2013)

*Re: mathematic related questions*

Absolutely right
great work

-----------------------------------------
*NEW OBJECTIVE QUESTION*


Also plz explain your answer

(Also you can ask your mathematics related questions or queries
 we will try to solve them)


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## Niilesh (Apr 19, 2013)

*Re: mathematic related questions*

Ans - 9
Sol.- use arithmetic mean of mth power is greater that mth power of arithmetic mean( if M belongs to real no - [0,1])


Spoiler



I know you will have to read it more than once to understand what's written


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## rst (Apr 19, 2013)

*Re: mathematic related questions*

Answer is correct

But i didn't understand the explanation
----------------------------------------
plz explain in simple way as you did in previous question


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## Niilesh (Apr 19, 2013)

*Re: mathematic related questions*

Alternate method for you  
I know my writing is not that good 
*i.imgur.com/RuGGjlT.jpg


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## rst (Apr 19, 2013)

*Re: mathematic related questions*

great !!
Thanks a lot 



Niilesh said:


> I know my writing is not that good


Its better than mine

-------------------------------------------------------
*NEW OBJECTIVE QUESTION*


Also plz explain your answer

(Also you can ask your mathematics related questions or queries
 we will try to solve them)


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## Niilesh (Apr 19, 2013)

*Re: mathematic related questions*

Answer - 100
Solution - 
diametre * length = 1(consider the string as a 2-D rectangle]
1/100 * length = 1
length = 100 m

BTW are they PSA questions?



rst said:


> Its better than mine


Its nice to know that they are people like me


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## rst (Apr 19, 2013)

*Re: mathematic related questions*



Niilesh said:


> Answer - 100
> Solution -
> diametre * length = 1(consider the string as a 2-D rectangle]
> 1/100 * length = 1
> length = 100 m



Your method is absolutely right
But as diameter is 1mm
so it will be
1/1000 * length = 1
length = 1000 m



Niilesh said:


> BTW are they PSA questions?



these are questions of various competitive examinations based on 10 th level


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## Niilesh (Apr 19, 2013)

*Re: mathematic related questions*

Silly me


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## rst (Apr 19, 2013)

*Re: mathematic related questions*

*NEW OBJECTIVE QUESTION*



 Also plz explain your answer
(Also you can ask your mathematics related questions or queries
 we will try to solve them)


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## Niilesh (Apr 19, 2013)

*Re: mathematic related questions*

Can you post the answer so i can guess what he means by 'between'(whether it is counting those days or not)
Is the ans (D)?


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## rst (Apr 19, 2013)

*Re: mathematic related questions*

answer is 91


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## Niilesh (Apr 19, 2013)

*Re: mathematic related questions*

Solution - 
Note: The answer should be a multiple of 7 so that that next day will be Friday
         No. of days possible between two 13ths of months in a leap year - 29,30,31
Now its basically hit and trial

Case 1 - in one month 
29,30,31 all are not divisible by 7

Case 2 - in two months
29+31= 60   not divisible by 7
30+31= 61   not divisible by 7
31+31= 62   not divisible by 7

Case 3- in three months
30+31+30= 91
31+30+31= 92 not divisible by 7
31+29+31= 91 
29+31+30= 90 not divisible by 7 

So ans - 91


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## rst (Apr 19, 2013)

*Re: mathematic related questions*

your answer is better than hit and trial method




Niilesh said:


> *i.imgur.com/RuGGjlT.jpg



But in your answer
I didn't understand last three lines or steps

for example
If A  ≥  R
B ≥ R

can we say A ≥ B ??


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## Niilesh (Apr 19, 2013)

*Re: mathematic related questions*



rst said:


> your answer is better than hit and trial method


Its just organized hit and trial



rst said:


> But in your answer
> I didn't understand last three lines or steps
> 
> for example
> ...


I think it is valid if A is variable and B is a constant( max value of R is B and since A  ≥  R=> A ≥ B )
I am not sure though
Alternate method -
A.M = GM when all the no.s are equal(let it be 'a')
now 7a=21 => a=3
So ans = 7*9/7 = 9


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## rst (Apr 19, 2013)

*Re: mathematic related questions*



Niilesh said:


> I think it is valid if A is variable and B is a constant( max value of R is B and since A  ≥  R=> A ≥ B )
> I am not sure though
> Alternate method -
> A.M = GM when all the no.s are equal(let it be 'a')
> ...


I am also not sure

------------------------------------------------
*NEW OBJECTIVE QUESTION*


 Also plz explain your answer

(Also you can ask your mathematics related questions or queries
 we will try to solve them)


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## darkv0id (Apr 19, 2013)

*Re: mathematic related questions*

^Umm, none of the above? Answer is (sqrt(2))*a [Pythagoras Theorem], which I don't see in the options.... (I think it's a typo with the radical sign)


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## rst (Apr 19, 2013)

*Re: mathematic related questions*



darkv0id said:


> ^Umm, none of the above? Answer is (sqrt(2))*a [Pythagoras Theorem), which I don't see in the options.... (I think it's a typo with the radical sign)



Yeah
Its √2 * a

But can you explain your answer ??


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## Niilesh (Apr 19, 2013)

*Re: mathematic related questions*

Answer - (B)(√2a)
Solution - Distance b/w vertex and centre will be equal to 'a'(connect the centre to two consecutive vertex, it will be isosceles triangle with one angle 60 degree, so it will be equilateral)
Now apply Pythagoras theorem

@rst 
Did you really understood the proof that vignesh quoted?


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## darkv0id (Apr 19, 2013)

*Re: mathematic related questions*

^ Haha, you beat me to it.


----------



## rst (Apr 19, 2013)

*Re: mathematic related questions*



Niilesh said:


> Answer - (B)(√2a)
> Solution - Distance b/w vertex and centre will be equal to 'a'(connect the centre to two consecutive vertex, it will be isosceles triangle with one angle 60 degree, so it will be equilateral)
> Now apply Pythagoras theorem



Great explanation



Niilesh said:


> @rst
> Did you really understood the proof that vignesh quoted?


his proof is same as I explained with example
I didn't understand much about both a & b are even
" *if you look at a/2, b/2 then one easily sees that both
 (a/2)^2 + 6(b/2)^2
 (b/2)^2 + 6(a/2)^2
 are squares (since the squares a^2 + 6b^2 and b^2 + 6a^2 are both even, hence divisible by 4)
This contradicts the minimality of our original solution.*"

I don't where is "vignesh" ??
He should explain this fact


----------



## Niilesh (Apr 19, 2013)

*Re: mathematic related questions*

^ i have confusion in the same lines,  (a/2)^2 + 6(b/2)^2 is not always a perfect square


----------



## rst (Apr 19, 2013)

*Re: mathematic related questions*



Niilesh said:


> ^ i have confusion in the same lines,  (a/2)^2 + 6(b/2)^2 is not always a perfect square



yeah
(a/2)^2 + 6(b/2)^2 is not always a perfect square
take example
a=2
b=2

then (a/2)^2 + 6(b/2)^2= 1+6=7
which is not perfect square

I hope   both *a & b are even case*  is now clear to you

------------------------------------------------
*NEW OBJECTIVE QUESTION*


 Also plz explain your answer

(Also you can ask your mathematics related questions or queries
 we will try to solve them)


----------



## Niilesh (Apr 19, 2013)

*Re: mathematic related questions*

no its not clear and I will not be satisfied with hit and trial  
answer to the above question should be infinity but I think in the key it would be given N+1


----------



## rst (Apr 19, 2013)

*Re: mathematic related questions*

Yeah
answer is N+1

-----------------------------------------------------------------
can you explain the answer ???


----------



## Niilesh (Apr 19, 2013)

*Re: mathematic related questions*

I read the question wrong the first time
,  n+1 is the right answer

max no of turns till one of each colour is taken out is N, in the next turn a pair will be found so ans is N+1


----------



## rst (Apr 19, 2013)

*Re: mathematic related questions*

ok

we can also solve it in simple way

 N is even( so socks are in pair)
 suppose we have 2 pairs of socks in different color(2 socks of white color and 2 socks of black color) in a drawer
 If we draw 3 socks from drawer then definitely, we will get a pair

 similarly if we have 3 pairs of socks in different color(2 socks of white color, 2 socks of black color and 2 socks of red color) a drawer
 If we draw 4 socks from drawer then definitely,we will get a pair

 so in general if we have N pairs of socks in different color
 then we have to draw N+1 socks to get a pair

----------------------------------------
*NEW OBJECTIVE QUESTION*



 Also plz explain your answer

(Also you can ask your mathematics related questions or queries
 we will try to solve them)


----------



## Niilesh (Apr 19, 2013)

*Re: mathematic related questions*

ans is 3
sol.  their will be 3 intersection points on each of the 4 lines(so they would be collinear), now every point will have 4 points on old lines(I think you will be able understand what I mean) so only one new line be formed form each point

so ans = 6/2 = 3


----------



## rst (Apr 19, 2013)

*Re: mathematic related questions*

absolutely right
-----------------------------------
*NEW OBJECTIVE QUESTION*


 Also plz explain your answer

(Also you can ask your mathematics related questions or queries
we will try to solve them)


----------



## Niilesh (Apr 19, 2013)

*Re: mathematic related questions*

is it 49/100? 

I just applied the unitary method so ans = ((100/100)/100)*7*7


----------



## rst (Apr 19, 2013)

*Re: mathematic related questions*

no
ans is 100 minutes


----------



## thetechfreak (Apr 20, 2013)

*Re: mathematic related questions*

100 minutes. Each one catches one in 100 minutes and hence 7 will take 7 minutes.


----------



## rst (Apr 20, 2013)

*Re: mathematic related questions*



thetechfreak said:


> 100 minutes. Each one catches one in 100 minutes and hence 7 will take 7 minutes.



yeah
Its looks like 7 minutes

But actual answer is "100 minutes"
--------------------------------------------------------
*here is the solution*
Let C1 be the number of cats who catch R1 rats in T1 minutes and

Let C2 be the number of cats who catch R2 rats in T2 minutes

C1T1/R1 = C2T2/R2

(100 x 100) /100 = (7 x T2) / 7
T2= 100

so time = 100 minutes
--------------------------------------------
Another method:-

let time be x

 Now number of cat and time are in indirect proportion
 Also number of rat and time are in direct proportion

 So 100 * 100 * 7= 7 * x *100
 x=100






Niilesh said:


> I read the question wrong the first time
> ,  n+1 is the right answer
> 
> max no of turns till one of each colour is taken out is N, in the next turn a pair will be found so ans is N+1


we can also solve it in simple way.

N is even( so socks are in pair)
suppose we have 2  pairs of socks in different color(2 socks of white color and 2 socks of black color) in a drawer
If we draw 3 socks from drawer then definitely, we will get a pair

similarly if we have 3  pairs of socks in different color(2 socks of white color, 2 socks of black color and 2 socks of red color) a drawer
If we draw 4 socks from drawer then definitely, we will get a pair

so in general if  we have N  pairs of socks in different color
then we have to draw N+1 socks to get a pair



Niilesh said:


> no its not clear and I will not be satisfied with hit and trial



OK
then here is the proof

*case 4: a and b are both even*
we want to prove that a^2 + 6b^2 and b^2 + 6a^2 are not perfect square 
here we will use contradiction method

If possible suppose that a^2 + 6b^2 and b^2 + 6a^2 is perfect square 
then  (a/2)^2 + 6(b/2)^2 and  (b/2)^2 + 6(a/2)^2 will also be perfect square 

But in  (a/2)^2 + 6(b/2)^2 and  (b/2)^2 + 6(a/2)^2
(a/2)^2 and (b/2)^2 can be even or odd

So by case 1,2 and 3 (which I discussed in detail in my previous post)
(a/2)^2 + 6(b/2)^2 and  (b/2)^2 + 6(a/2)^2 are not perfect square
which is a contradiction
Hence our assumption is wrong

 a^2 + 6b^2 and b^2 + 6a^2 are not perfect square


----------



## Niilesh (Apr 20, 2013)

*Re: mathematic related questions*



rst said:


> we can also solve it in simple way.
> 
> N is even( so socks are in pair)
> Whether no. of colours is odd or even it dosen't matter, What did you meant by "socks are in pair"
> ...







rst said:


> OK
> then here is the proof
> 
> *case 4: a and b are both even*
> ...


Why?
let 
a^2 + 6b^2 and b^2 + 6a^2 = k^2
=> (a/2)^2 + 6(b/2)^2=(k^2)/4 
Now since a square no. divided by 4 is not necessarily a square, therefore the statement is wrong

_______________________________________________________________________________________

*A Question - *
Find the smallest number such that if its rightmost digit is placed at its left end, the new number so formed is precisely 50% larger than the original number


----------



## rst (Apr 20, 2013)

*Re: mathematic related questions*



Niilesh said:


> Why?
> let
> a^2 + 6b^2 and b^2 + 6a^2 = k^2
> => (a/2)^2 + 6(b/2)^2=(k^2)/4
> Now since a square no. divided by 4 is not necessarily a square, therefore the statement is wrong



now k^2 is square 
then (k^2)/4 will also be square
how???
(k^2)/4= (k/2)^2

note :- even square no. divisible by 4 will always be square no.
If not, give an example


----------



## Niilesh (Apr 20, 2013)

*Re: mathematic related questions*

I said it is not 'ALWAYS' a square  
since you didn't prove that k is even,  hence the statement is wrong


----------



## rst (Apr 20, 2013)

*Re: mathematic related questions*



rst said:


> *N is even( so socks are in pair)*
> suppose we have 2  pairs of socks in different color(2 socks of white color and 2 socks of black color) in a drawer
> If we draw 3 socks from drawer then definitely, we will get a pair
> 
> ...


it means each color socks are in even pair(there will be 2 socks of black color,2 socks of white color(as I explained in my answer)]
for example :-2,4,6, or 8 etc pair of socks



Niilesh said:


> I said it is not 'ALWAYS' a square
> since you didn't prove that k is even,  hence the statement is wrong



if k^2 is odd
then also (k^2)/4= (k/2)^2 is perfect square

note :- when we divide two perfect square then result is always perfect square


----------



## Niilesh (Apr 20, 2013)

*Re: mathematic related questions*



rst said:


> it means each color socks are in even pair(there will be 2 socks of black color,2 socks of white color(as I explained in my answer)]
> for example :-2,4,6, or 8 etc pair of socks
> N is the no. of colours...
> It was given that it is even just because they wanted to prevent people from eliminating (D) option
> ...


 Chars


----------



## rst (Apr 20, 2013)

*Re: mathematic related questions*

they give even no. of socks
so that we start counting from 2 (i.e 2 pair of shocks)

In case of odd
counting starts from 1(i.e 1 pair of shocks)
which looks meaningless (as they are of same color)
-------------------------------------------------------
yeah

 perfect square is in fraction form (in case of odd perfect square)

Note :-  we require integral solution

don't relate it with perfect square
perfect square can be in fraction form


----------



## Niilesh (Apr 20, 2013)

*Re: mathematic related questions*

Read the question again

-------------------

I think perfect square means whose square root is a integer

if it can be fraction than almost every no. will be a perfect square . BTW in the solution it should be square of of a integer


----------



## rst (Apr 20, 2013)

*Re: mathematic related questions*



Niilesh said:


> Read the question again



In question they write *socks of N different color*

In case of odd number 
N can be 1

there is no sense *"socks of 1 different color"*


----------



## Niilesh (Apr 20, 2013)

*Re: mathematic related questions*

ya,  in the case of N=1 language seems funny but answer will be the same.


----------



## rst (Apr 20, 2013)

*Re: mathematic related questions*



Niilesh said:


> I think perfect square means whose square root is a integer
> 
> if it can be fraction than almost every no. will be a perfect square . BTW in the solution it should be square of of a integer



you are right

But in case 4

As both both a and b are even
so a^2 +6b^2 and b^2 +6a^2 are even perfect square (as we assume them to be perfect square)-------(1)

(a/2)^2 +6(b/2)^2 = (a^2 +6b^2)/4 
=(even perfect square)/4
=perfect square

similarly we can prove for (b/2)^2 +6(a/2)^2

hence (a/2)^2 +6(b/2)^2 and (b/2)^2 +6(a/2)^2 are perfect square


----------



## Niilesh (Apr 20, 2013)

*Re: mathematic related questions*

ohh... 
I shouldn't have missed it


----------



## rst (Apr 20, 2013)

*Re: mathematic related questions*



Niilesh said:


> *A Question - *
> Find the smallest number such that if its rightmost digit is placed at its left end, the new number so formed is precisely 50% larger than the original number



I think Its only possible with "hit and trial" method
If not, plz explain



Niilesh said:


> ans is 3
> sol.  their will be 3 intersection points on each of the 4 lines(so they would be collinear), now every point will have 4 points on old lines(I think you will be able understand what I mean) so only one new line be formed form each point
> 
> so ans = 6/2 = 3



can you explain your answer in more simple language ??
i.e  average student can also understand it.


----------



## dashing.sujay (Apr 20, 2013)

*Re: mathematic related questions*

I'm really enjoying this thread, makes me feel like going back to school days


----------



## rst (Apr 20, 2013)

*Re: mathematic related questions*

solving maths question is really a fun (if you get right answer)


----------



## Niilesh (Apr 20, 2013)

*Re: mathematic related questions*



rst said:


> I think Its only possible with "hit and trial" method
> If not, plz explain


If the answer would have been two or three digit maybe i could have solved(without hit and trial) it but the answer is in 6 digits!


If anyone can explain whats written here it would be appreciated(its the solution)


> After thinking about the problem for a couple of minutes (kinda embarrassed it took that long to get it), I went directly to the answer, no trial and error. Further, I can also tell you that the next lowest number with this property is 571428 and that there are no other numbers less than 10^50 with this property.
> 
> What I realized is that I needed to find a repeating fraction such that the number of repeating digits was one less than the number. This would be a prime number, and seven is the lowest number with this property. With this property, every digit in the repeating fraction appears in each place exactly once (i.e. every repeated digit appears as the first digit after the decimal exactly once for n/p where p > n > 0).
> 
> ...


----------



## rst (Apr 20, 2013)

*Re: mathematic related questions*



Niilesh said:


> If the answer would have been two or three digit maybe i could have solved(without hit and trial) it but the answer is in 6 digits!



i don't like "hit and trial" questions
we should be able to use maths concept in a question



Niilesh said:


> If anyone can explain whats written here it would be appreciated(its the solution)



Yeah

program is better option for such calculation

--------------------------------------------------------------------------------------------
*NEW OBJECTIVE QUESTION*



 Also plz explain your answer

(Also you can ask your mathematics related questions or queries
 we will try to solve them)


----------



## Niilesh (Apr 20, 2013)

*Re: mathematic related questions*



rst said:


> Yeah
> 
> program is better option for such calculation


Did you understand the proof above it?


rst said:


> *NEW OBJECTIVE QUESTION*
> View attachment 10109
> 
> Also plz explain your answer
> ...


Ans is 1?


----------



## dashing.sujay (Apr 20, 2013)

*Re: mathematic related questions*

@Niilesh - The solution is mind boggling. The question is really tough. I tried and left it 



> --------------------------------------------------------------------------------------------
> *NEW OBJECTIVE QUESTION*
> View attachment 10109
> 
> ...



One, and no explanation required I guess.


----------



## Niilesh (Apr 20, 2013)

*Re: mathematic related questions*

Another question - 
There is a 10-digit number, represented by ABCDEFGHIJ, where each numeral, 0 through 9, is used once. Given the following clues, what is the number?

1) A + D = a square number
2) G + J = a triangle number
3) B + I = an even number
4) E * F = an odd number
5) C * H = a prime number
6) A / G = G / C
7) E + I = B + H

Note: their are more than one answer


----------



## dashing.sujay (Apr 20, 2013)

*Re: mathematic related questions*

^If the values aren't in increasing order, then answers can be too many. For instance,
a) 0+1, 0+4, 1+3, 0+9, 1+8, 3+6, 4+5.
Am I getting it right?


----------



## Niilesh (Apr 20, 2013)

*Re: mathematic related questions*

^ All the conditions have to be satisfied together


----------



## dashing.sujay (Apr 21, 2013)

*Re: mathematic related questions*



Niilesh said:


> ^ All the conditions have to be satisfied together



I understood the question now.
But i'm not able to find a way other than hit and trial method.


----------



## Niilesh (Apr 21, 2013)

*Re: mathematic related questions*

^ Since the question has more than one solution, therefore you will have to apply Hit and Trial.
But conditions removes most of the combinations, 
For example C * H = a prime number, so one no. will be 1 and another will be prime.


----------



## dashing.sujay (Apr 21, 2013)

*Re: mathematic related questions*

Yes that prime thing removes most of the combinations.


----------



## rst (Apr 21, 2013)

*Re: mathematic related questions*



Niilesh said:


> Ans is 1?


yeah
 ans is 1
But I also need explanation



dashing.sujay said:


> One, and no explanation required I guess.



explanation is required

I know ,its more like a practical question
but you can explain
as length of wire and edges of tetrahedron is given



Niilesh said:


> Another question -
> There is a 10-digit number, represented by ABCDEFGHIJ, where each numeral, 0 through 9, is used once. Given the following clues, what is the number?
> 
> 1) A + D = a square number
> ...



5) C * H = a prime number
so either C or H= 1

6) A / G = G / C
AC= G^2

if *C=1*
A=G^2=4,9


possibilities are A=4, G=2, C=1 *or* A=9, G=3, C=1 

take *A=4, G=2, C=1*

2)G + J = a triangle number
G + J=2+8[as triangular number are 1,3,6,10,15]
clearly *J=8*

1) A + D = a square number
A + D = 4+0 or 4+5 [as square number are 1,4,9,16]
so D= 0 or 5.........(i)

4) E * F = an odd number
both E and F are odd number-----------(ii)

even number 6 can be only B or I [using i and ii]

3) B + I = an even number
 B + I = 6+0 or 0+6-----------------(iii)

using (i) and (iii)
*D=5*

Also H= 3 or 7 [as H is prime number (if C=1)]-------------------(iv)
7) E + I = B + H

As B, I can be 0 or 6
difference of B and I is 6
so difference of E and H is 6

only left numbers  are 3,7 and 9
E ,H can be 3,9
using (iv) *H =3*
hence *E= 9*

definitely* F=7*

Now E + I = B + H
9+I =B+3
6 =B - I

*So B= 6
I= 0*

Hence *ABCDEFGHIJ = 4615972308*

As there are more than one answer of the question
we can also try  H=1 condition [I get the above answer with condition C=1]
Also can try A=9, G=3, C=1



Niilesh said:


> ans is 3
> sol.  their will be 3 intersection points on each of the 4 lines(so they would be collinear), now every point will have 4 points on old lines(I think you will be able understand what I mean) so only one new line be formed form each point
> 
> so ans = 6/2 = 3



 plz explain in simple way
i.e how *there are 3 intersection points on each of the 4 lines*
*every point will have 4 points on old lines*


----------



## Niilesh (Apr 21, 2013)

*Re: mathematic related questions*



rst said:


> As there are more than one answer of the question
> we can also try  H=1 condition [I get the above answer with condition C=1]
> Also can try A=9, G=3, C=1


H cannot be 1 
Second answer is 9817453260



rst said:


> plz explain in simple way
> i.e how *there are 3 intersection points on each of the 4 lines*
> *every point will have 4 points on old lines*


Hopefully this image will solve your problem
*i.imgur.com/AKoD4m1.png


----------



## rst (Apr 21, 2013)

*Re: mathematic related questions*



Niilesh said:


> Second answer is 9817453260



Its look like mobile phone number
Its definitely someone mobile number



Niilesh said:


> H cannot be 1



Yeah
may be possible
I didn't try this case



Niilesh said:


> Hopefully this image will solve your problem
> *i.imgur.com/AKoD4m1.png



Definitely
Its really helpful

But I think that I even didn't understand the question
According to the question *lines are drawn joining the point of intersection of the previous four lines*
What is the meaning of this line ??
Can you elaborate it ??


----------



## Niilesh (Apr 21, 2013)

*Re: mathematic related questions*

Umm.. I am not sure how can i explain it more properly in words -   
new lines are drawn joining the point of intersections of the previous four lines


----------



## rst (Apr 21, 2013)

*Re: mathematic related questions*



Niilesh said:


> Umm.. I am not sure how can i explain it more properly in words -
> new lines are drawn joining the point of intersections of the previous four lines


no problem
I know its difficult to explain geometrical problem without figure
-----------------------------------------------------------------------------


----------



## Niilesh (Apr 21, 2013)

*Re: mathematic related questions*

The Red lines are the required lines 
*i.imgur.com/7CtOaVT.png


----------



## rst (Apr 21, 2013)

*Re: mathematic related questions*

great work
thanks a lot

Now the whole question and its answer is clear to me



rst said:


> *NEW OBJECTIVE QUESTION*
> View attachment 10109
> 
> Also plz explain your answer
> ...



ans is 1

Tetrahedron shape is given in the figure
It is 3 dimensional figure
Its a kind of pyramid

total length of wire is 6 m
Each edge of Tetrahedron is 1 m

With the wire first make equilateral triangle of each of side 1m
Then from one vertex of triangle raise the wire 1m  to the point O (say)then again join O to other vertex.
now we have to break wire first time
At last with remaining 1m wire join remaing third vertex of triangle to point O
Tetrahedron is formed after cutting wire 1 time

*----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION*


 Also plz explain your answer

 (Also you can ask your mathematics related questions or queries
 we will try to solve them)


----------



## Niilesh (Apr 21, 2013)

*Re: mathematic related questions*

Its 3.5

join the centre of the sphere to the point of contact then a right angled triangle will be formed
r/(h-r) = Sin(30)
R=H/3= 3.5

___________________________________________________________________________
There is a ten-digit mystery number (no leading 0), represented by ABCDEFGHIJ, where each numeral, 0 through 9, is used once. Given the following clues, what is the number?

1) A + B + C + D + E is a multiple of 6.
2) F + G + H + I + J is a multiple of 5.
3) A + C + E + G + I is a multiple of 9.
4) B + D + F + H + J is a multiple of 2.
5) AB is a multiple of 3.
6) CD is a multiple of 4.
7) EF is a multiple of 7.
8) GH is a multiple of 8.
9) IJ is a multiple of 10.
10) FE, HC, and JA are all prime numbers.


----------



## rst (Apr 22, 2013)

*Re: mathematic related questions*



Niilesh said:


> Its 3.5
> 
> join the centre of the sphere to the point of contact then a right angled triangle will be formed
> r/(h-r) = Sin(30)
> R=H/3= 3.5


Absolutely Right
----------------------------------
In detail, answer is:-
*img10.imageshack.us/img10/884/cone21.png
Here AOB is right angled triangle at B
let OB=r
OA=h-r

OB/OA = Sin(30)
r/ (h-r) =(1/2)
2r= h-r
3r=h
r=h/3
r=10.5/3
r= 3.5 cm



Niilesh said:


> There is a ten-digit mystery number (no leading 0), represented by ABCDEFGHIJ, where each numeral, 0 through 9, is used once. Given the following clues, what is the number?
> 
> 1) A + B + C + D + E is a multiple of 6.
> 2) F + G + H + I + J is a multiple of 5.
> ...



ans is 5736912480


----------



## Niilesh (Apr 22, 2013)

*Re: mathematic related questions*

Correct,  it took me more than half an hour to solve it...


----------



## rst (Apr 22, 2013)

*Re: mathematic related questions*

*----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION*

*img59.imageshack.us/img59/4996/volumeo.png

 Also plz explain your answer

 (Also you can ask your mathematics related questions or queries
 we will try to solve them)


----------



## Niilesh (Apr 22, 2013)

*Re: mathematic related questions*

^ Can you explain how are they arranged?


----------



## rst (Apr 22, 2013)

*Re: mathematic related questions*

I think they are arranged joining each other to form cuboid


----------



## Niilesh (Apr 22, 2013)

*Re: mathematic related questions*

Ans is 2?


----------



## rst (Apr 22, 2013)

*Re: mathematic related questions*

Can you show your work ???


----------



## Niilesh (Apr 22, 2013)

*Re: mathematic related questions*

A = na^2 (a= side length of cube)
V = na^3
=>A^3/V^2 = n


----------



## rst (Apr 22, 2013)

*Re: mathematic related questions*

yeah
your work looks fine

A^3/V^2 is the correct answer

*----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION*

*img834.imageshack.us/img834/8831/boxofstick.png

 Also plz explain your answer

 (Also you can ask your mathematics related questions or queries
 we will try to solve them)
*---------------------------------------------------------------------------------------------------------------------
Hint:-
answer is not 8 and 12*


----------



## Niilesh (Apr 23, 2013)

*Re: mathematic related questions*

the smallest enclosed 2D fig is triangle so probably tetrahedron will be the smallest 3D fig,  so ans is 6


----------



## rst (Apr 23, 2013)

*Re: mathematic related questions*

I also think the same
with 6 sticks tetrahedron is possible(which can enclose 3D volume)
*----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION*

*img9.imageshack.us/img9/6169/boysandcircle.png

 Also plz explain your answer

 (Also you can ask your mathematics related questions or queries
 we will try to solve them)


----------



## Niilesh (Apr 23, 2013)

*Re: mathematic related questions*

4 min(I Used the concept of relative velocity)


----------



## rst (Apr 23, 2013)

*Re: mathematic related questions*

Can you show your work ??


----------



## Niilesh (Apr 23, 2013)

*Re: mathematic related questions*

Let 'B' be at rest, then A is coming towards it with velocity '3v'.
At first distance between them is C/2(C=circumference) and to cross B again A will have to travel C distance, since Time is directly proportional to distance(when velocity is constant) so time=2*2= 4min


----------



## rst (Apr 23, 2013)

*Re: mathematic related questions*

Yeah
you are right

Also this method is shorter than mine


----------



## Niilesh (Apr 23, 2013)

*Re: mathematic related questions*

^ What was your method?
The only method i know other than the relative one is the 'imagination' one


----------



## rst (Apr 23, 2013)

*Re: mathematic related questions*

I also used relative velocity method 
But instead of 3v 
I used distance travel by A= 2v (time x velocity)
distance travel by B= 4v
total distance =6v 

In second time they have to move whole circumference
total distance =12v
vt +2vt =12v
t=4 min.

*----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION*

*img834.imageshack.us/img834/1553/fraction.png

 Also plz explain your answer

 (Also you can ask your mathematics related questions or queries
 we will try to solve them)


----------



## Niilesh (Apr 23, 2013)

*Re: mathematic related questions*

Ans is 3
Solution - 
1^13 < 2^13
3^13 < 5^13
.
.
.    and so on
.
.
.

so, C>B 
=>C/2 > B/2   ---(1)

Now, A= C/2 + B/2
=>C>A>B       (From equation 1)


----------



## rst (Apr 23, 2013)

*Re: mathematic related questions*

great!!
answer is absolutely right
Also with this method one can solve the question in short time

*----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION*

*img33.imageshack.us/img33/9770/puzzl.png

 Also plz explain your answer

 (Also you can ask your mathematics related questions or queries
 we will try to solve them)

*----------------------------------------------------------------------------------*
ans is (B) 3


----------



## rst (Apr 25, 2013)

*Re: mathematic related questions*

*----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION*

*img526.imageshack.us/img526/7576/profiti.png

 Also plz explain your answer

 (Also you can ask your mathematics related questions or queries
 we will try to solve them)

*----------------------------------------------------------------------------------*


----------



## rst (Apr 27, 2013)

*Re: mathematic related questions*

ans is 5
-----------------------------------------------------
Let number of units be x
S.P of one unit =60
S.P of x unit =60x
C.P=100

profit=500
S.P - C.P=500
60x- 100 =500
x=10
so number of units =10

C.P increase by 30%
New C.P=130
Let new S.P per unit be y
total S.P=10y (as number of units is 10)
As profit is same
new S.P - New C.P =500
10y - 130 =500
y=63
So new S.P per unit =63
increase percentage in S.P per unit = [(63-60)/60 ] x 100
=[3/60 ] x 100
=5


----------



## rst (Apr 28, 2013)

*Re: mathematic related questions*

*----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION*

*img191.imageshack.us/img191/6536/formulaj.png

 Also plz explain your answer

 (Also you can ask your mathematics related questions or queries
 we will try to solve them)
*----------------------------------------------------------------------------------------------------------------------*


----------



## avj (Apr 28, 2013)

*Re: mathematic related questions*



rst said:


> *----------------------------------------------------------------------------------------------------------------------
> NEW OBJECTIVE QUESTION*
> 
> *img191.imageshack.us/img191/6536/formulaj.png
> ...



it is option 1
take area of each small square as 1
total area is n(n+1)  (n rows and n+1 columns)
area of one half is 1+2+3...  (number of squares in each row(left)/column(right))
so total area is 2*(1+2+3....)


----------



## rst (Apr 28, 2013)

*Re: mathematic related questions*

answer is absolutely right
Also thanks for the explanation


----------



## rst (Apr 29, 2013)

*Re: mathematic related questions*

*----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION*

*img16.imageshack.us/img16/9886/easwari.png

 Also plz explain your answer

 (Also you can ask your mathematics related questions or queries
 we will try to solve them)
*----------------------------------------------------------------------------------------------------------------------*


----------



## rst (Apr 30, 2013)

*Re: mathematic related questions*

ans is *b. 10 %*

Can someone explain it ??


----------



## rst (May 1, 2013)

*Re: mathematic related questions*

Monthly deposit,P=350
Time,n=6 years
=72 months


total deposit money=350 x 72
=25200

total received money=32865

so interest= 32865- 25200
=7665

Now interest = PNr/100-----------------(1)
where N= n (n+1) /24
=(72 x 73) /24
=219

so 7665= (350 x 219 x r)/100
r= 10


----------



## rst (May 3, 2013)

*Re: mathematic related questions*

*----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION*

*img823.imageshack.us/img823/5362/worke.png

 Plz explain your answer

 (Also you can ask your mathematics related questions or queries
 we will try to solve them)
*----------------------------------------------------------------------------------------------------------------------*


----------



## Niilesh (May 3, 2013)

*Re: mathematic related questions*

24 Days

Solution - Let the work be equal to 'x' units
then work done by one man in 1 hour = x/(9*28*8)
Let the required no. of days be 'y'
Then 
x/(9*28*8) *(12*7*y) = x

=>Y=24 Days


----------



## rst (May 3, 2013)

*Re: mathematic related questions*

ans is correct
----------------------------------------
let number of days be x

Now number of men and no. of days are in indirect proportion
Also number of hours and no. of days are in indirect proportion

 So 8* 9 * 28= 12* 7 *x
 x=24


----------



## Niilesh (May 3, 2013)

*Re: mathematic related questions*



rst said:


> Monthly deposit,P=350
> Time,n=6 years
> =72 months
> 
> ...


Can you explain that?

also does the interest gets added monthly(not 10% each month but 1/12th part of that 10%) or yearly?



rst said:


> *----------------------------------------------------------------------------------------------------------------------
> NEW OBJECTIVE QUESTION*
> 
> *img33.imageshack.us/img33/9770/puzzl.png
> ...


Can you explain that?


----------



## rst (May 3, 2013)

*Re: mathematic related questions*



Niilesh said:


> Can you explain that?
> 
> also does the interest gets added monthly(not 10% each month but 1/12th part of that 10%) or yearly?


here r is annual rate of interest



rst said:


> *----------------------------------------------------------------------------------------------------------------------
> NEW OBJECTIVE QUESTION*
> 
> *img33.imageshack.us/img33/9770/puzzl.png
> ...



In first two figures there is difference of 3 between them
so answer is 3


----------



## Niilesh (May 3, 2013)

*Re: mathematic related questions*



rst said:


> here r is annual rate of interest


You got me wrong, how did you find the value of N?

also I wanted to know if does the interest gets added monthly(not 10% each month but 1/12th part of that 10%) or yearly
Like after 350 is added will it become 350* 100+r/( 100*12) after one month?                                                          



rst said:


> In first two figures there is difference of 3 between them
> so answer is 3


in the last two fig the difference is not equal to 5 in every case ( 6+5 =11)


----------



## rst (May 3, 2013)

*Re: mathematic related questions*



Niilesh said:


> in the last two fig the difference is not equal to 5 in every case ( 6+5 =11)


yeah
It should of 11 instead of 10
it is printing mistake in the question


----------



## Niilesh (May 3, 2013)

*Re: mathematic related questions*



rst said:


> yeah
> It should of 11 instead of 10
> it is printing mistake in the question


hmm..
So you must have got the solution after seeing the answer (?)


----------



## rst (May 3, 2013)

*Re: mathematic related questions*



Niilesh said:


> hmm..
> So you must have got the solution after seeing the answer (?)



you can say that





Niilesh said:


> You got me wrong, how did you find the value of N?
> 
> also I wanted to know if does the interest gets added monthly(not 10% each month but 1/12th part of that 10%) or yearly
> Like after 350 is added will it become 350* 100+r/( 100*12) after one month?


In monthly recurring deposit 
 interest = PNr/100
 where N= n (n+1) /24

Its the easy way to rember the formula
 Its like the simple interest formula
 -----------------------------------------------------------
 In given question
 Monthly deposit,P=350
 Time,n=6 years
 =72 months

In monthly recurring deposit, we deposit money each month
 so principal is changing per month


principal for 1st month = 350
 principal for 2nd month =2 x 350
 principal for 3rd month =3 x 350
 .
 .
 .
 principal for 72nd month =72x 350

Total principal or premium= 350+ 2 x 350 +3 x 350 +.....+72x 350
 =350 (1+2+......+72)
 in general
 Total principal or premium=P (1+2+............+n)
 =[P n (n+1) ]/2

interest = Total premium x rate/100 X 1/12
 =[Pn(n+1) ]/2 x r/100 X 1/12
 = PNr/100
 where N= n (n+1) /24


----------



## mastercool8695 (May 4, 2013)

*Re: mathematic related questions*

You guys have the potential to Launch 
"the solutions to 'TEST OF MATHEMATICAS AT 10+2 LEVEL'"

i'll join with questions from day..


----------



## rst (May 4, 2013)

*Re: mathematic related questions*

You are welcome


----------



## Niilesh (May 4, 2013)

*Re: mathematic related questions*

*@rst* Ohh.. i thought it was compound interest. thanks for the solution BTW


Spoiler



but i am more comfortable with the _Sum of GP_ method





mastercool8695 said:


> You guys have the potential to Launch
> "the solutions to 'TEST OF MATHEMATICAS AT 10+2 LEVEL'"
> 
> i'll join with questions from day..


Nah, I am in 12th so still can't solve 12th topics


----------



## mastercool8695 (May 4, 2013)

*Re: mathematic related questions*

no problem buddy..
i'hv (maybe) passed 12th still cant solve them..


----------



## rst (May 4, 2013)

*Re: mathematic related questions*



mastercool8695 said:


> no problem buddy..
> i'hv (maybe) passed 12th still cant solve them..


Then post the questions
I want to see such questions


----------



## mastercool8695 (May 6, 2013)

*Re: mathematic related questions*



Vignesh B said:


> Ok, my bad.
> Care to explain the answer  in brief?
> 
> One from my side too -
> ...



here you go..

*www.quora.com/Mathematics/How-many-pairs-of-integers-are-there-such-a-2-+-6b-2-and-b-2+6a-2-are-perfect-squares-from-Brilliant-org

this one : 
July 3, 1977 is a sunday, then July 3,1970 was ??

A) wednesday
B) friday
C) sunday
D) tuesday

anyways, found this..
*www-history.mcs.st-and.ac.uk/HistTopics/Pell.html


----------



## dashing.sujay (May 6, 2013)

*Re: mathematic related questions*

^Is it Sunday ?


----------



## mastercool8695 (May 6, 2013)

*Re: mathematic related questions*

no..
if i write the answer  so early, i cant have an explaination..
still its not sunday.
so its just three choices now..
i really dont know what theory will strike the question directly,  but some surely, will


----------



## rst (May 6, 2013)

*Re: mathematic related questions*



mastercool8695 said:


> this one :
> July 3, 1997 is a sunday, then July 3,1970 was ??
> 
> A) wednesday
> ...



It should be Monday But there is no such option
------------------------------------------------------------
There are 27 years from July 3,1997 to July 3, 1970
out of them 7 are leap years 
So 20 normal years and 7 leap years

In above question we are going bacward
In normal year ,we go 1 day backward
In leap year we go 2 days backward

total =(20 x 1) + (7 x 2)
=34

34 gives remainder 6 when divided by 7
So we go 6 days backward (or 1 day forward)

So answer is MONDAY


----------



## mastercool8695 (May 6, 2013)

*Re: mathematic related questions*

^^ sorry, posted 1977 as 1997 
sorry..

post edited now..

anyways, your explaination was great.. thanks..


----------



## rst (May 6, 2013)

*Re: mathematic related questions*



mastercool8695 said:


> this one :
> July 3, 1977 is a sunday, then July 3,1970 was ??
> 
> A) wednesday
> ...



Then ans is Friday

explanation is same as above
----------------------------------------
There are 7 years from July 3,1977 to July 3, 1970
 out of them 2 are leap years 
So 5 normal years and 2 leap years

In above question we are going bacward
 In normal year ,we go 1 day backward
 In leap year we go 2 days backward

total =(5 x 1) + (2 x 2)
 =9

9 gives remainder 2 when divided by 7
 So we go 2 days backward 

So answer is FRIDAY


----------



## mastercool8695 (May 6, 2013)

*Re: mathematic related questions*

yup..
thanks.
 check this and post your contribution..
*www.thinkdigit.com/forum/education/173487-informative-mathematical-theories.html


----------



## rst (May 6, 2013)

*Re: mathematic related questions*

But it doesnot look like 10+2 level question


----------



## mastercool8695 (May 6, 2013)

*Re: mathematic related questions*

yup..
i know..
the questions from the starting of that book are quite easy, 
but i couldn't fit the correct method..
thats why..


----------



## rst (May 6, 2013)

*Re: mathematic related questions*

*----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION*

*img836.imageshack.us/img836/1482/nextno.png

 Plz explain your answer

 (Also you can ask your mathematics related questions or queries
 we will try to solve them)
*----------------------------------------------------------------------------------------------------------------------*


----------



## mastercool8695 (May 6, 2013)

*Re: mathematic related questions*

*i.imgur.com/FMKAwY1.png


----------



## rst (May 6, 2013)

*Re: mathematic related questions*



mastercool8695 said:


> *i.imgur.com/FMKAwY1.png



Ans is B
----------------------------------------------------------------------------------------

 A) It is not true
 Counter example :-
 Let a=2 ,b=3 and n=4
 aⁿ +bⁿ =97 ( prime)


----------



## darkv0id (May 6, 2013)

*Re: mathematic related questions*

^ I don't think you understood the question; it's not asking whether a^n+b^n will be prime when one the 4 four choices is true, but rather which of the choices will be true *if we already know* that a^n+b^n is prime.


----------



## Niilesh (May 6, 2013)

*Re: mathematic related questions*



rst said:


> Ans is (D)
> ----------------------------------------------------------------------------------------
> 
> A) It is not true
> ...


The answer did not say that it will be true for all values of a,b if the conditions on 'n' are satisfied. So you need to give am example in which the condition for n is not satisfied and it is a prime number to prove that the option is wrong


----------



## rst (May 6, 2013)

*Re: mathematic related questions*



darkv0id said:


> ^ I don't think you understood the question; it's not asking whether a^n+b^n will be prime when one the 4 four choices is true, but rather which of the choices will be true *if we already know* that a^n+b^n is prime.



Yeah
you are right

I have edited my previous answer


----------



## mastercool8695 (May 6, 2013)

*Re: mathematic related questions*

so you guys suggest hit and trial ??


----------



## rst (May 6, 2013)

*Re: mathematic related questions*

Other method
------------------------------------------------------------------
aⁿ+bⁿ is prime number 
where n is positive integer other than 1
here n can be even or odd

case 1:- If n is odd
 aⁿ+bⁿ =prime number
(a+b) (aⁿ⁻¹  - aⁿ⁻² b +................+(-1)ⁿ⁻¹  bⁿ⁻¹ ) =prime number 
Not possible as here prime number is product of two numbers

case 2:- If n is even

I)Let n =2m where m=odd 
aⁿ+bⁿ = prime number
a^2m+ b^2m = prime number 
(a² )^m+(b² )^m = prime number
so aⁿ+bⁿ can't be prime number (by case 1)

Hence aⁿ+bⁿ = prime number is only possible if n=2m where m is even
i.e n= 4,8,12,16,20,24,………………………………

C) possible n=12,20,24,.......................
They can be expressed as 
12=3 x 8
20= 5x 4
24=3 x 8
.
.
similar to case 2(I) not prime number

B)here possible values of n =4,8,16,...................
if aⁿ+bⁿ = prime number
then it will definitely be prime number for these values of n


----------



## mastercool8695 (May 6, 2013)

*Re: mathematic related questions*

@rst : for a=2 , b=4, n= 4

2[SUP]4[/SUP] + 4[SUP]4[/SUP] = 272 (not prime)


----------



## rst (May 6, 2013)

*Re: mathematic related questions*

if aⁿ+bⁿ = prime number
then n will power of 2

don't do it reverse manner


----------



## mastercool8695 (May 6, 2013)

*Re: mathematic related questions*

ok...
forgot that..
thanks..

any example ?


----------



## rst (May 6, 2013)

*Re: mathematic related questions*

take example which I gave in previous post

 2 ⁴  +3 ⁴  =97 ( prime) 
it is possible for n=4


----------



## mastercool8695 (May 6, 2013)

*Re: mathematic related questions*

ok..
got that now..

do you know the answer for your "whats next in series" one ?

cuz, i cant decipher the relation..


----------



## rst (May 6, 2013)

*Re: mathematic related questions*



rst said:


> *----------------------------------------------------------------------------------------------------------------------
> NEW OBJECTIVE QUESTION*
> 
> *img836.imageshack.us/img836/1482/nextno.png
> ...



ans is 55
--------------------------------------------------------------------
Difference between 1st and 3rd number =4(2² )
Difference between 3rd and 5th number =9(3² )
Difference between 5th and 7th number =16(4² )

so Difference between 7th and 9th number =25(5² )

hence  number= 30+25
=55


----------



## mastercool8695 (May 6, 2013)

*Re: mathematic related questions*

55 is the *9*th number,

anyways, 
i was at some time, looking at the even and odd placed numbers together, but couldn't find the link..


----------



## rst (May 6, 2013)

*Re: mathematic related questions*

ok, edited.


----------



## mastercool8695 (May 6, 2013)

*Re: mathematic related questions*

x[SUP]2[/SUP]  + y[SUP]2[/SUP] = 2007
how many solutions (x,y) exist such that x and y are positive integers

A)NONE
B)EXACTLY 2
C)>2 BUT FINITELY MANY
D)INFINITELY MANY


----------



## rst (May 7, 2013)

*Re: mathematic related questions*



mastercool8695 said:


> x[SUP]2[/SUP]  + y[SUP]2[/SUP] = 2007
> how many solutions (x,y) exist such that x and y are positive integers
> 
> A)NONE
> ...



Ans is A)NONE
------------------------------------------------------------------------
x ²  + y ²   = 2007
y=√ (2007-x²)

It is defined for (2007-x² ) ≥ 0
2007 ≥ x²
√ 2007 ≥ x
44.79 ≥ x

As x is positive integer
x ∈ {1,2,..............,44}

these finite values of x will give finite values of y
So option (D) is rejected

As y is also positive integer
So in y=√ (2007-x²)
(2007-x²) should give square number
put finite values of x and check it

(2007-x²) will not give square value for any value of x from 1 to 44
------------------------------------------------------------------------------------
If you find this solution hard
then also there is another( easy) method of solving this question


----------



## Niilesh (May 7, 2013)

*Re: mathematic related questions*



rst said:


> Other method
> ------------------------------------------------------------------
> aⁿ+bⁿ is prime number
> where n is positive integer other than 1
> ...



Can you explain this expansion? -  aⁿ+bⁿ = (a+b) (aⁿ⁻¹  - aⁿ⁻² b² +................+(-1)ⁿ⁻¹  bⁿ⁻¹ )    when 'n' is odd


----------



## rst (May 7, 2013)

*Re: mathematic related questions*

This is an identity
It is valid when n is odd

For example
a³+b³ =(a+b) (a²-ab+b²)

EDIT : aⁿ+bⁿ =(a+b) (aⁿ⁻¹ - aⁿ⁻²* b* +................+(-1)ⁿ⁻¹ bⁿ⁻¹ )


----------



## Niilesh (May 7, 2013)

*Re: mathematic related questions*



rst said:


> Ans is A)NONE
> ------------------------------------------------------------------------
> x ²  + y ²   = 2007
> y=√ (2007-x²)
> ...


That's a little unconvincing statement 
Isn't there any way to prove that statement, except putting all the no.s? 



BTW My method :
x²  + y² = 2007 = 2004 + 3 = 4K +3 (where k = 2004/4)
Now we know that any perfect square when divided by 4 gives remainder either 0 or 1 
So 
     x²  + y² = 4K
or  x²  + y² = 4K +1
or  x²  + y² = 4K +2
Hence, no integral values of (x,y) will be possible.



rst said:


> This is an identity
> It is valid when n is odd
> 
> For example
> ...


It is derived from binomial expansion/equation or it was also proved with PMI ?


----------



## rst (May 7, 2013)

*Re: mathematic related questions*



Niilesh said:


> It is derived from binomial expansion/equation or it was also proved with PMI ?



I used the identity directly (as we do with the formulas)


Anyway we can derive this identity


----------



## mastercool8695 (May 7, 2013)

*Re: mathematic related questions*

^^ @nillesh : couldn't get the point buddy..


----------



## rst (May 7, 2013)

*Re: mathematic related questions*



Niilesh said:


> BTW My method :
> x²  + y² = 2007 = 2004 + 3 = 4K +3 (where k = 2004/4)
> Now we know that any perfect square when divided by 4 gives remainder either 0 or 1
> So
> ...


I think you mean
----------------------------------
2007 will give remainder 3 when divided by 4
 x²  + y² will give remainder 0,1,2 when divided by 4
As 2007 give remainder other than 0,1,2 so there is no solution
---------------------------------------------
But if we take example 
x²  + y² = 102
here 102 give remainder 2 when divided by 4
x²  + y² will give remainder 0,1,2 when divided by 4 
But even then there is no solution


----------



## Niilesh (May 7, 2013)

*Re: mathematic related questions*



rst said:


> I think you mean
> ----------------------------------
> 2007 will give remainder 3 when divided by 4
> x²  + y² will give remainder 0,1,2 when divided by 4
> ...



you got it wrong, I didn't said that if the remainder is 0, 1, 2 then we will get solutions. I said that if the remainder is 3 then their will be no solution



mastercool8695 said:


> ^^ @nillesh : couldn't get the point buddy..


It's Niilesh(or Nilesh whatever you like) 
I will explain the solution tomorrow when i will be online from PC


----------



## rst (May 7, 2013)

*Re: mathematic related questions*



rst said:


> I think you mean
> ----------------------------------
> 2007 will give remainder 3 when divided by 4
> x²  + y² will give remainder 0,1,2 when divided by 4
> As 2007 give *remainder other than 0,1,2* so there is no solution


I also meant the same
bolded line means remainder is 3


----------



## Niilesh (May 7, 2013)

*Re: mathematic related questions*



rst said:


> I also meant the same
> bolded line means remainder is 3


that line is correct 
I was replying to your example


----------



## rst (May 7, 2013)

*Re: mathematic related questions*



Niilesh said:


> that line is correct
> I was replying to your example



yeah

If a number number gives remainder 3 when divided by 4
then we have no solution in * "x² + y²=number"*

we can use this method with number giving remainder 3

But there are other number (as I mentioned in my example) which give no solution in * "x² + y²=number"*


----------



## Niilesh (May 8, 2013)

*Re: mathematic related questions*



mastercool8695 said:


> ^^ @nillesh : couldn't get the point buddy..


I will re frame my answer
Note: Any perfect square when divided by 4 gives remainder either 0 or 1
Proof:
We know that all numbers can be in the of form 4k or 4k+1 or 4k+2 or 4k+3 (where K is some integer)
Case 1: No. is of the form 4K
=>(4K)²= 4a + 0                        Remainder is 0

Case 2: No. is of the form 4K+1 
=>(4K + 1)²= 4b + 1                  Remainder is 1

Case 3: No. is of the form 4K+2
=>(4K+2)²= 4c + 2² = 4d            Remainder is 0

Case 4: No. is of the form 4K+3
=>(4K+3)² = 4e +3² = 4f +1        Remainder is 1
Hence, Proved.
----------------------------------------------------------------------------------------
Now it is given that 
x² + y² = 2007 = 2004 + 3 = 4K +3 (where k = 2004/4)
We know that any perfect square when divided by 4 gives remainder either 0 or 1
So
    x² + y² = 4K                    (if Both are x and y are even)
or x² + y² = 4K +1                (if one is even and other is odd) 
or x² + y² = 4K +2                (if Both are odd)

As 2007 give remainder other than 0,1,2 so no integral values of (x,y) will be possible




rst said:


> yeah
> 
> If a number number gives remainder 3 when divided by 4
> then we have no solution in * "x² + y²=number"*
> ...


ya, your method is better but a little lengthy.
BTW if I had to solve such questions i would have probably applied hit and trial using the following conditions :


> x² + y² = 4K                    (if Both are x and y are even)
> or x² + y² = 4K +1                (if one is even and other is odd)
> or x² + y² = 4K +2                (if Both are odd)


----------



## rst (May 8, 2013)

*Re: mathematic related questions*



Niilesh said:


> I will re frame my answer
> Note: Any perfect square when divided by 4 gives remainder either 0 or 1
> Proof:
> We know that all numbers can be in the of form 4k or 4k+1 or 4k+2 or 4k+3 (where K is some integer)
> ...



Now your proof looks nice


----------



## mastercool8695 (May 8, 2013)

*Re: mathematic related questions*

The number of different positive integral solutions (x,y,z) such that * x+y+z = 10*

A) 36
B) 121
C) 10[SUP]3[/SUP] -10
D) [SUP]10[/SUP]C[SUB]3[/SUB] - [SUP]10[/SUP]C[SUB]2[/SUB]

offtopic : do you guys know what is the shortcut key for subscript and superscript here in this forum ?
have to go advanced everytime


----------



## rst (May 8, 2013)

*Re: mathematic related questions*



Niilesh said:


> BTW if I had to solve such questions i would have probably applied hit and trial using the following conditions :



But you can use this method only with number giving remainder 3 (when divided by 4)
If a number gives remainder 0,1,2 ,then *x² + y²=number* may have solution or not
Also with this method we can't find number of solutions of *x² + y²=number*  (in case it has solution)



mastercool8695 said:


> offtopic : do you guys know what is the shortcut key for subscript and superscript here in this forum ?
> have to go advanced everytime



see my first post (page 1)


----------



## mastercool8695 (May 8, 2013)

*Re: mathematic related questions*

^^ so we are supposed to copy and paste ???


----------



## rst (May 8, 2013)

*Re: mathematic related questions*

Yeah
I use copy and paste


----------



## Niilesh (May 8, 2013)

*Re: mathematic related questions*



mastercool8695 said:


> The number of different positive integral solutions (x,y,z) such that * x+y+z = 10*
> 
> A) 36
> B) 121
> ...


Is it 9C2 = 36?


----------



## rst (May 8, 2013)

*Re: mathematic related questions*

yeah
it is the correct answer


----------



## Niilesh (May 8, 2013)

*Re: mathematic related questions*



rst said:


> But you can use this method only with number giving remainder 3 (when divided by 4)
> If a number gives remainder 0,1,2 ,then *x² + y²=number* may have solution or not
> Also with this method we can't find number of solutions of *x² + y²=number*  (in case it has solution)


I was referening to questions in which remainder is other than 3 only
You can get solutions with hit and trial and also prove that their will be no solution with logic


----------



## rst (May 8, 2013)

*Re: mathematic related questions*



Niilesh said:


> I was referening to questions in which remainder is other than 3 only
> You can get solutions with hit and trial and also prove that their will be no solution with logic


OK
No doubt,this is better method in case number has remainder 3



Niilesh said:


> Is it 9C2 = 36?



Can you explain your work ??


----------



## mastercool8695 (May 8, 2013)

*Re: mathematic related questions*

yup..
guys please dont just give answers.
give method and theories too..


----------



## rst (May 8, 2013)

*Re: mathematic related questions*



mastercool8695 said:


> The number of different positive integral solutions (x,y,z) such that * x+y+z = 10*
> 
> A) 36
> B) 121
> ...



* x+y+z = 10*
possibilities are
 1) 1,1,8 
2) 1,2,7
 3)1,3,6
 .
 .
 .
 8) 1,8,1

here I fixed 1 in first place 

similarly I can fix 2 in first place
 1) 2,1,7
 2) 2,2,6
 3)2,3,5
 .
 .
 .
 7) 2,7,1

similarly I can fix 3 in first place
 1) 3,1,6
 2) 3,2,5
 3)3,3,4
 .
 .
 .
 6) 3,6,1

so in this way 
we get = 8+7+6...+1
 =36ways

*-------------------------------------------------------------------------------------------------------------*
Another method

This question is same as no. of ways of distributing 10 items among 3 persons such that each receive atleast one item
So  ⁹C ₂  = 36

[As no. of ways of distributing n items among r persons such that each receive atleast one item=ⁿ⁻¹ Cr-1 ]


----------



## Niilesh (May 8, 2013)

*Re: mathematic related questions*



rst said:


> OK
> No doubt,this is better method in case number has remainder 3
> 
> 
> ...


I just realised we can also use multinomial theorem(also known as coefficient method) 



Not now, i am online from mobile.  BTW i had just recalled general result which is (n-1) C(r-1). It is useful to solve a lot of questions.


----------



## mastercool8695 (May 9, 2013)

*Re: mathematic related questions*

^^ can you guys post as many results relating to P&M like the one you gave..


----------



## rst (May 9, 2013)

*Re: mathematic related questions*

1) No. of ways of distributing n items among r persons such that one can receive 0 item = n+r-1 C r-1

Example : Find  number of different *non negative *integral solutions (x,y,z) such that x+y+z = 10
Solution:- This question is same as no. of ways of distributing 10 items among 3 persons such that one can receive 0 item 
 So we get ¹² C ₂


----------



## Amit Digga (May 9, 2013)

*Re: mathematic related questions*

What will be the remainder when
3^2002  +7^2002 + 2002 is divided by 29?


----------



## mastercool8695 (May 9, 2013)

*Re: mathematic related questions*

found this  :
MBA|CAT|CAT 2011|CAT 2012|CAT Online|MBA 2012|MBA Entrance Exams|CAT Test|Preparation|CAT Questions


> 3.3^2001 + 25^1001
> =>3.27^667 + 25^1001
> =>3.1^667  + (-1)^1001
> => 3-1
> ...



its related to the above one.. 
but i still cannot get the explaination in that forum (answer in Quote tags)


----------



## ©mß (May 9, 2013)

*Re: mathematic related questions*

I think remainder is 1.


----------



## mastercool8695 (May 9, 2013)

*Re: mathematic related questions*

^^ Explain..


----------



## rst (May 9, 2013)

*Re: mathematic related questions*

remainder is 1


----------



## TheSloth (May 9, 2013)

*Re: mathematic related questions*

damn. you guys are too much. I am very poor in these stuff. If someone explain the solution of above question i'll be thankful.


----------



## mastercool8695 (May 9, 2013)

*Re: mathematic related questions*

^^ exactly..
just writing the answer doesn't do anything


----------



## ©mß (May 9, 2013)

*Re: mathematic related questions*

I just used Python to solve it.
I don't know the explanation.


----------



## rst (May 9, 2013)

*Re: mathematic related questions*

remainder is 1
-----------------------------------------
3^2002 +7^2002 + 2002



3^2002 =(1+2)^2001  x 3 
=[2001C0 + 2001C1 x 2 + 2001C2 x 2^2 ..... +2001 C 2001 x 2^2001 ] x3
=[1 + 2001C1 x 2 + 2001C2 x 2^2 ..... +2^2001 ] x3

when divided by 29
remainder = (1 +0 +0 +.............+2) x3
=3 x 3
=9

7^2002 =(1+6)^2001  x 7
=[2001C0 + 2001C1 x 6+ 2001C2 x 6^2 ..... +2001 C 2001 x 6^2001 ] x7
=[1 + 2001C1 x 6 + 2001C2 x 6^2 ..... +6^2001 ] x7


when divided by 29
remainder = (1 +0 +0 +.............+*6*) x7
=7 x 7
=49
or remainder =20

2002 give remainder 1 when divided by 29

total remainder = 9+20+1
=30
which give remainder 1


----------



## Amit Digga (May 9, 2013)

*Re: mathematic related questions*



rst said:


> =[1 + 2001C1 x 6 + 2001C2 x 6^2 ..... +2^2001 x 3^2001 ] x7
> 
> remainder = (1 +0 +0 +.............+2 x3) x7




Can u please explain that ...2x3 in (1 +0 +0 +.............+2 x3) x7


----------



## mastercool8695 (May 9, 2013)

*Re: mathematic related questions*



©mß said:


> I just used Python to solve it.
> I don't know the explanation.



this is cheating... 
anyways redirect me to one of the download links .. 



rst said:


> remainder is 1
> -----------------------------------------
> 3^2002 +7^2002 + 2002
> 
> ...



couldn't get this.. is this any theory/theorem??


----------



## rst (May 9, 2013)

*Re: mathematic related questions*



mastercool8695 said:


> couldn't get this.. is this any theory/theorem??


When we divide 1 by 29 then remainder is 1
2001C1 x 2 = 2001 x 2 ,this will give remainder 0 when divisible by 29
2001C2 x 2^2 = [(2001 x 2000)/2] x 4 ,this will give remainder 0 when divisible by 29
.
.
.
2^2001 =(1+1)^2001 
 =[2001C0 + 2001C1  + 2001C2 ..... +2001 C 2001  ] 
=[1 + 2001C1  + 2001C2 ..... +1 ] 
When  divisible by 29
Then remainder = (1+0+0……+1)
=2



Amit Digga said:


> Can u please explain that ...2x3 in (1 +0 +0 +.............+2 x3) x7



Edited


----------



## mastercool8695 (May 9, 2013)

*Re: mathematic related questions*

^^ thanks. got it now.


----------



## ©mß (May 9, 2013)

*Re: mathematic related questions*



mastercool8695 said:


> anyways redirect me to one of the download links ..


Python is used for programming,it's not a calculator.(Think you know that.)


----------



## mastercool8695 (May 9, 2013)

yup..
buddy..
i know that 

ok leave it, i found the link

*www.python.org/download/

found this too..   *www.learnpython.org

will learn prog lingis after june 2


----------



## rst (May 9, 2013)

*----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION*

*img62.imageshack.us/img62/1227/scooterd.png

 Plz explain your answer

 (Also you can ask your mathematics related questions or queries
 we will try to solve them)
*----------------------------------------------------------------------------------------------------------------------*[/QUOTE]


----------



## Niilesh (May 9, 2013)

can someone give me the solution to the following question? 

Find range of (2-x)^½ + (1+x)^½.


----------



## rst (May 9, 2013)

Let f(x)=(2-x)^½ + (1+x)^½
it is defined for 2-x≥ 0 and 1+x ≥ 0
 2≥ x and x ≥ -1
-1 ≤x≤ 2
domain =[-1,2]

Put values in function and find range


----------



## Niilesh (May 9, 2013)

rst said:


> Let f(x)=(2-x)^½ + (1+x)^½
> it is defined for 2-x≥ 0 and 1+x ≥ 0
> 2≥ x and x ≥ -1
> -1 ≤x≤ 2
> ...



I have already tried it,  couldn't get the answer 
answer is [3^½, 6^½] , Can you justify it?

Finaly got it,  i just made a rough graph from sign of f'(x)


----------



## rst (May 10, 2013)

let f(x)=y
y=(2-x)^½ + (1+x)^½
squaring both side we ,get
y^2 = (2-x) +(1+x) + 2(2-x)^½(1+x)^½
y^2 = 3+ 2(2-x)^½(1+x)^½

Now 2(2-x)^½(1+x)^½ ≥ 0 for all x ∈ [-1,2]
 3+ 2(2-x)^½(1+x)^½  ≥ 3 
y^2 ≥ 3 
y  ≥ 3^½

So minimum value of y is  3^½

y has maximum value if (2-x)(1+x) is maximum
i.e if -x^2+x+2  is maximum
Let z= -x^2+x+2
Differentiating we get
z' =-2x+1

For max/min z'=0
-2x+1=0
x=1/2


At x= 1/2 
z"=-2 (negative)
Hence z is maximum at x= 1/2 


max value of  -x^2+x+2 = 9/4

max value of y^2 =3+ 2(2-x)^½(1+x)^½

y^2 =3+ 2 (9/4)^1/2
=3+2 (3/2)
y^2=6
y= 6 ^(1/2)

Hence range =[3^½,6 ^1/2 ]


----------



## rst (May 10, 2013)

rst said:


> *----------------------------------------------------------------------------------------------------------------------
> NEW OBJECTIVE QUESTION*
> 
> *img62.imageshack.us/img62/1227/scooterd.png
> ...



Ans is 5400
-------------------------------------------------
down payment =88800
Money in 60 instalment =60 x 7520
=451200 

Total money =451200 +88800
=540000

So money in each 100 instalment =540000/100
=5400


----------



## Niilesh (May 10, 2013)

*i.imgur.com/Tlmth3m.jpg
please explain your answer


----------



## rst (May 11, 2013)

Niilesh said:


> View attachment 10434
> please explain your answer



√ [1- f(x)] is defined for 1- f(x) ≥ 0
 1 ≥ f(x)
 cubing both sides
 1 ≥ f ³ (x)
 1 ≥ 1 - x³ -3x
 0 ≥ - x³ -3x
 x³ +3x ≥0
 x(x ² +3) ≥0
 x ≥ 0 [As (x ² +3) is always positive]

Hence domain = [0, ∞ )
 0r Domain = R + U {0}


----------



## Niilesh (May 11, 2013)

rst said:


> √ [1- f(x)] is defined for 1- f(x) ≥ 0
> 1 ≥ f(x)
> cubing both sides
> 1 ≥ f ³ (x)
> ...


you wrote the expression for f^3(x) wrong,  it also contains f(x) at the last


----------



## rst (May 11, 2013)

Anyway, what is the answer (A,B,C,or D) of this question


----------



## Niilesh (May 11, 2013)

Its C.
BTW i tried hit and trial and saw that it was satisfied for x=0, 1, 2 and it was not satisfied at x = -1 so by elimination i got the answer but cannot find a proper method


----------



## rst (May 11, 2013)

Then I think you misunderstood the question
f(x) is not part of f ³ (x)


----------



## Niilesh (May 11, 2013)

Maybe,  but there must have been a reason that it is mentioned in the question that f(1) ≠ -1


----------



## rst (May 12, 2013)

I think its just a condition
Also its looks true
example :-
f ³ (x) = 1 - x³ -3x
At x=1
f ³ (1) = 1 - (1)³ -3(1)

If f(1)= -1
-1 = 1-1 -3
-1 = -3
which is not true
So f(1) ≠ -1


----------



## rst (May 12, 2013)

*----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION*

*img819.imageshack.us/img819/3645/perimeter.png

 Plz explain your answer

 (Also you can ask your mathematics related questions or queries
 we will try to solve them)
*----------------------------------------------------------------------------------------------------------------------*


----------



## Niilesh (May 13, 2013)

Its too easy,  ans is A) 34 (7*4+3*2)


----------



## rst (May 13, 2013)

How did you get side 7 and 3 ??


----------



## darkv0id (May 13, 2013)

^He must have got it via trial and error, I guess.

Only two whole numbers have the sum of their squares equal 58, i.e. 7 and 3.


----------



## rst (May 13, 2013)

--------------------------------------
Let side of small square be x
Let side of big square be y
Area of figure =58
x ² + y ² = 58
y=√ (58-x²)

It is defined for (58-x² ) ≥ 0
58 ≥ x²
√ 58 ≥ x
7.6 ≥ x

As x is whole number
x ∈ {0,1,2,..........,7}


As y is also whole number
So in y=√ (58-x²)
(58-x²) should give square number

Possible solutions are (3,7), (7,3)
As x<y
So only solution (3,7)
Side of small square=3
Side of big square =7
Perimeter = 7+7+7+3+3+3+4
=34 cm


----------



## darkv0id (May 13, 2013)

^What's up with "2007" ?


----------



## rst (May 13, 2013)

edited
Actually ,this question is similar to x ² + y ² = 2007 [page 7,post 185]
So I did copy and paste


----------



## darkv0id (May 13, 2013)

rst said:


> edited
> Actually ,this question is similar to x ² + y ² = 2007 [page 7,post 185]
> So I did copy and paste



Ah ok.


----------



## rst (May 13, 2013)

*----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION*

*img560.imageshack.us/img560/8773/thyroid.png

 Plz explain your answer

 (Also you can ask your mathematics related questions or queries
 we will try to solve them)
*----------------------------------------------------------------------------------------------------------------------*


----------



## Niilesh (May 14, 2013)

rst said:


> How did you get side 7 and 3 ??


The Closest perfect sqaure no. less than 58 was 49  and 58 - 49= 9 = 3[sup]2[/sup]


rst said:


> *----------------------------------------------------------------------------------------------------------------------
> NEW OBJECTIVE QUESTION*
> 
> *img560.imageshack.us/img560/8773/thyroid.png
> ...



Ans is (4-3)*9 = 9


----------



## rst (May 14, 2013)

Niilesh said:


> Ans is (4-3)*9 = 9



Both ans and explanation are correct
well done !


----------



## TheSloth (May 14, 2013)

@rst: Are you doing B.Tech?


----------



## rst (May 14, 2013)

*----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION*

*img832.imageshack.us/img832/1183/trianglex.png

 Plz explain your answer

 (Also you can ask your mathematics related questions or queries
 we will try to solve them)
*----------------------------------------------------------------------------------------------------------------------*


----------



## Niilesh (May 14, 2013)

If its single correct than ans will clearly be (D) . it can be proved with cosine law BTW.

Edit: Alternate method
first draw ab=6 then make a line at an angle of 30° from it at B.  Now the least value of side AC will be the perpendicular distance of A from that line 
AC≥6 Sin 30°
AC≥3


----------



## rst (May 14, 2013)

Niilesh said:


> If its single correct than ans will clearly be (D) . it can be proved with cosine law BTW.



Yeah 
With cosine law ,we will get answer
cosB =(a² +c² -b² )/2ac



Niilesh said:


> Edit: Alternate method
> first draw ab=6 then make a line at an angle of 30° from it at B.  Now the least value of side AC will be the perpendicular distance of A from that line
> AC≥6 Sin 30°
> AC≥3



It is also valid method


----------



## Niilesh (May 15, 2013)

rst said:


> Then I think you misunderstood the question
> f(x) is not part of f ³ (x)


The Question was right 



Niilesh said:


> *i.imgur.com/Tlmth3m.jpg
> please explain your answer



We know that if a[sup]3[/sup] + b[sup]3[/sup] + c[sup]3[/sup] = 3abc then a + b + c = 0 
Now,
f³(x) = 1 - x³ -3xf(x)
=> f³(x) + (-1)³ + x³ = -3xf(x)
=> f(x) -1 + x = 0
=> x= 1-f(x) 

Now √(1- f(x)) is defined for 1- f(x) ≥ 0 
=>x≥ 0 
Domain --> R U {0}


----------



## rst (May 15, 2013)

Niilesh said:


> The Question was right
> 
> 
> 
> ...



It is not necessary
for example
if a=1,b=1,c=1
then a[sup]3[/sup] + b[sup]3[/sup] + c[sup]3[/sup] = 3abc 
but a + b + c = 3


----------



## Niilesh (May 15, 2013)

^ Ohh you are right actually it is the reverse


----------



## rst (May 15, 2013)

*----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION*

*img855.imageshack.us/img855/8792/installh.png

 Plz explain your answer

 (Also you can ask your mathematics related questions or queries
 we will try to solve them)
*----------------------------------------------------------------------------------------------------------------------*


----------



## Niilesh (May 15, 2013)

is it 1680? 
i am not sure how these transactions work


----------



## rst (May 15, 2013)

Answer is 1640


----------



## rst (May 16, 2013)

rst said:


> *----------------------------------------------------------------------------------------------------------------------
> NEW OBJECTIVE QUESTION*
> 
> *img855.imageshack.us/img855/8792/installh.png
> ...



If I =each  annual installment
P = total money borrowed
r = annual compound interest
n=no. of years

Then formula of annual installment is

[I/(1+r/100)]+[I/(1+r/100) ² ]+[I/(1+r/100)³]+-----+[I/(1+r/100)ⁿ ]= P

Here I =882
r = 5
n=2
Using in above formula

[I/(1+r/100)]+[I/(1+r/100) ² ]= P
[882/(1+5/100)]+[882/(1+5/100) ² ]= P

solving we get
P=1640


----------



## rst (May 17, 2013)

*----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION*

*img694.imageshack.us/img694/6659/principala.png

 Plz explain your answer

 (Also you can ask your mathematics related questions or queries
 we will try to solve them)
*----------------------------------------------------------------------------------------------------------------------*


----------



## Niilesh (May 18, 2013)

rst said:


> It is not necessary
> for example
> if a=1,b=1,c=1
> then a[sup]3[/sup] + b[sup]3[/sup] + c[sup]3[/sup] = 3abc
> but a + b + c = 3



Can you prove that (a[SUP]2[/SUP] + b[SUP]2[/SUP] + c[SUP]2[/SUP] − ab − bc − ca) is non zero? where  a,b,c = f(x),x,-1?
i.e. f(x)[SUP]2[/SUP] + x[SUP]2[/SUP] + 1 + f(x) + x -f(x)x ≠ 0


----------



## rst (May 18, 2013)

Niilesh said:


> Can you prove that (a[SUP]2[/SUP] + b[SUP]2[/SUP] + c[SUP]2[/SUP] − ab − bc − ca) is non zero? where  a,b,c = f(x),x,-1?
> i.e. f(x)[SUP]2[/SUP] + x[SUP]2[/SUP] + 1 + f(x) + x -f(x)x ≠ 0



Yeah
I can prove above thing
-----------------------------------------
if a[sup]3[/sup] + b[sup]3[/sup] + c[sup]3[/sup] = 3abc and (a[SUP]2[/SUP] + b[SUP]2[/SUP] + c[SUP]2[/SUP] − ab − bc − ca) ≠ 0
then a + b + c = 0 

In such case you will get your answer


----------



## Niilesh (May 18, 2013)

Can you care  to show your proof?


----------



## rst (May 20, 2013)

Suppose f(x)[sup]2[/sup] + x[sup]2[/sup] + 1 + f(x) + x -f(x)x =0
let f(x)=y
then
y[sup]2[/sup] + x[sup]2[/sup] + 1 + y + x -yx =0
y[sup]2[/sup] + y  -yx+ x[sup]2[/sup] + 1+ x  =0
y[sup]2[/sup] + y(1  -x)+ x[sup]2[/sup] + 1+ x  =0

Here a=1 , b=(1  -x) and c=x[sup]2[/sup] + 1+ x 

D=b[sup]2[/sup] -4ac
=(1  -x)[sup]2[/sup] -4(x[sup]2[/sup] + 1+ x )
= -(3 x[sup]2[/sup]+6x+3)
= -3(x[sup]2[/sup]+2x+1)
= -3 (x+1)[sup]2[/sup]
= - ve [as (x+1)[sup]2[/sup] is always positive]

So there is no real value of y [or f(x)] which satisfy this equation

Hence  f(x)[sup]2[/sup] + x[sup]2[/sup] + 1 + f(x) + x -f(x)x =0 is not possible for any real fuction f(x)


----------



## rst (May 20, 2013)

rst said:


> *----------------------------------------------------------------------------------------------------------------------
> NEW OBJECTIVE QUESTION*
> 
> *img694.imageshack.us/img694/6659/principala.png
> ...



S.I for 2 years =6800 -6080
=720

So S.I for 3 years =(720/2) x 3
=1080

Amount for 3 years =6080

Hence Principal= 6080 - 1080
=5000


----------



## Niilesh (May 20, 2013)

rst said:


> Suppose f(x)[sup]2[/sup] + x[sup]2[/sup] + 1 + f(x) + x -f(x)x =0
> let f(x)=y
> then
> y[sup]2[/sup] + x[sup]2[/sup] + 1 + y + x -yx =0
> ...


 Thanks


----------



## rst (May 20, 2013)

*----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION*

*img577.imageshack.us/img577/204/histogram.png

 Plz explain your answer

 (Also you can ask your mathematics related questions or queries
 we will try to solve them)
*----------------------------------------------------------------------------------------------------------------------*


----------



## rst (May 21, 2013)

Ans is *Median* (I think no explanation is required)


----------



## rst (May 22, 2013)

*----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION*

*img812.imageshack.us/img812/4310/circlef.png

 Plz explain your answer

 (Also you can ask your mathematics related questions or queries
 we will try to solve them)
*----------------------------------------------------------------------------------------------------------------------*


----------



## darkv0id (May 22, 2013)

My answer is coming out to be [2*(a^2)*(b^2)]/[sqrt(a^2 + b^2)].

First we find the equation of the radical axis, i.e. S-S'= 0

=>ax -by = 0

Then we find distance of the above line from any one centre, and then use Pyth. Theorem to find the length of the chord. 

Alternatively, we can simply find the points of intersection of the common chord with any one circle, then use distance formula to calculate the length.

Please confirm if my answer is correct.


----------



## rst (May 23, 2013)

darkv0id said:


> My answer is coming out to be [2*(a^2)*(b^2)]/[sqrt(a^2 + b^2)].
> 
> First we find the equation of the radical axis, i.e. S-S'= 0
> 
> ...



Yeah
your method is correct
-----------------------------------------
equation of common chord,S-S'= 0
ax -by = 0

Distance of centre (a,0) from common chord,
d= a[sup]2[/sup] / √ (a[sup]2[/sup]+b[sup]2[/sup])
Radius of circle,r =a

By Pythagoras th,
d[sup]2[/sup] +p[sup]2[/sup] =r[sup]2[/sup]
 [a[sup]4[/sup] / (a[sup]2[/sup]+b[sup]2[/sup])]+p[sup]2[/sup] =a[sup]2[/sup]
solving we get
p= ab/√ (a[sup]2[/sup]+b[sup]2[/sup]

Length of chord =2p
=2ab/√ (a[sup]2[/sup]+b[sup]2[/sup])


----------



## Niilesh (May 23, 2013)

@rst Can you post some good question on P&C? most of them are interesting


----------



## rst (May 23, 2013)

This thread is only for maths questions

For P&C questions ,we have to start new thread


----------



## nisargshah95 (May 23, 2013)

rst said:


> This thread is only for maths questions
> 
> For P&C questions ,we have to start new thread


P&C comes under maths.


----------



## rst (May 23, 2013)

I thought he mean "physics and chemistry"


----------



## Niilesh (May 23, 2013)

I mean Permutation and combinations


----------



## rst (May 24, 2013)

*----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION*

*img27.imageshack.us/img27/2843/penzr.png

 Plz explain your answer

 (Also you can ask your mathematics related questions or queries
 we will try to solve them)
*----------------------------------------------------------------------------------------------------------------------*


----------



## Niilesh (May 24, 2013)

^ 329 ( [sup]11[/sup]c[sub]4[/sub] - 1 )


----------



## rst (May 24, 2013)

Niilesh said:


> ^ 329 ( [sup]11[/sup]c[sub]4[/sub] - 1 )



total pen =11

So you use  [sup]11[/sup]c[sub]4[/sub]

Then why you subract 1

plz explain ans in some more detail


----------



## Niilesh (May 24, 2013)

rst said:


> total pen =11
> 
> So you use  [sup]11[/sup]c[sub]4[/sub]
> 
> ...


TBH i found that to be a little too easy that's why....

Let us assume that i take all 11 pens from both of them and then divide it into two groups of 4 & 7(pens are different). No. of ways dividing the the pens would be [sup]11[/sup]c[sub]4[/sub] = [sup]11[/sup]c[sub]7[/sub]. (we then give the new group of pens to both of them) 
Now we subtract 1 because we also counted the initial configuration. 
So ans is [sup]11[/sup]c[sub]4[/sub]


----------



## rst (May 24, 2013)

*----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION*

*img850.imageshack.us/img850/7996/roser.png

 Plz explain your answer

 (Also you can ask your mathematics related questions or queries
 we will try to solve them)
*----------------------------------------------------------------------------------------------------------------------*


----------



## Niilesh (May 24, 2013)

^ If we consider all roses of same colour to be identical then ans will be 1.

If we consider all of them distinct then my ans is coming 3600((6-1)! * [sup]5[/sup]c[sub]1[/sub] * 3!).
Can you give the answer in terms of factorial and/or C(n,r)?


----------



## rst (May 24, 2013)

rst said:


> *----------------------------------------------------------------------------------------------------------------------
> NEW OBJECTIVE QUESTION*
> 
> *img850.imageshack.us/img850/7996/roser.png
> ...


As white roses come together ,so we consider them as one flower
 so we have 7 flowers
 they can form garland in 6! ways
 Also white flowers can be arranged in 3! ways
 total ways =6! x 3!

As  there is no difference between clockwise and anticlockwise permutation
So total ways =(6! x 3! )/2
=4320/2
=2160


----------



## Niilesh (May 24, 2013)

rst said:


> I think we have to consider all of them distinct
> ------------------------------------------------------------------------------------
> As white roses come together ,so we consider them as one flower
> Now total flowers = 7
> ...



They can be arranged in 6! in a garland not 7! as only order matter there is no positions in a circle. I also did it first by same method but i think some cases will be repeated in it.


----------



## rst (May 24, 2013)

Yeah
I missed to edit it due to slow Internet (5 KBps)


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## Niilesh (May 24, 2013)

Point out mistake in this method

First i made a garland of red flowers in 5! ways. Then i selected 1 gap out of 5 gaps between the red flowers to put white flowers then arranged those flowers in 3! ways


----------



## rst (May 24, 2013)

Niilesh said:


> Point out mistake in this method
> 
> First i made a garland of red flowers in 5! ways. Then i selected 1 gap out of 5 gaps between the red flowers to put white flowers then arranged those flowers in 3! ways



your method is correct
 It will also give same result


----------



## Niilesh (May 24, 2013)

rst said:


> your method is correct
> It will also give same result



6! x 3! =/= 5! x 5c1 x3!
4320 =/= 3600


----------



## rst (May 24, 2013)

Niilesh said:


> 6! x 3! =/= 5! x 5c1 x3!
> 4320 =/= 3600



According to your method
Total ways = 5! x 6 x3!

=6! 3!
=4320

note:-gap should 6 (as there are 6 gaps if we arrange them in linear way)


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## Niilesh (May 24, 2013)

rst said:


> According to your method
> Total ways = 5! x 5 x3! +5!3!
> =5!3! (5+1)
> =6x 5!3!
> =6! x 3!


What's that extra 5!3! for?


----------



## hari1 (May 24, 2013)

2^(x)=0
Find x. This is surely going to scratch your brains if you don't read it properly.


----------



## Niilesh (May 24, 2013)

I know this is not the answer but still : x is tending to minus infinity.


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## rst (May 25, 2013)

hari1 said:


> 2^(x)=0
> Find x. This is surely going to scratch your brains if you don't read it properly.



Range of 2[sup]x[/sup]  function is (0,∞)
So 2[sup]x[/sup]=0 is not possible for any real value of x
Hence 2[sup]x[/sup]=0 has no solution

In other words, there is no such x in R
---------------------------------------------
However, It is possible in extended real line i.e [-∞,∞]
2[sup]x[/sup]=0 is possible for x= -∞


----------



## Niilesh (May 25, 2013)

Niilesh said:


> What's that extra 5!3! for?


ohh.. I realised that there were 6 spaces not 5


----------



## rst (May 25, 2013)

*----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION*

*img15.imageshack.us/img15/2318/probability.png

 Plz explain your answer

 (Also you can ask your mathematics related questions or queries
 we will try to solve them)
*----------------------------------------------------------------------------------------------------------------------*


----------



## Niilesh (May 25, 2013)

^ No. of equilateral triangle possible = 6/3 = 2(6 vertices, 3 vertices have the same triangle)
Total No. of triangles possible 6C3 = 5*4 = 20
So probability = 2/20 = 1/10


----------



## rst (May 25, 2013)

*----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION*

*img191.imageshack.us/img191/8311/comkl.png

 Plz explain your answer

 (Also you can ask your mathematics related questions or queries
 we will try to solve them)
*----------------------------------------------------------------------------------------------------------------------*


----------



## rst (May 26, 2013)

rst said:


> *----------------------------------------------------------------------------------------------------------------------
> NEW OBJECTIVE QUESTION*
> 
> *img191.imageshack.us/img191/8311/comkl.png
> ...


C(n+1,5) -C(n,4) = 56

solving we get
n (n-1)(n-2)(n-3)(n-4)= 6720

1) At most 7

not possible

2)Atleast 10
not possible

3) n=8 will give the correct ans

Ans is (3)


----------



## nisargshah95 (May 26, 2013)

Here is a really good question of functions I found in one of Arihant's books -

 If f(x) = 4[SUP]x[/SUP]/(4[SUP]x[/SUP] +2), then [ f(1/2008) + f(2/2008) + ... + f(2007/2008) ] - 1000 is equal to _  (where [.] Denotes the greatest integer function)


Answer -


Spoiler



3


----------



## rst (May 27, 2013)

nisargshah95 said:


> Here is a really good question of functions I found in one of Arihant's books -
> 
> If f(x) = 4[SUP]x[/SUP]/(4[SUP]x[/SUP] +2), then [ f(1/2008) + f(2/2008) + ... + f(2007/2008) ] - 1000 is equal to _  (where [.] Denotes the greatest integer function)
> 
> ...



f(x) = 4[SUP]x[/SUP]/(4[SUP]x[/SUP] +2)

f(1-x)=2/(4[SUP]x[/SUP] +2)

so f(x) + f(1-x)=1

f(1/2008) + f(2007/2008)=1
f(2/2008) + f(2006/2008)=1
f(3/2008) + f(2005/2008)=1
.
.
.
f(1003/2008) + f(1005/2008)=1
f(1004/2008)=1/2

Adding above ,we get
f(1/2008) + f(2/2008)+f(3/2008)+............ +f(2007/2008)=2007/2
=1003.5

Hence [ f(1/2008) + f(2/2008) + ... + f(2007/2008) ] - 1000 =[1003.5]-1000
=1003-1000
=3


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## nisargshah95 (May 27, 2013)

Dude did you solve this on your own or just copied the solution? Its amazing if you did it on your own. Really needs a lot of brains to find that f(x) + f(1-x) = 1. Never occurred to me.


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## rst (May 27, 2013)

*----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION*

*img163.imageshack.us/img163/9162/combau.png

 Plz explain your answer

 (Also you can ask your mathematics related questions or queries
 we will try to solve them)
*----------------------------------------------------------------------------------------------------------------------*
*you can use following mathematic symbols(just do copy and paste)*
 £ ω ∴ ∂ Φ γ δ μ σ Є Ø Ω ∩ ≈ ≡ ≈ ≅ ≠ ≤ ≥ • ~ ± ∓ ∤ ◅∈ ∉ ⊆ ⊂ ∪ ⊥ ô ∫ Σ → ∞ Π Δ Ψ Γ ∮ ∇∂ √ °α β γ δ ε ζ η θ ι κ λ ν ξ ο π ρ σ τ υ φ χ ψ 
x⁰ ¹ ² ³ ⁴ ⁵ ⁶ ⁷ ⁸ ⁹ ⁺ ⁻ ⁿ 
 x₀ ₁ ₂ ₃ ₄ ₅ ₆ ₇ ₈ ₉ ₊ ₋


----------



## rst (May 28, 2013)

rst said:


> *----------------------------------------------------------------------------------------------------------------------
> NEW OBJECTIVE QUESTION*
> 
> *img163.imageshack.us/img163/9162/combau.png
> ...



total people =15
7 people can be selected in ¹⁵C₇ ways
7 people can be seated around round table in 6! ways

remaining 8 people can be  seated around round  table in 7! ways

total ways =¹⁵C₇ x 6! x 7!


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## rst (May 29, 2013)

*----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION*

*img90.imageshack.us/img90/51/workpaint.png

 Plz explain your answer

 (Also you can ask your mathematics related questions or queries
 we will try to solve them)
*----------------------------------------------------------------------------------------------------------------------*
*you can use following mathematic symbols(just do copy and paste)*
 £ ω ∴ ∂ Φ γ δ μ σ Є Ø Ω ∩ ≈ ≡ ≈ ≅ ≠ ≤ ≥ • ~ ± ∓ ∤ ◅∈ ∉ ⊆ ⊂ ∪ ⊥ ô ∫ Σ → ∞ Π Δ Ψ Γ ∮ ∇∂ √ °α β γ δ ε ζ η θ ι κ λ ν ξ ο π ρ σ τ υ φ χ ψ 
x⁰ ¹ ² ³ ⁴ ⁵ ⁶ ⁷ ⁸ ⁹ ⁺ ⁻ ⁿ 
 x₀ ₁ ₂ ₃ ₄ ₅ ₆ ₇ ₈ ₉ ₊ ₋


----------



## hari1 (May 29, 2013)

Evaluate sin(cot^(-1)x)


----------



## rst (May 29, 2013)

hari1 said:


> Evaluate sin(cot^(-1)x)


sin(cot^(-1)x)

sin [sin⁻¹(1/√(1+x² )] 
=1/√(1+x² )
----------------------------------------------
long method:-

sin(cot^(-1)x)---(1)

let cot^(-1)x =y----------(2)
then cot y = x

sin y = 1/√(1+x² ) 

y= sin⁻¹(1/√(1+x² )

using in (2) we get,
cot^(-1)x =sin⁻¹(1/√(1+x² )

hence (1) becomes
sin [sin⁻¹(1/√(1+x² )] 
=1/√(1+x² )


----------



## rst (May 30, 2013)

rst said:


> *----------------------------------------------------------------------------------------------------------------------
> NEW OBJECTIVE QUESTION*
> 
> *img90.imageshack.us/img90/51/workpaint.png
> ...



let number of days be x

If one person works 4 hours daily
Then each person will work (8+8+8+8+4)/5
=36/5 hours each day

Now number of persons and no. of days are in indirect proportion
Also number of hours and no. of days are in indirect proportion

So 5 * 3 * 8= 5 *  x  *(36/5)
x=10/3

(i think there is misprinting in option D )
(It should be 10/3)


----------



## rst (May 31, 2013)

*----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION*

*img407.imageshack.us/img407/9670/limitd.png

 Plz explain your answer

 (Also you can ask your mathematics related questions or queries
 we will try to solve them)
*----------------------------------------------------------------------------------------------------------------------*
*you can use following mathematic symbols(just do copy and paste)*
 £ ω ∴ ∂ Φ γ δ μ σ Є Ø Ω ∩ ≈ ≡ ≈ ≅ ≠ ≤ ≥ • ~ ± ∓ ∤ ◅∈ ∉ ⊆ ⊂ ∪ ⊥ ô ∫ Σ → ∞ Π Δ Ψ Γ ∮ ∇∂ √ °α β γ δ ε ζ η θ ι κ λ ν ξ ο π ρ σ τ υ φ χ ψ 
x⁰ ¹ ² ³ ⁴ ⁵ ⁶ ⁷ ⁸ ⁹ ⁺ ⁻ ⁿ 
 x₀ ₁ ₂ ₃ ₄ ₅ ₆ ₇ ₈ ₉ ₊ ₋


----------



## Thetrueblueviking (Jun 1, 2013)

Here solve the limit in two parts - 

part 1 - sinx^1/x - you can easily see it is approaching zero because sinx tends to 0 and you are raising a number tending to 0 to a number tending to infinity - so the obvious result = 0.

part 2 - [1/x]^sinx = 1^sinx/x^sinx = 1/x^sinx since 1^y = 1 where y ∈ R. 
So part 2 = 1/x^sinx. 
lets solve for x^sinx -
x^sinx = y
take logarithm - sinx log x = log y
∴ logx / cosec x = log y
now the lhs is turned into an infinity/infinty limit - so apply l'hospital's rule - 
lhs = -[1/x]/cosecx.cotx = -sinx.sinx/x.cosx = -sinx/cosx = 0
∴log y = 0
∴ y = x^sinx = 1
∴1/x^sinx = 1

now final answer - sinx^1/x + [1/x]^sinx = 0 + 1 = 1 = D. NONE OF THESE

Why is it that every time you your-self answer your question - are you trying to take a test of the TD members asking questions to which you already know the answers or do you seriously dont have the answers to them when you ask the qts //


----------



## rst (Jun 1, 2013)

Thetrueblueviking said:


> Here solve the limit in two parts -
> 
> part 1 - sinx^1/x - you can easily see it is approaching zero because sinx tends to 0 and you are raising a number tending to 0 to a number tending to infinity - so the obvious result = 0.
> 
> ...



Great explanation

Thanks a lot



Thetrueblueviking said:


> Why is it that every time you your-self answer your question - are you trying to take a test of the TD members asking questions to which you already know the answers or do you seriously dont have the answers to them when you ask the qts //



I dont have the answers to them when I ask the qts
But If don't get answer then I try myself to give ans.


----------



## Thetrueblueviking (Jun 1, 2013)

You're welcome - btw I love maths.


----------



## rst (Jun 1, 2013)

*----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION*

*img713.imageshack.us/img713/5994/limitint.png

 Plz explain your answer

 (Also you can ask your mathematics related questions or queries
 we will try to solve them)
*----------------------------------------------------------------------------------------------------------------------*
*you can use following mathematic symbols(just do copy and paste)*
 £ ω ∴ ∂ Φ γ δ μ σ Є Ø Ω ∩ ≈ ≡ ≈ ≅ ≠ ≤ ≥ • ~ ± ∓ ∤ ◅∈ ∉ ⊆ ⊂ ∪ ⊥ ô ∫ Σ → ∞ Π Δ Ψ Γ ∮ ∇∂ √ °α β γ δ ε ζ η θ ι κ λ ν ξ ο π ρ σ τ υ φ χ ψ 
x⁰ ¹ ² ³ ⁴ ⁵ ⁶ ⁷ ⁸ ⁹ ⁺ ⁻ ⁿ 
 x₀ ₁ ₂ ₃ ₄ ₅ ₆ ₇ ₈ ₉ ₊ ₋


----------



## Thetrueblueviking (Jun 1, 2013)

Is the answer - b ? If so, then I would post my explanation.


----------



## rst (Jun 1, 2013)

Well, I don't know ans


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## Thetrueblueviking (Jun 1, 2013)

I got 1 but through a very long method - It'd be painful to type.


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## rst (Jun 1, 2013)

Give ans in short

Write only main steps or just give hint


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## Thetrueblueviking (Jun 1, 2013)

Lol - I committed a blunder so 1 is not the ans.

I could tell you how to go about it but its a tedious task.

First take the x-sinx out of the integral.

Then integrate the function in t by some method - I used substitution. 

Then Substitute the limits and you have function in x.

Now solve for final integral/x-sin x (x tends to 0) = 1.

Apply LH a couple of times and use the condition that if limit is to exist and since denom is 0 - then num should as well be 0 and find 'a'  OR  apply LH till the 0 in the denom vanishes and equate to 1 to get a.


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## rst (Jun 1, 2013)

Thetrueblueviking said:


> Lol - I committed a blunder so 1 is not the ans.
> 
> I could tell you how to go about it but its a tedious task.
> 
> ...


You put t = ?? 
i.e you put what value to t (in sustitution method)


----------



## Thetrueblueviking (Jun 1, 2013)

Take a new variable say P=sqrt(a+t)

So P^2 = a + t 
So 2PdP = dt
and also t^2 = (P^2-a)^2

so integral t^2 dt /sqrt (a+t) = [(P^2-a)^2.2P.dP]/P = 2(P^2-a)^2.dP

Note after integrating in P - the limits will be changed.


----------



## rst (Jun 1, 2013)

rst said:


> *----------------------------------------------------------------------------------------------------------------------
> NEW OBJECTIVE QUESTION*
> 
> *img713.imageshack.us/img713/5994/limitint.png
> ...


We can also solve this question by putting P = a + t
then it will give ∫(x+a,a) [(P-a)² / √P ] dP  {here x+a is upper limit and a is lower limit }

∫(x+a,a) [P^(3/2)  +a² P^ (-1/2) + 2a √P] dP 

Solving we ,get
2/5 (x+a)^ (5/2) +2a² √(x+a) +(4/3) a * (x+a)^(3/2) - [2/5 (a)^ (5/2)] -[2 a² √a] -[(4/3) a * (a)^(3/2) ]

This is easier to do in LH rule

By solving with LH rule ,we get

1/ √a =1

which is possible for a=1


----------



## Thetrueblueviking (Jun 1, 2013)

rst said:


> We can also solve this question by putting P = a + t
> 
> By solving with LH rule ,we get
> 
> ...



You get 1 even by doing what I said 
A bit tedious though


----------



## rst (Jun 2, 2013)

*----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION*

*img545.imageshack.us/img545/2017/intexp.png

 Plz explain your answer

 (Also you can ask your mathematics related questions or queries
 we will try to solve them)
*----------------------------------------------------------------------------------------------------------------------*
*you can use following mathematic symbols(just do copy and paste)*
 £ ω ∴ ∂ Φ γ δ μ σ Є Ø Ω ∩ ≈ ≡ ≈ ≅ ≠ ≤ ≥ • ~ ± ∓ ∤ ◅∈ ∉ ⊆ ⊂ ∪ ⊥ ô ∫ Σ → ∞ Π Δ Ψ Γ ∮ ∇∂ √ °α β γ δ ε ζ η θ ι κ λ ν ξ ο π ρ σ τ υ φ χ ψ 
x⁰ ¹ ² ³ ⁴ ⁵ ⁶ ⁷ ⁸ ⁹ ⁺ ⁻ ⁿ 
 x₀ ₁ ₂ ₃ ₄ ₅ ₆ ₇ ₈ ₉ ₊ ₋


----------



## rst (Jun 3, 2013)

rst said:


> *----------------------------------------------------------------------------------------------------------------------
> NEW OBJECTIVE QUESTION*
> 
> *img545.imageshack.us/img545/2017/intexp.png
> ...



Put x= t²

Then integral will become
2 ∫(1,2) t e^t dt {here 1 is lower limit}

Then by using integration by parts, we get

ans 2e² (option 4)


----------



## rst (Jun 4, 2013)

*----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION*

*img14.imageshack.us/img14/8696/kvalue.png

 Plz explain your answer

 (Also you can ask your mathematics related questions or queries
 we will try to solve them)
*----------------------------------------------------------------------------------------------------------------------*
*you can use following mathematic symbols(just do copy and paste)*
 £ ω ∴ ∂ Φ γ δ μ σ Є Ø Ω ∩ ≈ ≡ ≈ ≅ ≠ ≤ ≥ • ~ ± ∓ ∤ ◅∈ ∉ ⊆ ⊂ ∪ ⊥ ô ∫ Σ → ∞ Π Δ Ψ Γ ∮ ∇∂ √ °α β γ δ ε ζ η θ ι κ λ ν ξ ο π ρ σ τ υ φ χ ψ 
x⁰ ¹ ² ³ ⁴ ⁵ ⁶ ⁷ ⁸ ⁹ ⁺ ⁻ ⁿ 
 x₀ ₁ ₂ ₃ ₄ ₅ ₆ ₇ ₈ ₉ ₊ ₋


----------



## rst (Jun 5, 2013)

rst said:


> *----------------------------------------------------------------------------------------------------------------------
> NEW OBJECTIVE QUESTION*
> 
> *img14.imageshack.us/img14/8696/kvalue.png
> ...



In ∫(0 to 1) (tan⁻¹ x /x )dx 

Put x =tanθ

Then integral will become
∫(0 to π/4 ) [ (θ sec² θ) /tanθ ]dθ
∫(0 to π/4 ) [ (2θ /sin2θ ]dθ

Put 2θ =t ,we get
∫(0 to π/2 ) [ (t/2sint )dt

Using in given question ,we get
k=1/2


----------



## rst (Jun 6, 2013)

*----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION*

*img811.imageshack.us/img811/6514/domaint.png

 Plz explain your answer

 (Also you can ask your mathematics related questions or queries
 we will try to solve them)
*----------------------------------------------------------------------------------------------------------------------*
*you can use following mathematic symbols(just do copy and paste)*
 £ ω ∴ ∂ Φ γ δ μ σ Є Ø Ω ∩ ≈ ≡ ≈ ≅ ≠ ≤ ≥ • ~ ± ∓ ∤ ◅∈ ∉ ⊆ ⊂ ∪ ⊥ ô ∫ Σ → ∞ Π Δ Ψ Γ ∮ ∇∂ √ °α β γ δ ε ζ η θ ι κ λ ν ξ ο π ρ σ τ υ φ χ ψ 
x⁰ ¹ ² ³ ⁴ ⁵ ⁶ ⁷ ⁸ ⁹ ⁺ ⁻ ⁿ 
 x₀ ₁ ₂ ₃ ₄ ₅ ₆ ₇ ₈ ₉ ₊ ₋


----------



## mastercool8695 (Jun 7, 2013)

Solve these (all are ISI interview Questions) : 


1.  Suppose n lines are drawn on a plane. Some of them can be concurrent (pass through same point). How many different regions created in this process? (find the least and the greatest number of regions that can be created).

2.  Suppose there are 'n' circles no three of which pass through the same point and all of which intersect every other circle at two points. How many regions are created?

3.  Suppose there are 100 points on a plane no three of which are on the same straight line. Can you draw a line on the plane such that 50 points are on one side of it?

4.  A game is played between two players. There is a round table and unlimited supply of stones (dimension of the stone is unimportant). In each turn of the game a player can put one stone on the table. Whoever fails to find space one the table looses the game. Find a winning strategy for the first player.

5.  On a 20 by 20 board a special knight is moving. In each turn the knight moves 1 step in a direction and 5 steps in a direction perpendicular to it. The knight is allowed to take as many turns as required. Can it come back to any of it's four adjacent squares of the square from which it started moving?

6.  We often say that log (1+x) = x - x^2/2 + x^3/3 etc. Is the statement true for x = 500 (or may be larger values?) If not, why?

7.  Find a point on the plane of a triangle such that the sum of it's distances from three vertices is minimum.


----------



## rst (Jun 7, 2013)

mastercool8695 said:


> 6.  We often say that log (1+x) = x - x^2/2 + x^3/3 etc. Is the statement true for x = 500 (or may be larger values?) If not, why?



log (1+x) = x - x^2/2 + x^3/3 -......

It is valid for |x|<1
As 500 >1
So it is not true for x=500



mastercool8695 said:


> 7.  Find a point on the plane of a triangle such that the sum of it's distances from three vertices is minimum.



Such point is known as fermat point.
For solution here is the link; *en.wikipedia.org/wiki/Fermat_point


----------



## mastercool8695 (Jun 7, 2013)

thanks..
forwarding it to him..
great knowledge for me too.

thanks..


----------



## rst (Jun 8, 2013)

rst said:


> *----------------------------------------------------------------------------------------------------------------------
> NEW OBJECTIVE QUESTION*
> 
> *img811.imageshack.us/img811/6514/domaint.png
> ...



Function is defined for |[ |x|-1]|-5 >0  {As denominator can't be 0}
|[ |x|-1]| >5

=> [ |x|-1] >5 or [ |x|-1] < -5
=> [ |x|]-1>5 or [ |x|]-1 < -5  {As [x-1]=[x]-1 }
=> [ |x|]>6 or [ |x|]< -4
=> |x|≥ 7 or  |x|< -4
=> |x|≥7  {here |x|< -4 rejected as modulus function can't be negative}
=> x≥ 7 or  x≤  -7 

So ans is option (1)


----------



## Vignesh B (Jun 8, 2013)

Prove 0! = 1.
Quite easy one though, but I found the proof quite interesting.


----------



## rst (Jun 8, 2013)

Vignesh B said:


> Prove 0! = 1.
> Quite easy one though, but I found the proof quite interesting.



n! =n (n-1)!

n! /n =(n-1)!

put n=1 ,then we get
1! /1 =0!
1= 0!


----------



## Vignesh B (Jun 8, 2013)

rst said:


> n! =n (n-1)!
> 
> n! /n =(n-1)!
> 
> ...


Correct! 
I had seen this on numberphile. They themselves gave the solution that you have give, but later stated that by this formula, -1! = 0!/0 = 1/0 which negates mathematic's rule. They have give another solution too.


----------



## rst (Jun 8, 2013)

*----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION*

*img19.imageshack.us/img19/7158/inverseq.png

 Plz explain your answer

 (Also you can ask your mathematics related questions or queries
 we will try to solve them)
*----------------------------------------------------------------------------------------------------------------------*
*you can use following mathematic symbols(just do copy and paste)*
 £ ω ∴ ∂ Φ γ δ μ σ Є Ø Ω ∩ ≈ ≡ ≈ ≅ ≠ ≤ ≥ • ~ ± ∓ ∤ ◅∈ ∉ ⊆ ⊂ ∪ ⊥ ô ∫ Σ → ∞ Π Δ Ψ Γ ∮ ∇∂ √ °α β γ δ ε ζ η θ ι κ λ ν ξ ο π ρ σ τ υ φ χ ψ 
x⁰ ¹ ² ³ ⁴ ⁵ ⁶ ⁷ ⁸ ⁹ ⁺ ⁻ ⁿ 
 x₀ ₁ ₂ ₃ ₄ ₅ ₆ ₇ ₈ ₉ ₊ ₋


----------



## rst (Jun 9, 2013)

rst said:


> *----------------------------------------------------------------------------------------------------------------------
> NEW OBJECTIVE QUESTION*
> 
> *img19.imageshack.us/img19/7158/inverseq.png
> ...



We can rewrite above question as
S= Σ r (1 to ∞) tan⁻¹ [(2r )/(2+r² +r⁴)]

S= lim (n→ ∞)  Σ r (1 to n) tan⁻¹ [(2r )/(2+r² +r⁴)]

S= lim (n→ ∞)  Σ r (1 to n) tan⁻¹ {[(1+r² +r )-(1+r² -r )] / [1+(1+r² +r )(1+r² -r )]}

S= lim (n→ ∞)  Σ r (1 to n) tan⁻¹ (1+r² +r ) - tan⁻¹ (1+r² -r )

S= lim (n→ ∞)  tan⁻¹ (1+n² +n )  - tan⁻¹ (1)

S= lim (n→ ∞)  tan⁻¹ [(n² +n )/(n² +n+2)]

S=tan⁻¹ (1)

S=  π /4


----------



## rst (Jun 11, 2013)

*----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION*

*img827.imageshack.us/img827/8467/rangeb.png

 Plz explain your answer

 (Also you can ask your mathematics related questions or queries
 we will try to solve them)
*----------------------------------------------------------------------------------------------------------------------*
*you can use following mathematic symbols(just do copy and paste)*
 £ ω ∴ ∂ Φ γ δ μ σ Є Ø Ω ∩ ≈ ≡ ≈ ≅ ≠ ≤ ≥ • ~ ± ∓ ∤ ◅∈ ∉ ⊆ ⊂ ∪ ⊥ ô ∫ Σ → ∞ Π Δ Ψ Γ ∮ ∇∂ √ °α β γ δ ε ζ η θ ι κ λ ν ξ ο π ρ σ τ υ φ χ ψ 
x⁰ ¹ ² ³ ⁴ ⁵ ⁶ ⁷ ⁸ ⁹ ⁺ ⁻ ⁿ 
 x₀ ₁ ₂ ₃ ₄ ₅ ₆ ₇ ₈ ₉ ₊ ₋


----------



## rst (Jun 13, 2013)

rst said:


> *----------------------------------------------------------------------------------------------------------------------
> NEW OBJECTIVE QUESTION*
> 
> *img827.imageshack.us/img827/8467/rangeb.png
> ...



Let f(x)=y
Then y= 9^x-3^x+1
y= 3^2x-3^x+1

Put 3^x =u
Then y= u²-u+1
y= u²-u+(1/4)+1-(1/4) {for making it perfect square}
y= [u-(1/2)]²+ (3/4) 
As [u-(1/2)]² can't be negative,so its least value is 0
Hence y≥ (3/4) 

Ans is option (4)


----------



## rst (Jun 14, 2013)

*----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION*

*img42.imageshack.us/img42/9306/z0l.png

 Plz explain your answer

 (Also you can ask your mathematics related questions or queries
 we will try to solve them)
*----------------------------------------------------------------------------------------------------------------------*
*you can use following mathematic symbols(just do copy and paste)*
 £ ω ∴ ∂ Φ γ δ μ σ Є Ø Ω ∩ ≈ ≡ ≈ ≅ ≠ ≤ ≥ • ~ ± ∓ ∤ ◅∈ ∉ ⊆ ⊂ ∪ ⊥ ô ∫ Σ → ∞ Π Δ Ψ Γ ∮ ∇∂ √ °α β γ δ ε ζ η θ ι κ λ ν ξ ο π ρ σ τ υ φ χ ψ 
x⁰ ¹ ² ³ ⁴ ⁵ ⁶ ⁷ ⁸ ⁹ ⁺ ⁻ ⁿ 
 x₀ ₁ ₂ ₃ ₄ ₅ ₆ ₇ ₈ ₉ ₊ ₋


----------



## Niilesh (Jun 14, 2013)

^ using all the conditions(for log to be defined) from the left we will get
18x-x²-77 >3
x²-18x+8*10< 0
So 8< x <10


----------



## rst (Jun 15, 2013)

Niilesh said:


> ^ using all the conditions(for log to be defined) from the left we will get
> 18x-x²-77 >3
> x²-18x+8*10< 0
> So 8< x <10



Great explanation
Thanks a lot, Niilesh



rst said:


> *----------------------------------------------------------------------------------------------------------------------
> NEW OBJECTIVE QUESTION*
> 
> *img42.imageshack.us/img42/9306/z0l.png
> ...


log_4(log_5(log_3(18x-x^2-77))) is defined if log_5(log_3(18x-x^2-77))>0

 log_3(18x-x^2-77)>5^0
 log_3(18x-x^2-77)>1
 (18x-x^2-77)>3^1
  18x-x²-77 >3


----------



## rst (Jun 17, 2013)

*----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION*

*img4.imageshack.us/img4/7743/h3x.png

 Plz explain your answer

 (Also you can ask your mathematics related questions or queries
 we will try to solve them)
*----------------------------------------------------------------------------------------------------------------------*
*you can use following mathematic symbols(just do copy and paste)*
 £ ω ∴ ∂ Φ γ δ μ σ Є Ø Ω ∩ ≈ ≡ ≈ ≅ ≠ ≤ ≥ • ~ ± ∓ ∤ ◅∈ ∉ ⊆ ⊂ ∪ ⊥ ô ∫ Σ → ∞ Π Δ Ψ Γ ∮ ∇∂ √ °α β γ δ ε ζ η θ ι κ λ ν ξ ο π ρ σ τ υ φ χ ψ 
x⁰ ¹ ² ³ ⁴ ⁵ ⁶ ⁷ ⁸ ⁹ ⁺ ⁻ ⁿ 
 x₀ ₁ ₂ ₃ ₄ ₅ ₆ ₇ ₈ ₉ ₊ ₋


----------



## Niilesh (Jun 17, 2013)

^ 
We know that always in the above mentioned functions
f(g(x)) = x
Differentiating both sides wrt x
f'(g(x)) * g'(x) = 1
g'(x)=1/f'(g(x))
Now since f'(x) = 1/1+x³
So, g'(x)= 1+(g(x))³


----------



## rst (Jun 17, 2013)

Niilesh said:


> ^
> We know that always in the above mentioned functions
> f(g(x)) = x
> Differentiating both sides wrt x
> ...



Nice method

Can you prove f(g(x)) = x if g is inverse function of f


----------



## Niilesh (Jun 17, 2013)

umm....
i can explain it in words but it would be better to prove it mathematically. Give me any relation of a function and its inverse that you know(if any) and i will try to prove it .


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## rst (Jun 18, 2013)

Niilesh said:


> umm....
> i can explain it in words but it would be better to prove it mathematically. Give me any relation of a function and its inverse that you know(if any) and i will try to prove it .



If g is inverse function of f
Then f(x)=y and g(y)=x 

So g(f(x))=g(y)=x
---------------------------------------------------

If g is inverse function of f

Then f(x)=y and g(y)=x

 So f(g(y))=f(x)=y



or f(g(x))=x


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## Niilesh (Jun 18, 2013)

rst said:


> If g is inverse function of f
> Then f(x)=y and g(y)=x not valid always
> 
> So g(f(x))=g(y)=x   not valid always
> ...



Reason - If f(x) is not one-one then there will be more than one value of g(f(x)) but since g(x) is a function we take a interval in f(x) in which it is bijective


----------



## rst (Jun 18, 2013)

Niilesh said:


> Reason - If f(x) is not one-one then there will be more than one value of g(f(x)) but since g(x) is a function we take a interval in f(x) in which it is bijective



g is inverse of function f
So f is invertible function

Inverse exists only for bijective function
Hence f is bijective (one -one and onto)
So f is always one one


----------



## Niilesh (Jun 19, 2013)

rst said:


> g is inverse of function f
> So f is invertible function
> 
> Inverse exists only for bijective function
> ...


Technically that is right but sometimes we just make non bijective function( f(x) ) bijective by adjusting their domain and co-domain so that it becomes invertible and find its inverse( g(x) ) but when we refer to f(x) we mean the original function
Example - Sin x and Sin⁻¹x

Sin(Sin⁻¹(x)) = x

Sin⁻¹(Sin (x)) = many different values including 'x' depending upon its value

*www.goiit.com/upload/2009/7/18/ab59e6078e1f4a31a4c0e6e644cde7f7_1408226.gif


----------



## rst (Jun 19, 2013)

Niilesh said:


> Technically that is right but sometimes we just make non bijective function( f(x) ) bijective by adjusting their domain and co-domain so that it becomes invertible and find its inverse( g(x) ) but when we refer to f(x) we mean the original function
> Example - Sin x and Sin⁻¹x
> 
> Sin(Sin⁻¹(x)) = x
> ...



For sine function
domain = R
range=[-1,1]
So sine function is many one function
That why sine is not invertible function

sinx becomes invertible if its domain is [- π/2, π/2] {its range is [-1,1]}
For Sin⁻¹(x) domain becomes [-1,1] and range [- π/2, π/2] 

now Sin⁻¹(Sin (x)) =x (always)


----------



## Niilesh (Jun 19, 2013)

rst said:


> For sine function
> domain = R
> range=[-1,1]
> So sine function is many one function
> ...



Try to read and understand my post again -



> Technically that is right but sometimes we just make non bijective function( f(x) ) bijective by adjusting their domain and co-domain so that it becomes invertible and find its inverse( g(x) ) *but generally when we refer to f(x) we mean the original function*


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## rst (Jun 19, 2013)

Niilesh said:


> Try to read and understand my post again -


I understand your point

But as *f is invertible function and g is its inverse *
so f(x)=y and g(y)=x will always be true (according to the given hypothesis)

*Sin⁻¹(Sin (x)) = many different values including 'x' depending upon its value* 
Above statement is true in general
But in the above question or proof we are talking about invertible function 
So we have to put values in interval  [- π/2, π/2]
otherwise it will not be invertible (this will be contradiction to our hypothesis]


----------



## rst (Jun 21, 2013)

*----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION*

*img515.imageshack.us/img515/5681/cxn.png

 Plz explain your answer

 (Also you can ask your mathematics related questions or queries
 we will try to solve them)
*----------------------------------------------------------------------------------------------------------------------*
*you can use following mathematic symbols(just do copy and paste)*
 £ ω ∴ ∂ Φ γ δ μ σ Є Ø Ω ∩ ≈ ≡ ≈ ≅ ≠ ≤ ≥ • ~ ± ∓ ∤ ◅∈ ∉ ⊆ ⊂ ∪ ⊥ ô ∫ Σ → ∞ Π Δ Ψ Γ ∮ ∇∂ √ °α β γ δ ε ζ η θ ι κ λ ν ξ ο π ρ σ τ υ φ χ ψ 
x⁰ ¹ ² ³ ⁴ ⁵ ⁶ ⁷ ⁸ ⁹ ⁺ ⁻ ⁿ 
 x₀ ₁ ₂ ₃ ₄ ₅ ₆ ₇ ₈ ₉ ₊ ₋


----------



## Niilesh (Jun 21, 2013)

rst said:


> I understand your point
> 
> But as *f is invertible function and g is its inverse *
> so f(x)=y and g(y)=x will always be true (according to the given hypothesis)
> ...


Fair enough
BTW sin x will not be only invertible in [- π/2, π/2]


----------



## axelzdly1 (Jun 21, 2013)

Nice thread guys! A good way to recall mathematics

Given f''(x)= -f(x) .., consider f(x)= sin x
and then f(x)= - g(x) .., so g(x)= -sin x     limits?

So.., as h(x) = {f(x)^2} + {g(x)^2} = (sinx)^2 + (-sinx)^2 = 2[(sinx)^2]
and given h(5)=3  and req is h(10) i.e h(2x)

So the new eq will be h(2x) = 2(sin2x)^2..

we need to write 2[sin²(2x)] in terms of 2[sin²x] an we will get the ans..

Im hopeless from here..
enough brain excercise for today


----------



## darkv0id (Jun 21, 2013)

^ Such a solution is okay for competitive exam purposes, but how do you "assume" that f(x) = sinx ?

EDIT: I think you read the question wrong... h(x)= (f(x)^2 + g(x))^2 and not f(x)^2 + g(x)^2


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## rohanz (Jun 21, 2013)

axelzdly1 said:


> Nice thread guys! A good way to recall mathematics
> 
> Given f''(x)= -f(x) .., consider f(x)= sin x
> and then f(x)= - g(x) .., so g(x)= -sin x     limits?
> ...



Can't assume x to be sinx
Sin x comes to be (3/2)^1/2 which can never be true. (greater than 1)


----------



## axelzdly1 (Jun 21, 2013)

@rohanz.. hmm.. huge mistake.. Im sorry.. 

@darkvoid.. h(x)= ( f ( x )² + g( x ) )²  not ( f ( x )² ) + ( g ( x )² )   ?? 

aren't there 4 parenthesis around each of the f,g functions..?


----------



## darkv0id (Jun 22, 2013)

axelzdly1 said:


> @darkvoid.. h(x)= ( f ( x )² + g( x ) )²  not ( f ( x )² ) + ( g ( x )² )   ??
> 
> aren't there 4 parenthesis around each of the f,g functions..?



Now I feel like an idiot.....

My apologies.


----------



## axelzdly1 (Jun 22, 2013)

Just the same feeling when I thought I was wrong.> 
Nah,its okay..

About the problem.,can you solve from there ?
or do you have another solution?


----------



## Niilesh (Jun 22, 2013)

axelzdly1 said:


> Nice thread guys! A good way to recall mathematics
> 
> Given f''(x)= -f(x) .., consider f(x)= sin x
> and then f(x)= - g(x) .., so g(x)= -sin x     limits?
> ...


For competitive exam we can use this 
Let f(x) = a * sinx (where a ∈ R )
=>h(x) = 2a²sin²x
Now, h(5) = 3
=>3/sin²5 = 2a²
Now h(x) becomes = (3/sin²5)*sin²x
=>h(10) = (3/sin²5) * sin²10 = 3*4*cos²5 = 0.96557
So ans is (D)


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## rohanz (Jun 22, 2013)

Niilesh said:


> For competitive exam we can use this
> Let f(x) = a * sinx (where a ∈ R )
> =>h(x) = 2a²sin²x
> Now, h(5) = 3
> ...



Seems right but we won't get calculators during the exam :O


----------



## Niilesh (Jun 22, 2013)

rohanz said:


> Seems right but we won't get calculators during the exam :O



You know that cos5 should be a little greater than cos(3π/2) so their is no way 3*4*cos²5 will be equal to an integer so.......


----------



## rst (Jun 23, 2013)

*----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION*

*img94.imageshack.us/img94/8084/wyzg.png

 Plz explain your answer

 (Also you can ask your mathematics related questions or queries
 we will try to solve them)
*----------------------------------------------------------------------------------------------------------------------*
*you can use following mathematic symbols(just do copy and paste)*
 £ ω ∴ ∂ Φ γ δ μ σ Є Ø Ω ∩ ≈ ≡ ≈ ≅ ≠ ≤ ≥ • ~ ± ∓ ∤ ◅∈ ∉ ⊆ ⊂ ∪ ⊥ ô ∫ Σ → ∞ Π Δ Ψ Γ ∮ ∇∂ √ °α β γ δ ε ζ η θ ι κ λ ν ξ ο π ρ σ τ υ φ χ ψ 
x⁰ ¹ ² ³ ⁴ ⁵ ⁶ ⁷ ⁸ ⁹ ⁺ ⁻ ⁿ 
 x₀ ₁ ₂ ₃ ₄ ₅ ₆ ₇ ₈ ₉ ₊ ₋


----------



## axelzdly1 (Jun 23, 2013)

Assume the equation to be  a+b=0 for a= |2x²-5x+3|  and b= x-1
to satisfy a+b=0 there are two conditions.. either <  a=b=0 >  or  <  a= -b >

case 1:   a= |2x²-5x+3| = 0   
=> 2x²-5x+3=0   x=(1.5,1)

case 2: b= x-1 =0 => x=1 

common solution i.e, x=1 ∩ x=1,1.5   is 1..

for second condition apply 
| 2x²-5x+3 | => ± (2x²-5x+3) = (1-x) and you will get the same result i.e, x= 1 

so the solution is 1.., unique solution

Corrections? any other simple methods?


----------



## Niilesh (Jun 24, 2013)

rst said:


> *----------------------------------------------------------------------------------------------------------------------
> NEW OBJECTIVE QUESTION*
> 
> 
> ...


Ummm... Sometimes your questions are too easy 

Case 1: 2x²-5x+3≥0 =>x∉(1,3/2)
=>2x²-5x+3 + x-1 = 0
=>x = 1,1

Common - x=1

Case 2: 2x²-5x+3≤0 x∈(1,3/2)
=> -2x²+5x-3 + x-1 =0
=> x² -3x + 2 = 0
x= 1,2

Common - x∈Φ

So only solution is 1(only one solution)



axelzdly1 said:


> Assume the equation to be  a+b=0 for a= |2x²-5x+3|  and b= x-1
> to satisfy a+b=0 there are two conditions.. *either <  a=b=0 >  or  <  a= -b >* There is no need of the first condition, 2nd already covers it
> 
> case 1:   a= |2x²-5x+3| = 0
> ...


----------



## rst (Jun 24, 2013)

Niilesh said:


> Ummm... Sometimes your questions are too easy
> 
> Case 1: 2x²-5x+3≥0 =>x∉(1,3/2)
> =>2x²-5x+3 + x-1 = 0
> ...



How is it -1  ??


----------



## Niilesh (Jun 24, 2013)

rst said:


> How is it -1  ??


It's 1, typo


----------



## rst (Jun 24, 2013)

Niilesh said:


> It's 1, typo


So only one solution is 1 or 2



Niilesh said:


> Ummm... Sometimes your questions are too easy
> 
> Case 1: 2x²-5x+3≥0 =>x∉(1,3/2)
> =>2x²-5x+3 + x-1 = 0
> ...



I mean only solution is not x=2
as at x=2,  2x²-5x+3=1>0 which contradicts case (2)


----------



## Niilesh (Jun 24, 2013)

rst said:


> So only one solution is 1 or 2
> 
> 
> 
> ...


wth happened to me 
another typo, its 1 only


----------



## rst (Jun 24, 2013)

*----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION*

*img843.imageshack.us/img843/3268/lmuo.png

 Plz explain your answer

 (Also you can ask your mathematics related questions or queries
 we will try to solve them)
*----------------------------------------------------------------------------------------------------------------------*
*you can use following mathematic symbols(just do copy and paste)*
 £ ω ∴ ∂ Φ γ δ μ σ Є Ø Ω ∩ ≈ ≡ ≈ ≅ ≠ ≤ ≥ • ~ ± ∓ ∤ ◅∈ ∉ ⊆ ⊂ ∪ ⊥ ô ∫ Σ → ∞ Π Δ Ψ Γ ∮ ∇∂ √ °α β γ δ ε ζ η θ ι κ λ ν ξ ο π ρ σ τ υ φ χ ψ 
x⁰ ¹ ² ³ ⁴ ⁵ ⁶ ⁷ ⁸ ⁹ ⁺ ⁻ ⁿ 
 x₀ ₁ ₂ ₃ ₄ ₅ ₆ ₇ ₈ ₉ ₊ ₋


----------



## Niilesh (Jun 25, 2013)

rst said:


> *----------------------------------------------------------------------------------------------------------------------
> NEW OBJECTIVE QUESTION*
> 
> 
> ...


I think this was the first question given in our class to tell how to solve this kind of questions 
BTW i know two methods i will use the general one

f(x+y) = f(x) . f(y)
Now partially differentiating wrt x we get

f'(x+y) = f'(x) . f(y)
Now at x = 0
f'(y)=2f(y)
=>f'(x)=2f(x)

we can also get f(x) by integration.....


----------



## rst (Jun 25, 2013)

Niilesh said:


> I think this was the first question given in our class to tell how to solve this kind of questions
> BTW i know two methods i will use the general one
> 
> f(x+y) = f(x) . f(y)
> ...



f '(x) represent ordinary diff.

But you used it for partial diff.


----------



## the_conqueror (Jun 25, 2013)

rst said:


> f '(x) represent ordinary diff.
> 
> But you used it for partial diff.



No, he did it correctly AFAIK.


----------



## Niilesh (Jun 26, 2013)

rst said:


> f '(x) represent ordinary diff.
> 
> But you used it for partial diff.


ohhh... diffrentiation is not taught to us till now on maths so.....


the_conqueror said:


> No, he did it correctly AFAIK.


i think he objected to only the use of f'(x) for patial diff.


----------



## rst (Jun 26, 2013)

rst said:


> *----------------------------------------------------------------------------------------------------------------------
> NEW OBJECTIVE QUESTION*
> 
> *img843.imageshack.us/img843/3268/lmuo.png
> ...



 f(x+y)=f(x)f(y)
 f(0+0)=f(0)f(0)
=>f(0)=f²(0)
=> f(0)=0,1
 but f(x)≠0 for any xЄ R => f(0)=1


Then, f'(x)= lim(h  → 0) [f(x+h)-f(x)]/h 
f'(x)= lim(h  → 0) [f(x)f(h)-f(x)]/h 
f'(x)= f(x)lim(h  → 0) [f(h)-1]/h --------(1)

putting x=0 in (1),we get
f'(0)= f(0)lim(h  → 0) [f(h)-1]/h
f'(0)/ f(0)=lim(h  → 0) [f(h)-1]/h
2=lim(h  → 0) [f(h)-1]/h--------------------(2)

using (2) in (1),we get
f'(x)= 2f(x)


----------



## rst (Jun 27, 2013)

*----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION*

*img266.imageshack.us/img266/5720/y32v.png

 Plz explain your answer

 (Also you can ask your mathematics related questions or queries
 we will try to solve them)
*----------------------------------------------------------------------------------------------------------------------*
*you can use following mathematic symbols(just do copy and paste)*
 £ ω ∴ ∂ Φ γ δ μ σ Є Ø Ω ∩ ≈ ≡ ≈ ≅ ≠ ≤ ≥ • ~ ± ∓ ∤ ◅∈ ∉ ⊆ ⊂ ∪ ⊥ ô ∫ Σ → ∞ Π Δ Ψ Γ ∮ ∇∂ √ °α β γ δ ε ζ η θ ι κ λ ν ξ ο π ρ σ τ υ φ χ ψ 
x⁰ ¹ ² ³ ⁴ ⁵ ⁶ ⁷ ⁸ ⁹ ⁺ ⁻ ⁿ 
 x₀ ₁ ₂ ₃ ₄ ₅ ₆ ₇ ₈ ₉ ₊ ₋


----------



## Vignesh B (Jun 27, 2013)

135 min.
I don't know if this question is really this easy. My experience with Maths tells me that, if it is too easy, then probably my answer is wrong. 
Anyway, by applying pythagoras theorem, we can find the third route back to hive which comes to 50min. √(30² + 40²) = 50
Then by adding all the times taken, 30+5+40+10+50 = 135


----------



## rst (Jun 27, 2013)

Vignesh B said:


> 135 min.
> I don't know if this question is really this easy. My experience with Maths tells me that, if it is too easy, then probably my answer is wrong.
> Anyway, by applying pythagoras theorem, we can find the third route back to hive which comes to 50min. √(30² + 40²) = 50
> Then by adding all the times taken, 30+5+40+10+50 = 135



I also think the same
But ans is "less than 1 hour"


----------



## Vignesh B (Jun 27, 2013)

^^ Can you verify the answer from somewhere else also?
Because the total flying time of 40+30 min cannot be neglected by any means which itself amounts for more than an hour. :/


----------



## axelzdly1 (Jun 27, 2013)

How about ignoring the waiting time...? Then the answer will be < an hour 

The question is 'How long was the bee away from its hive?' not 'How much time does the bee take to travel the whole distance?'

If a friend asks you 'How far is your home from the XY centre..?' , he meant the distance/time taken for him, or in general...not the time/distance taking for you to return home, during which you take several halts for buying stuff, looking at women.,


----------



## rst (Jun 27, 2013)

Vignesh B said:


> ^^ Can you verify the answer from somewhere else also?
> Because the total flying time of 40+30 min cannot be neglected by any means which itself amounts for more than an hour. :/



I think, there is some problem in ans key
Answer is definitely 135 mins
It will be clear in a week time(when original result will out)


----------



## rst (Jun 28, 2013)

*----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION*

*img824.imageshack.us/img824/5669/3cwi.png

 Plz explain your answer

 (Also you can ask your mathematics related questions or queries
 we will try to solve them)
*----------------------------------------------------------------------------------------------------------------------*
*you can use following mathematic symbols(just do copy and paste)*
 £ ω ∴ ∂ Φ γ δ μ σ Є Ø Ω ∩ ≈ ≡ ≈ ≅ ≠ ≤ ≥ • ~ ± ∓ ∤ ◅∈ ∉ ⊆ ⊂ ∪ ⊥ ô ∫ Σ → ∞ Π Δ Ψ Γ ∮ ∇∂ √ °α β γ δ ε ζ η θ ι κ λ ν ξ ο π ρ σ τ υ φ χ ψ 
x⁰ ¹ ² ³ ⁴ ⁵ ⁶ ⁷ ⁸ ⁹ ⁺ ⁻ ⁿ 
 x₀ ₁ ₂ ₃ ₄ ₅ ₆ ₇ ₈ ₉ ₊ ₋


----------



## Niilesh (Jun 29, 2013)

is it the 2nd option?

i used angle of incident = angle of reflection = 45°


so in the isocelecs triangle formed by joining the points of contact and cente of hemisphere the value of base = 2 * R/cos45° = R*2^	(1/2)


----------



## rst (Jun 29, 2013)

Niilesh said:


> is it the 2nd option?
> 
> i used angle of incident = angle of reflection = 45°
> 
> ...


Absolutely right


----------



## rst (Jun 30, 2013)

*----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION*

*img404.imageshack.us/img404/83/iecq.png

 Plz explain your answer

 (Also you can ask your mathematics related questions or queries
 we will try to solve them)
*----------------------------------------------------------------------------------------------------------------------*
*you can use following mathematic symbols(just do copy and paste)*
 £ ω ∴ ∂ Φ γ δ μ σ Є Ø Ω ∩ ≈ ≡ ≈ ≅ ≠ ≤ ≥ • ~ ± ∓ ∤ ◅∈ ∉ ⊆ ⊂ ∪ ⊥ ô ∫ Σ → ∞ Π Δ Ψ Γ ∮ ∇∂ √ °α β γ δ ε ζ η θ ι κ λ ν ξ ο π ρ σ τ υ φ χ ψ 
x⁰ ¹ ² ³ ⁴ ⁵ ⁶ ⁷ ⁸ ⁹ ⁺ ⁻ ⁿ 
 x₀ ₁ ₂ ₃ ₄ ₅ ₆ ₇ ₈ ₉ ₊ ₋


----------



## rohanz (Jul 1, 2013)

rst said:


> *----------------------------------------------------------------------------------------------------------------------
> NEW OBJECTIVE QUESTION*
> 
> *img404.imageshack.us/img404/83/iecq.png
> ...



120 ants?


----------



## rst (Jul 1, 2013)

rohanz said:


> 120 ants?



May be possible.
I don't know the answer.
Can you explain your answer ??


----------



## Niilesh (Jul 2, 2013)

Hmmm... that's a question of kinematics not of pure maths. But I have forgotten how to solve that kind of questions only thing i can remember is that we use relative motion in it. I think its time to take out my 11th notes......


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## rst (Jul 3, 2013)

*----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION*

*img585.imageshack.us/img585/9980/t2ts.png

 Plz explain your answer

 (Also you can ask your mathematics related questions or queries
 we will try to solve them)
*----------------------------------------------------------------------------------------------------------------------*
*you can use following mathematic symbols(just do copy and paste)*
 £ ω ∴ ∂ Φ γ δ μ σ Є Ø Ω ∩ ≈ ≡ ≈ ≅ ≠ ≤ ≥ • ~ ± ∓ ∤ ◅∈ ∉ ⊆ ⊂ ∪ ⊥ ô ∫ Σ → ∞ Π Δ Ψ Γ ∮ ∇∂ √ °α β γ δ ε ζ η θ ι κ λ ν ξ ο π ρ σ τ υ φ χ ψ 
x⁰ ¹ ² ³ ⁴ ⁵ ⁶ ⁷ ⁸ ⁹ ⁺ ⁻ ⁿ 
 x₀ ₁ ₂ ₃ ₄ ₅ ₆ ₇ ₈ ₉ ₊ ₋


----------



## Niilesh (Jul 4, 2013)

^ 17.5
I think this expression would be enough for explanation  ->  (360/12) * 35/60 = 17.5



rst said:


> *----------------------------------------------------------------------------------------------------------------------
> NEW OBJECTIVE QUESTION*
> 
> *img404.imageshack.us/img404/83/iecq.png
> ...


IMO the question is wrong as length of ant is not given and length of a point is undefined also a line consists of infinite no. of points so already infinite no. of ants would be present on the line AB

BTW I got answer = 240 [((60*2)*2] with some assumptions


----------



## Chaitanya (Jul 4, 2013)

rst said:


> *----------------------------------------------------------------------------------------------------------------------
> NEW OBJECTIVE QUESTION*
> 
> 
> ...





Ans is 50 mins. (<1 hr)
explanation


no by pythagoras theorem 3 rd side is 50 meters so it means 50 min..


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## Niilesh (Jul 4, 2013)

Chaitanya said:


> Ans is 50 mins. (<1 hr)
> explanation
> View attachment 11193
> no by pythagoras theorem 3 rd side is 50 meters so it means 50 min..



Actually its a problem of language.... When he said long we understood the usual meaning(for how much time he was away) but he actually(probably) asked how much far away he was(in min.)


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## rst (Jul 4, 2013)

Niilesh said:


> ^ 17.5
> I think this expression would be enough for explanation  ->  (360/12) * 35/60 = 17.5


Ans and explanation are correct



Niilesh said:


> ^ 17.5
> 
> 
> 
> ...



Ans is correct
What are the assumptions ?


----------



## rst (Jul 5, 2013)

*----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION*

*img849.imageshack.us/img849/2030/6ka.png

 Plz explain your answer

 (Also you can ask your mathematics related questions or queries
 we will try to solve them)
*----------------------------------------------------------------------------------------------------------------------*
*you can use following mathematic symbols(just do copy and paste)*
 £ ω ∴ ∂ Φ γ δ μ σ Є Ø Ω ∩ ≈ ≡ ≈ ≅ ≠ ≤ ≥ • ~ ± ∓ ∤ ◅∈ ∉ ⊆ ⊂ ∪ ⊥ ô ∫ Σ → ∞ Π Δ Ψ Γ ∮ ∇∂ √ °α β γ δ ε ζ η θ ι κ λ ν ξ ο π ρ σ τ υ φ χ ψ 
x⁰ ¹ ² ³ ⁴ ⁵ ⁶ ⁷ ⁸ ⁹ ⁺ ⁻ ⁿ 
 x₀ ₁ ₂ ₃ ₄ ₅ ₆ ₇ ₈ ₉ ₊ ₋


----------



## Chaitanya (Jul 5, 2013)

rst said:


> *----------------------------------------------------------------------------------------------------------------------
> NEW OBJECTIVE QUESTION*
> 
> *img849.imageshack.us/img849/2030/6ka.png
> ...



Ans : 20

1 term for each of x1.. x20


----------



## rst (Jul 6, 2013)

*----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION*

*img69.imageshack.us/img69/2893/80a.png

 Plz explain your answer

 (Also you can ask your mathematics related questions or queries
 we will try to solve them)
*----------------------------------------------------------------------------------------------------------------------*
*you can use following mathematic symbols(just do copy and paste)*
 £ ω ∴ ∂ Φ γ δ μ σ Є Ø Ω ∩ ≈ ≡ ≈ ≅ ≠ ≤ ≥ • ~ ± ∓ ∤ ◅∈ ∉ ⊆ ⊂ ∪ ⊥ ô ∫ Σ → ∞ Π Δ Ψ Γ ∮ ∇∂ √ °α β γ δ ε ζ η θ ι κ λ ν ξ ο π ρ σ τ υ φ χ ψ 
x⁰ ¹ ² ³ ⁴ ⁵ ⁶ ⁷ ⁸ ⁹ ⁺ ⁻ ⁿ 
 x₀ ₁ ₂ ₃ ₄ ₅ ₆ ₇ ₈ ₉ ₊ ₋


----------



## Niilesh (Jul 6, 2013)

rst said:


> *----------------------------------------------------------------------------------------------------------------------
> NEW OBJECTIVE QUESTION*
> 
> *img69.imageshack.us/img69/2893/80a.png
> ...


V ∝ r²*h
Now using trigonometry/similar triangle we get new radius to be half of the original
So V' = V*(1/2)²(1/2) = V/8


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## rst (Jul 6, 2013)

Niilesh said:


> V ∝ r²*h
> Now using trigonometry/similar triangle we get new radius to be half of the original
> So V' = V*(1/2)²(1/2) = V/8



Absolutely right


----------



## rst (Jul 7, 2013)

*----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION*

*img42.imageshack.us/img42/3598/n7en.png

 Plz explain your answer

 (Also you can ask your mathematics related questions or queries
 we will try to solve them)
*----------------------------------------------------------------------------------------------------------------------*
*you can use following mathematic symbols(just do copy and paste)*
 £ ω ∴ ∂ Φ γ δ μ σ Є Ø Ω ∩ ≈ ≡ ≈ ≅ ≠ ≤ ≥ • ~ ± ∓ ∤ ◅∈ ∉ ⊆ ⊂ ∪ ⊥ ô ∫ Σ → ∞ Π Δ Ψ Γ ∮ ∇∂ √ °α β γ δ ε ζ η θ ι κ λ ν ξ ο π ρ σ τ υ φ χ ψ 
x⁰ ¹ ² ³ ⁴ ⁵ ⁶ ⁷ ⁸ ⁹ ⁺ ⁻ ⁿ 
 x₀ ₁ ₂ ₃ ₄ ₅ ₆ ₇ ₈ ₉ ₊ ₋


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## Niilesh (Jul 7, 2013)

^ and is the third option(find ratio of areas)


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## rst (Jul 7, 2013)

Niilesh said:


> ^ and is the third option(find ratio of areas)



correct


----------



## rst (Jul 9, 2013)

*----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION*

*img163.imageshack.us/img163/365/76bn.png

 Plz explain your answer

 (Also you can ask your mathematics related questions or queries
 we will try to solve them)
*----------------------------------------------------------------------------------------------------------------------*
*you can use following mathematic symbols(just do copy and paste)*
 £ ω ∴ ∂ Φ γ δ μ σ Є Ø Ω ∩ ≈ ≡ ≈ ≅ ≠ ≤ ≥ • ~ ± ∓ ∤ ◅∈ ∉ ⊆ ⊂ ∪ ⊥ ô ∫ Σ → ∞ Π Δ Ψ Γ ∮ ∇∂ √ °α β γ δ ε ζ η θ ι κ λ ν ξ ο π ρ σ τ υ φ χ ψ 
x⁰ ¹ ² ³ ⁴ ⁵ ⁶ ⁷ ⁸ ⁹ ⁺ ⁻ ⁿ 
 x₀ ₁ ₂ ₃ ₄ ₅ ₆ ₇ ₈ ₉ ₊ ₋


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## Niilesh (Jul 9, 2013)

^ 9 m i will let someone else explain it as your questions are getting easy


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## rst (Jul 9, 2013)

Niilesh said:


> ^ 9 m i will let someone else explain it as your questions are getting easy



Ans is correct

Can you explain your ans ?


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## Niilesh (Jul 9, 2013)

rst said:


> Ans is correct
> 
> Can you explain your ans ?


*www.thinkdigit.com/forum/attachmen...athematic-related-questions-here-untitled.png
apply Pythagoras theorem to find the value of 'x'


----------



## rst (Jul 11, 2013)

*----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION*

*img5.imageshack.us/img5/898/jbzj.png

 Plz explain your answer

 (Also you can ask your mathematics related questions or queries
 we will try to solve them)
*----------------------------------------------------------------------------------------------------------------------*
*you can use following mathematic symbols(just do copy and paste)*
 £ ω ∴ ∂ Φ γ δ μ σ Є Ø Ω ∩ ≈ ≡ ≈ ≅ ≠ ≤ ≥ • ~ ± ∓ ∤ ◅∈ ∉ ⊆ ⊂ ∪ ⊥ ô ∫ Σ → ∞ Π Δ Ψ Γ ∮ ∇∂ √ °α β γ δ ε ζ η θ ι κ λ ν ξ ο π ρ σ τ υ φ χ ψ 
x⁰ ¹ ² ³ ⁴ ⁵ ⁶ ⁷ ⁸ ⁹ ⁺ ⁻ ⁿ 
 x₀ ₁ ₂ ₃ ₄ ₅ ₆ ₇ ₈ ₉ ₊ ₋


----------



## Chaitanya (Jul 11, 2013)

^^154


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## rst (Jul 11, 2013)

Chaitanya said:


> ^^154


Yeah,its ans is 154

But I also need explanation


----------



## Chaitanya (Jul 11, 2013)

rst said:


> Yeah,its ans is 154
> 
> But I also need explanation



Center of an equilateral trig divides it's median in 2:1 ratio..

so it divides it into ratio 7 : 3.5 cm

now that 7 is radius of circle..
rest u kno


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## rst (Jul 11, 2013)

*----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION*

*img33.imageshack.us/img33/6712/0j4g.png
*img89.imageshack.us/img89/4672/3lz8.png
 Plz explain your answer

 (Also you can ask your mathematics related questions or queries
 we will try to solve them)
*----------------------------------------------------------------------------------------------------------------------*
*you can use following mathematic symbols(just do copy and paste)*
 £ ω ∴ ∂ Φ γ δ μ σ Є Ø Ω ∩ ≈ ≡ ≈ ≅ ≠ ≤ ≥ • ~ ± ∓ ∤ ◅∈ ∉ ⊆ ⊂ ∪ ⊥ ô ∫ Σ → ∞ Π Δ Ψ Γ ∮ ∇∂ √ °α β γ δ ε ζ η θ ι κ λ ν ξ ο π ρ σ τ υ φ χ ψ 
x⁰ ¹ ² ³ ⁴ ⁵ ⁶ ⁷ ⁸ ⁹ ⁺ ⁻ ⁿ 
 x₀ ₁ ₂ ₃ ₄ ₅ ₆ ₇ ₈ ₉ ₊ ₋


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## Chaitanya (Jul 11, 2013)

r/12


vol of hemisphere = (4*pi*(r/2)^3)/6

vol of water raised = pi*r^2*x

pi*r^2*x = (4*pi*(r/2)^3)/6

solving for x..


----------



## rst (Jul 11, 2013)

*----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION*

*img580.imageshack.us/img580/3396/g8l.png

 Plz explain your answer

 (Also you can ask your mathematics related questions or queries
 we will try to solve them)
*----------------------------------------------------------------------------------------------------------------------*
*you can use following mathematic symbols(just do copy and paste)*
 £ ω ∴ ∂ Φ γ δ μ σ Є Ø Ω ∩ ≈ ≡ ≈ ≅ ≠ ≤ ≥ • ~ ± ∓ ∤ ◅∈ ∉ ⊆ ⊂ ∪ ⊥ ô ∫ Σ → ∞ Π Δ Ψ Γ ∮ ∇∂ √ °α β γ δ ε ζ η θ ι κ λ ν ξ ο π ρ σ τ υ φ χ ψ 
x⁰ ¹ ² ³ ⁴ ⁵ ⁶ ⁷ ⁸ ⁹ ⁺ ⁻ ⁿ 
 x₀ ₁ ₂ ₃ ₄ ₅ ₆ ₇ ₈ ₉ ₊ ₋


----------



## Chaitanya (Jul 11, 2013)

*i.imgur.com/aZV9YSD.png


----------



## rst (Jul 11, 2013)

*----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION*

*img692.imageshack.us/img692/1927/6l0.png

 Plz explain your answer

 (Also you can ask your mathematics related questions or queries
 we will try to solve them)
*----------------------------------------------------------------------------------------------------------------------*
*you can use following mathematic symbols(just do copy and paste)*
 £ ω ∴ ∂ Φ γ δ μ σ Є Ø Ω ∩ ≈ ≡ ≈ ≅ ≠ ≤ ≥ • ~ ± ∓ ∤ ◅∈ ∉ ⊆ ⊂ ∪ ⊥ ô ∫ Σ → ∞ Π Δ Ψ Γ ∮ ∇∂ √ °α β γ δ ε ζ η θ ι κ λ ν ξ ο π ρ σ τ υ φ χ ψ 
x⁰ ¹ ² ³ ⁴ ⁵ ⁶ ⁷ ⁸ ⁹ ⁺ ⁻ ⁿ 
 x₀ ₁ ₂ ₃ ₄ ₅ ₆ ₇ ₈ ₉ ₊ ₋


----------



## Niilesh (Jul 12, 2013)

Answer would be '4'

Explanation - it is given that 20 is the HCF so let the no.'s be of form
N₁=20*K₁
N₂=20*K₂
where K₂ and K₁ don't have any common factor(other than 1)
also it is given that LCM = 600 = 20*K₂K₁ 
=>K₁K₂ = 30 = 2*3*5
So answer would be ⁴C₁(selecting one from 1,2,3,5) = 4


----------



## rst (Jul 15, 2013)

*----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION*

*img69.imageshack.us/img69/9241/33se.png

 Plz explain your answer

 (Also you can ask your mathematics related questions or queries
 we will try to solve them)
*----------------------------------------------------------------------------------------------------------------------*
*you can use following mathematic symbols(just do copy and paste)*
 £ ω ∴ ∂ Φ γ δ μ σ Є Ø Ω ∩ ≈ ≡ ≈ ≅ ≠ ≤ ≥ • ~ ± ∓ ∤ ◅∈ ∉ ⊆ ⊂ ∪ ⊥ ô ∫ Σ → ∞ Π Δ Ψ Γ ∮ ∇∂ √ °α β γ δ ε ζ η θ ι κ λ ν ξ ο π ρ σ τ υ φ χ ψ 
x⁰ ¹ ² ³ ⁴ ⁵ ⁶ ⁷ ⁸ ⁹ ⁺ ⁻ ⁿ 
 x₀ ₁ ₂ ₃ ₄ ₅ ₆ ₇ ₈ ₉ ₊ ₋


----------



## rst (Jul 16, 2013)

rst said:


> *----------------------------------------------------------------------------------------------------------------------
> NEW OBJECTIVE QUESTION*
> 
> *img69.imageshack.us/img69/9241/33se.png
> ...



Domain of sin⁻¹x is [-1,1]

So -1≤log₂(x/2) ≤ 1

=> 2⁻¹≤(x/2) ≤ 2¹

=> 1≤ x ≤ 4


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## nisargshah95 (Jul 16, 2013)

Couldn't solve this one -
∫ (tan(2x)/√(cos[SUP]6[/SUP]x+sin[SUP]6[/SUP]x))dx


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## rst (Jul 17, 2013)

nisargshah95 said:


> Couldn't solve this one -
> ∫ (tan(2x)/√(cos[SUP]6[/SUP]x+sin[SUP]6[/SUP]x))dx



∫ (tan(2x)/√(cos[SUP]6[/SUP]x+sin[SUP]6[/SUP]x))dx
∫ (sin(2x)/cos2x √(cos[SUP]6[/SUP]x+sin[SUP]6[/SUP]x))dx---------------------------(1)

Now cos[SUP]6[/SUP]x+sin[SUP]6[/SUP]x=(cos² x+sin² x)³ -3cos² xsin² x(cos² x+sin² x)
=1-(3cos² xsin² x)
=1-(3/4)(sin2x)² =1-3/4(1-cos² (2x)
=[1+3cos² (2x)]/4

put cos 2x =t
using in (1) ,we get

∫ dt/t√(1+3t² )

Now it is easy to solve


----------



## rst (Jul 18, 2013)

*----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION*

*img51.imageshack.us/img51/1486/fpt9.png

 Plz explain your answer

 (Also you can ask your mathematics related questions or queries
 we will try to solve them)
*----------------------------------------------------------------------------------------------------------------------*
*you can use following mathematic symbols(just do copy and paste)*
 £ ω ∴ ∂ Φ γ δ μ σ Є Ø Ω ∩ ≈ ≡ ≈ ≅ ≠ ≤ ≥ • ~ ± ∓ ∤ ◅∈ ∉ ⊆ ⊂ ∪ ⊥ ô ∫ Σ → ∞ Π Δ Ψ Γ ∮ ∇∂ √ °α β γ δ ε ζ η θ ι κ λ ν ξ ο π ρ σ τ υ φ χ ψ 
x⁰ ¹ ² ³ ⁴ ⁵ ⁶ ⁷ ⁸ ⁹ ⁺ ⁻ ⁿ 
 x₀ ₁ ₂ ₃ ₄ ₅ ₆ ₇ ₈ ₉ ₊ ₋


----------



## Chaitanya (Jul 18, 2013)

^^by mod you mean remainder after division ri8??


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## rst (Jul 18, 2013)

Chaitanya said:


> ^^by mod you mean remainder after division ri8??



Yeah, you got it right


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## Chaitanya (Jul 18, 2013)

then there seems 2 be a problem how can 587 be remainder for any possible division of 167 ??

Edit : may be x is negative??

Edit 2 : I am an idiot.. there is no -ve no in option...


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## rst (Jul 19, 2013)

rst said:


> *----------------------------------------------------------------------------------------------------------------------
> NEW OBJECTIVE QUESTION*
> 
> *img51.imageshack.us/img51/1486/fpt9.png
> ...



a≡ b(mod x) means x divides a-b (or b-a)
In terms of remainder, a≡ b(mod x) means remainder will be same if we divide "a by x" or "b by x"

According to the question, 
x divides 385 - 21 =364 and x divides 587 - 167 =420
H.C.F of 364 and 420 is 28

So 28 is the highest number which divides both 364 and 420


----------



## rst (Jul 20, 2013)

*----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION*

*img16.imageshack.us/img16/7392/o2p8.png

 Plz explain your answer

 (Also you can ask your mathematics related questions or queries
 we will try to solve them)
*----------------------------------------------------------------------------------------------------------------------*
*you can use following mathematic symbols(just do copy and paste)*
 £ ω ∴ ∂ Φ γ δ μ σ Є Ø Ω ∩ ≈ ≡ ≈ ≅ ≠ ≤ ≥ • ~ ± ∓ ∤ ◅∈ ∉ ⊆ ⊂ ∪ ⊥ ô ∫ Σ → ∞ Π Δ Ψ Γ ∮ ∇∂ √ °α β γ δ ε ζ η θ ι κ λ ν ξ ο π ρ σ τ υ φ χ ψ 
x⁰ ¹ ² ³ ⁴ ⁵ ⁶ ⁷ ⁸ ⁹ ⁺ ⁻ ⁿ 
 x₀ ₁ ₂ ₃ ₄ ₅ ₆ ₇ ₈ ₉ ₊ ₋


----------



## rst (Jul 22, 2013)

rst said:


> *----------------------------------------------------------------------------------------------------------------------
> NEW OBJECTIVE QUESTION*
> 
> *img16.imageshack.us/img16/7392/o2p8.png
> ...



Diff. e^(ax) cosbx w.r.t x we get
ae^(ax) cosbx - be^(ax) sinbx
e^(ax)[a cosbx - bsinbx]-----------------(1)

Now r e^(ax) cos[bx+tan⁻¹(b/a)] =r e^(ax) [cosbx sin{tan⁻¹(b/a)} - sinbx cos{tan⁻¹(b/a)} ]
=r e^(ax) [cosbx {a /√ (a² +b²)} - sinbx {b /√ (a² +b²)}  ]
=r e^(ax) /√ (a² +b²)  [acosbx  - bsinbx ]-------------(2)

Comparing (1) and (2)
r=√ (a² +b²)


----------



## rst (Jul 23, 2013)

*----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION*

*img33.imageshack.us/img33/8869/vtgv.png

 Plz explain your answer

 (Also you can ask your mathematics related questions or queries
 we will try to solve them)
*----------------------------------------------------------------------------------------------------------------------*
*you can use following mathematic symbols(just do copy and paste)*
 £ ω ∴ ∂ Φ γ δ μ σ Є Ø Ω ∩ ≈ ≡ ≈ ≅ ≠ ≤ ≥ • ~ ± ∓ ∤ ◅∈ ∉ ⊆ ⊂ ∪ ⊥ ô ∫ Σ → ∞ Π Δ Ψ Γ ∮ ∇∂ √ °α β γ δ ε ζ η θ ι κ λ ν ξ ο π ρ σ τ υ φ χ ψ 
x⁰ ¹ ² ³ ⁴ ⁵ ⁶ ⁷ ⁸ ⁹ ⁺ ⁻ ⁿ 
 x₀ ₁ ₂ ₃ ₄ ₅ ₆ ₇ ₈ ₉ ₊ ₋


----------



## rst (Jul 24, 2013)

rst said:


> *----------------------------------------------------------------------------------------------------------------------
> NEW OBJECTIVE QUESTION*
> 
> *img33.imageshack.us/img33/8869/vtgv.png
> ...



Let f(x)=y,then we have
∫ y sinx cosx dx =[1/ 2(b²-a²)] logy +c

Differentiating both sides we,get
 y sinx cosx  =[y'/ 2(b²-a²)y] 

2y (b²-a²)sinx cosx  = y'/ y

2  (b²-a²)sinx cosx  =[y'/ y²]

(b²-a²)sin2x dx =[dy/ y²]
integrating we get
 -[(b²-a²)cos2x] /2 = -(1/ y )

y=2/[(b²-a²)cos2x]


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## rst (Jul 30, 2013)

*----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION*

*img441.imageshack.us/img441/6261/cgyq.png

 Plz explain your answer

 (Also you can ask your mathematics related questions or queries
 we will try to solve them)
*----------------------------------------------------------------------------------------------------------------------*
*you can use following mathematic symbols(just do copy and paste)*
 £ ω ∴ ∂ Φ γ δ μ σ Є Ø Ω ∩ ≈ ≡ ≈ ≅ ≠ ≤ ≥ • ~ ± ∓ ∤ ◅∈ ∉ ⊆ ⊂ ∪ ⊥ ô ∫ Σ → ∞ Π Δ Ψ Γ ∮ ∇∂ √ °α β γ δ ε ζ η θ ι κ λ ν ξ ο π ρ σ τ υ φ χ ψ 
x⁰ ¹ ² ³ ⁴ ⁵ ⁶ ⁷ ⁸ ⁹ ⁺ ⁻ ⁿ 
 x₀ ₁ ₂ ₃ ₄ ₅ ₆ ₇ ₈ ₉ ₊ ₋


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## Niilesh (Jul 31, 2013)

^ cot⁻¹(√2+x/2-x) = cos⁻¹[(√2+x)/2)]                  
so the original expression becomes (2+x)/4           [ cos(cos⁻¹x) = x ]
so ans is 1/4

or one can always apply chain rule...


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## rst (Aug 1, 2013)

Niilesh said:


> ^ cot⁻¹(√2+x/2-x) = cos⁻¹[(√2+x)/2)]
> so the original expression becomes (2+x)/4           [ cos(cos⁻¹x) = x ]
> so ans is 1/4
> 
> or one can always apply chain rule...



Absolutely right


----------



## rst (Aug 3, 2013)

*----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION*

*img10.imageshack.us/img10/4633/dvr7.png

 Plz explain your answer

 (Also you can ask your mathematics related questions or queries
 we will try to solve them)
*----------------------------------------------------------------------------------------------------------------------*
*you can use following mathematic symbols(just do copy and paste)*
 £ ω ∴ ∂ Φ γ δ μ σ Є Ø Ω ∩ ≈ ≡ ≈ ≅ ≠ ≤ ≥ • ~ ± ∓ ∤ ◅∈ ∉ ⊆ ⊂ ∪ ⊥ ô ∫ Σ → ∞ Π Δ Ψ Γ ∮ ∇∂ √ °α β γ δ ε ζ η θ ι κ λ ν ξ ο π ρ σ τ υ φ χ ψ 
x⁰ ¹ ² ³ ⁴ ⁵ ⁶ ⁷ ⁸ ⁹ ⁺ ⁻ ⁿ 
 x₀ ₁ ₂ ₃ ₄ ₅ ₆ ₇ ₈ ₉ ₊ ₋


----------



## Niilesh (Aug 3, 2013)

It's easy π² ≈ g ≈ 9.8
[π²] = 9
[-π²] = -10
so ans will be (2)


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## rst (Aug 3, 2013)

Niilesh said:


> It's easy π² ≈ g ≈ 9.8
> [π²] = 9
> [-π²] = -10
> so ans will be (2)



Correct


----------



## rst (Aug 6, 2013)

*----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION*

*img202.imageshack.us/img202/149/oj3b.png

 Plz explain your answer

 (Also you can ask your mathematics related questions or queries
 we will try to solve them)
*----------------------------------------------------------------------------------------------------------------------*
*you can use following mathematic symbols(just do copy and paste)*
 £ ω ∴ ∂ Φ γ δ μ σ Є Ø Ω ∩ ≈ ≡ ≈ ≅ ≠ ≤ ≥ • ~ ± ∓ ∤ ◅∈ ∉ ⊆ ⊂ ∪ ⊥ ô ∫ Σ → ∞ Π Δ Ψ Γ ∮ ∇∂ √ °α β γ δ ε ζ η θ ι κ λ ν ξ ο π ρ σ τ υ φ χ ψ 
x⁰ ¹ ² ³ ⁴ ⁵ ⁶ ⁷ ⁸ ⁹ ⁺ ⁻ ⁿ 
 x₀ ₁ ₂ ₃ ₄ ₅ ₆ ₇ ₈ ₉ ₊ ₋


----------



## rst (Aug 9, 2013)

rst said:


> *----------------------------------------------------------------------------------------------------------------------
> NEW OBJECTIVE QUESTION*
> 
> *img202.imageshack.us/img202/149/oj3b.png
> ...



principal value of cos⁻¹ x lies in [0,π ]

principal value of sin⁻¹ x lies in [-π/2,π/2 ]

Intersection of region is [0,π/2 ]

So 0≤ 2cos⁻¹ x ≤ π/2
0≤ cos⁻¹ x ≤ π/4
cos0≥ x ≥ cosπ/4
1≥ x ≥ 1/√2


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## rst (Aug 11, 2013)

*----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION*

*img32.imageshack.us/img32/905/kxef.png

 Plz explain your answer

 (Also you can ask your mathematics related questions or queries
 we will try to solve them)
*----------------------------------------------------------------------------------------------------------------------*
*you can use following mathematic symbols(just do copy and paste)*
 £ ω ∴ ∂ Φ γ δ μ σ Є Ø Ω ∩ ≈ ≡ ≈ ≅ ≠ ≤ ≥ • ~ ± ∓ ∤ ◅∈ ∉ ⊆ ⊂ ∪ ⊥ ô ∫ Σ → ∞ Π Δ Ψ Γ ∮ ∇∂ √ °α β γ δ ε ζ η θ ι κ λ ν ξ ο π ρ σ τ υ φ χ ψ 
x⁰ ¹ ² ³ ⁴ ⁵ ⁶ ⁷ ⁸ ⁹ ⁺ ⁻ ⁿ 
 x₀ ₁ ₂ ₃ ₄ ₅ ₆ ₇ ₈ ₉ ₊ ₋


----------



## Ironman (Aug 12, 2013)

i am facing problems in learning vedic maths 
is there any forum where i can discuss it ??


----------



## rst (Aug 12, 2013)

Post here 
We will try to solve them


----------



## rst (Aug 15, 2013)

rst said:


> *----------------------------------------------------------------------------------------------------------------------
> NEW OBJECTIVE QUESTION*
> 
> *img32.imageshack.us/img32/905/kxef.png
> ...



Put x=0,then we get
e=|0  -1  3|
   |1  0   -4|
   |-3  4   0|

e=0


----------



## rst (Aug 22, 2013)

*----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION*

*img13.imageshack.us/img13/3600/s327.png

 Plz explain your answer

 (Also you can ask your mathematics related questions or queries
 we will try to solve them)
*----------------------------------------------------------------------------------------------------------------------*
*you can use following mathematic symbols(just do copy and paste)*
 £ ω ∴ ∂ Φ γ δ μ σ Є Ø Ω ∩ ≈ ≡ ≈ ≅ ≠ ≤ ≥ • ~ ± ∓ ∤ ◅∈ ∉ ⊆ ⊂ ∪ ⊥ ô ∫ Σ → ∞ Π Δ Ψ Γ ∮ ∇∂ √ °α β γ δ ε ζ η θ ι κ λ ν ξ ο π ρ σ τ υ φ χ ψ 
x⁰ ¹ ² ³ ⁴ ⁵ ⁶ ⁷ ⁸ ⁹ ⁺ ⁻ ⁿ 
 x₀ ₁ ₂ ₃ ₄ ₅ ₆ ₇ ₈ ₉ ₊ ₋


----------



## krazylearner (Aug 30, 2013)

I luv dis section !!


----------



## rst (Aug 30, 2013)

rst said:


> *----------------------------------------------------------------------------------------------------------------------
> NEW OBJECTIVE QUESTION*
> 
> *img13.imageshack.us/img13/3600/s327.png
> ...



cosα+2cosβ+3cos γ=0-----(1)

sinα+2sinβ+3sinγ=0
isinα+2isinβ+3isinγ=0(multiplying both sides by iota)--------(2)

Adding (1) and (2),we get
 e^iα+2e^iβ+3e^iγ=0.

using a³+b³+c³−3abc=(a+b+c)(a²+b²+c²−ab−bc−ca).
e^i3α+8e^i3β+27e^i3γ=6e^i(α+β+γ)=−6

Comparing imaginary parts,we get
sin3α+8sin3β+27sin3γ=0


----------



## theterminator (Sep 2, 2013)

I have one confusion: 
In the question 
48÷2*12. 

will we do the division first or multiplication first & by what rule??

being from comp sci background , i know that multiplication & division are in the same level of priority but the operation will be performed first when encountered upon going from left to right. Some say to follow BODMAS rule to solve...in that case , what will be the answer to this question
48*2÷12


----------



## rst (Sep 2, 2013)

We will solve this question by BODMAS rule
So 48÷2*12 = 24*12=288


----------



## theterminator (Sep 2, 2013)

rst said:


> We will solve this question by BODMAS rule
> So 48÷2*12 = 24*12=288



& this  48*2÷12


----------



## rst (Sep 2, 2013)

theterminator said:


> & this  48*2÷12



48*2÷12 = 48 *(1/6)=8


----------



## theterminator (Sep 2, 2013)

rst said:


> 48*2÷12 = 48 *(1/6)=8


but in computer program, multiplication will be performed first coz order of precedence is from left to right in case of operators of same level (  * and / are of same level ) 



Spoiler



Did you first looked for ÷ sign in your keyboard?


----------



## rst (Sep 3, 2013)

theterminator said:


> but in computer program, multiplication will be performed first coz order of precedence is from left to right in case of operators of same level (  * and / are of same level )



Well,its also correct
By doing so,you will  get the same answer
48*2÷12 =96 ÷12 =8
---------------------------------------------
From left to right you can do any "D" or "M" 
Similar is for addition and subtraction


----------



## the_conqueror (Sep 6, 2013)

∫sec²x dx/(sec x + tan x )^9/2


----------



## rst (Sep 6, 2013)

the_conqueror said:


> ∫sec²x dx/(sec x + tan x )^9/2



put tanx =t
then sex²x dx= dt

∫dt/[t+ √ (1+t²)]^9/2--------(1)

Now let t+ √ (1+t²) =u
clearly 1/u =√ (1+t²) -t

So t =1/2 [u - (1/u)]
dt =1/2 [1+ (1/u²)]du

Using in (1),we get
(1/2)∫du [u ^(-9/2) + u ^(-13/2)]du

Now its easy to solve


----------



## Niilesh (Sep 6, 2013)

theterminator said:


> I have one confusion:
> In the question
> 48÷2*12.
> 
> ...



Since there is no bracket 48÷2*12 = 24*12 = 288 = 48*12÷2 = 12÷2*48


----------



## harshilsharma63 (Sep 8, 2013)

View attachment 12070
With AO || CD, find area of shaded region.


----------



## rst (Sep 8, 2013)

harshilsharma63 said:


> View attachment 12070
> With AO || CD, find area of shaded region.



 ∠O=<D=θ (corresponding angles)
Then <ODC= π-θ
The area of the shaded region = the area of the sector OCB - the area of the 
triangle OCD

Let <BOC =x and <OCD =y

sin(π−θ)/R=siny/ℓ
sinθ/R=siny/ℓ
y= arcsin(ℓsinθ/R)

x+y+(π−θ)=π
x=θ-y
x=θ-arcsin(ℓsinθ/R)

A(ΔOCD)=(1/2)Rℓsinx


----------



## rst (Oct 8, 2013)

*----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION*

*img834.imageshack.us/img834/9262/exgv.png

 Plz explain your answer

 (Also you can ask your mathematics related questions or queries
 we will try to solve them)
*----------------------------------------------------------------------------------------------------------------------*
*you can use following mathematic symbols(just do copy and paste)*
 £ ω ∴ ∂ Φ γ δ μ σ Є Ø Ω ∩ ≈ ≡ ≈ ≅ ≠ ≤ ≥ • ~ ± ∓ ∤ ◅∈ ∉ ⊆ ⊂ ∪ ⊥ ô ∫ Σ → ∞ Π Δ Ψ Γ ∮ ∇∂ √ °α β γ δ ε ζ η θ ι κ λ ν ξ ο π ρ σ τ υ φ χ ψ 
x⁰ ¹ ² ³ ⁴ ⁵ ⁶ ⁷ ⁸ ⁹ ⁺ ⁻ ⁿ 
 x₀ ₁ ₂ ₃ ₄ ₅ ₆ ₇ ₈ ₉ ₊ ₋


----------



## Mr.Kickass (Oct 8, 2013)

A gift for all members

Math Help Forum


----------



## ACidBaseD (Oct 22, 2013)

rst said:


> *----------------------------------------------------------------------------------------------------------------------
> NEW OBJECTIVE QUESTION*
> 
> *img834.imageshack.us/img834/9262/exgv.png
> ...



I am getting ans = -2
By using L'Hospital Rule then opening the Determinant and using limits respectively on each summation term


----------



## rst (Nov 13, 2013)

*----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION*

*img10.imageshack.us/img10/5821/ygvl.png

 Plz explain your answer

 (Also you can ask your mathematics related questions or queries
 we will try to solve them)
*----------------------------------------------------------------------------------------------------------------------*
*you can use following mathematic symbols(just do copy and paste)*
 £ ω ∴ ∂ Φ γ δ μ σ Є Ø Ω ∩ ≈ ≡ ≈ ≅ ≠ ≤ ≥ • ~ ± ∓ ∤ ◅∈ ∉ ⊆ ⊂ ∪ ⊥ ô ∫ Σ → ∞ Π Δ Ψ Γ ∮ ∇∂ √ °α β γ δ ε ζ η θ ι κ λ ν ξ ο π ρ σ τ υ φ χ ψ 
x⁰ ¹ ² ³ ⁴ ⁵ ⁶ ⁷ ⁸ ⁹ ⁺ ⁻ ⁿ 
 x₀ ₁ ₂ ₃ ₄ ₅ ₆ ₇ ₈ ₉ ₊ ₋


----------



## rst (Nov 14, 2013)

*----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION*

*img818.imageshack.us/img818/1681/f3ry.png

 Plz explain your answer

 (Also you can ask your mathematics related questions or queries
 we will try to solve them)
*----------------------------------------------------------------------------------------------------------------------*
*you can use following mathematic symbols(just do copy and paste)*
 £ ω ∴ ∂ Φ γ δ μ σ Є Ø Ω ∩ ≈ ≡ ≈ ≅ ≠ ≤ ≥ • ~ ± ∓ ∤ ◅∈ ∉ ⊆ ⊂ ∪ ⊥ ô ∫ Σ → ∞ Π Δ Ψ Γ ∮ ∇∂ √ °α β γ δ ε ζ η θ ι κ λ ν ξ ο π ρ σ τ υ φ χ ψ 
x⁰ ¹ ² ³ ⁴ ⁵ ⁶ ⁷ ⁸ ⁹ ⁺ ⁻ ⁿ 
 x₀ ₁ ₂ ₃ ₄ ₅ ₆ ₇ ₈ ₉ ₊ ₋


----------



## rst (Apr 9, 2014)

if f(x)=|x-2| + |x+1|-x then find value of f ' (-10)


----------



## Superayush (Apr 10, 2014)

Is it given x belong to which domain?


----------



## rst (Apr 10, 2014)

Superayush said:


> Is it given x belong to which domain?



domain is not given

options are 
(a) -3 (b) -2 (c) -1 (d) 0


----------



## avj (Apr 10, 2014)

(a) -3
d/dx of |x| is |x|/x and obviously x != 0

- - - Updated - - -



Superayush said:


> Is it given x belong to which domain?



x=-10 so in the domain (-inf,-1)

- - - Updated - - -



rst said:


> *----------------------------------------------------------------------------------------------------------------------
> NEW OBJECTIVE QUESTION*
> 
> *img818.imageshack.us/img818/1681/f3ry.png
> ...



the function is 1 at x=0 and 2/pi at x=pi/2

and we know that def. integrals is nothing but area under the graph

so consider 2 rectangles with width pi/2 and height 1 and 2/pi.
it is obvious that (sin x)/x integral lies between area of these 2 rectangles since (sin x )/x is monotonically decreasing function in this range(1st derivative test).

so option 1 is right.


----------



## rst (Apr 11, 2014)

^^ Thanks a lot avg


----------



## rst (Jun 5, 2014)

What will be remainder  if (16 ! +1) is divided by 17 ?


----------



## rst (Jun 15, 2014)

rst said:


> What will be remainder  if (16 ! +1) is divided by 17 ?



We have that 16!≡−1(mod 17) by Wilson's Theorem.

therefore 16! +1 ≡ 0 (mod 17) 

So remainder is 0


----------



## rst (Jun 16, 2014)

what is inverse of y = 5 ^ logx


----------



## powerhoney (Jun 17, 2014)

rst said:


> what is inverse of y = 5 ^ logx


y=Antilog((log x)/(log 5))???


----------



## powerhoney (Jun 17, 2014)

Or is it Y=5^x???


----------



## seamon (Jun 17, 2014)

^^Asking or answering?


----------



## powerhoney (Jun 17, 2014)

seamon said:


> ^^Asking or answering?


Answering but sorry, y=5^x is wrong...  You could do???


----------



## powerhoney (Jun 17, 2014)

I could only do till log x Base 10= log y Base 5


----------



## powerhoney (Jun 17, 2014)

Is the answer y=10^(log x base 5)???


----------



## powerhoney (Jun 17, 2014)

powerhoney said:


> Answering but sorry, y=5^x is wrong...
> Could you solve???



.....


----------



## rst (Jun 17, 2014)

rst said:


> what is inverse of y = 5 ^ logx



y = 5 ^ log x

log_5 y =log x ,( Here log_5 y is  log y Base 5)

e^(log_5 y)  = x  

e^(log y /log 5)  = x  

(e^log y ) ^(1/log 5) = x  

y ^(1/log 5) = x


----------



## powerhoney (Jun 17, 2014)

rst said:


> y = 5 ^ log x
> 
> log_5 y =log x ,( Here log_5 y is  log y Base 5)
> 
> ...


Er, I think I got the same answer... You took base e and I took base 10... Also, I represented inverse as we were taught by switching 'x' and 'y'...


----------



## powerhoney (Jun 17, 2014)

powerhoney said:


> Is the answer y=10^(log x base 5)???


Rewriting:

y=10^(log_5 x)

Changing the x to y and vice versa 

x=10^(log_5 y)
x=[10^(log_10 y)]^[1/(log_10 5)]
x=y^[1/(log_10 5)]

Which is the same...


----------



## rst (Jun 17, 2014)

If ax + by = c is tangent to the circle x^2 + y^2 = 16 then which of the following is correct option

(A)16 ( a^2  + b^2) = c ^2    
(B)16 ( a^2  - b^2) = c ^2 
(C)16 ( a^2  +b^2) = - c^2    
(D)16 ( a^2  - b^2) = - c^2


----------



## nomad47 (Jun 17, 2014)

rst said:


> If ax + by = c is tangent to the circle x^2 + y^2 = 16 then which of the following is correct option
> 
> (A)16 ( a^2  + b^2) = c ^2
> (B)16 ( a^2  - b^2) = c ^2
> ...



option A.
As the line is a tangent there will be a single solution for the set of equation. Substitute either x or y from the linear equation in the quadratic equation and use condition both the roots are equal. You will arrive at A.


----------



## rst (Jun 18, 2014)

nomad47 said:


> option A.
> As the line is a tangent there will be a single solution for the set of equation. Substitute either x or y from the linear equation in the quadratic equation and use condition both the roots are equal. You will arrive at A.



Answer is correct

Roots are equal i.e D=0

But it will form equation of degree 4 either in a or b

How will we get option (A)


----------



## powerhoney (Jun 18, 2014)

[MENTION=158560]rst[/MENTION]

You got PM...


----------



## rst (Jun 18, 2014)

powerhoney said:


> [MENTION=158560]rst[/MENTION]
> 
> You got PM...



Yeah, I got it


----------



## nomad47 (Jun 18, 2014)

rst said:


> Answer is correct
> 
> Roots are equal i.e D=0
> 
> ...



a or b is not equal to zero. Clear?


----------



## rst (Jun 18, 2014)

rst said:


> If ax + by = c is tangent to the circle x^2 + y^2 = 16 then which of the following is correct option
> 
> (A)16 ( a^2  + b^2) = c ^2
> (B)16 ( a^2  - b^2) = c ^2
> ...



x^2 + y^2 = 16

Here centre =(0,0)
r=4

Perpendicular distance from centre (0,0) to the tangent  = r

c / sq rt ( a^2 + b^2) =4

squaring both sides, we get

16 ( a^2  + b^2) = c ^2


----------



## nomad47 (Jun 18, 2014)

rst said:


> x^2 + y^2 = 16
> 
> Here centre =(0,0)
> r=4
> ...



That's one way to do that. And am really bad with all these formulas.


----------



## rst (Jun 18, 2014)

The distance between the directrices of the ellipse 9 x^2 + 4 y^2 = 36 is  which of the following option  
(A) 2 √5   
(B) √5   
(C) 9/√5   
(D) 18/ √5


----------



## powerhoney (Jun 18, 2014)

rst said:


> The distance between the directrices of the ellipse 9 x^2 + 4 y^2 = 36 is  which of the following option
> (A) 2 √5
> (B) √5
> (C) 9/√5
> (D) 18/ √5


D. 18/(Square Root of 5)


----------



## rst (Jun 18, 2014)

powerhoney said:


> D. 18/(Square Root of 5)



Answer is correct

Can you explain it ?


----------



## powerhoney (Jun 18, 2014)

rst said:


> Answer is correct
> 
> Can you explain it ?


Solved using formula...


----------



## rst (Jun 19, 2014)

rst said:


> The distance between the directrices of the ellipse 9 x^2 + 4 y^2 = 36 is  which of the following option
> (A) 2 √5
> (B) √5
> (C) 9/√5
> (D) 18/ √5



9 x^2 + 4 y^2 = 36 

 (x^2) /4 +  (y^2 )/9 = 1

Here a^2 =9, b^2 =4

a=3 , b= 2

As a^2 = b^2 + c^2
9= 4 + c^2
√5 =c
√5 =a e
√5/3 =e

The distance between the directrices =2a/e
= 18/ √5


----------



## powerhoney (Jun 20, 2014)

Hey, how to find limit of ((2^x-1)/x)^2 as x tends to zero???


----------



## powerhoney (Jun 20, 2014)

Btw, your total posts have reached 666!!!


----------



## nomad47 (Jun 20, 2014)

powerhoney said:


> Hey, how to find limit of ((2^x-1)/x)^2 as x tends to zero???



Take m=xln2. As x tends to 0 m tends to 0. Go by the formula if (e^x-1)x you will get your answer


----------



## rst (Jun 20, 2014)

powerhoney said:


> Hey, how to find limit of ((2^x-1)/x)^2 as x tends to zero???



lim (x->0) ((2^x-1)/x)^2

lim (x->0) (2^x-1)^2  / x^2

lim (x->0) (2^2x+1-2.2^x ) / x^2
Using L' Hospital rule

lim (x->0) (2 log2 2^2x -2.2^x log2) / 2x
lim (x->0) ( log2 2^2x -2^x log2) / x

Using L' Hospital rule

lim (x->0) [ 2 (log2)^2 2^2x -2^x  (log2)^2]/ 1

=(log2)^2


----------



## powerhoney (Jul 13, 2014)

Find the total number of squares and rectangles in an NxN chessboard


----------



## rst (Jul 14, 2014)

powerhoney said:


> Find the total number of squares and rectangles in an NxN chessboard



In 8x8 chessboard we have only ONE 8x8 square. 
There are FOUR 7x7 squares.  There are NINE 6x6 squares, and so on

In 8x8 chessboard we have 1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2 + 7^2 + 8^2 square

Hence in  NxN chessboard chessboard  
the total number of squares = 1^2 + 2^2 + 3^2 + ... + N^2

 Also NxN chessboard chessboard  
the total number of rectangle  = [C(n+1,2) ] ^2


----------



## powerhoney (Jul 14, 2014)

rst said:


> In 8x8 chessboard we have only ONE 8x8 square.
> There are FOUR 7x7 squares.  There are NINE 6x6 squares, and so on
> 
> In 8x8 chessboard we have 1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2 + 7^2 + 8^2 square
> ...


Good... There were no new questions being posted so I posted that one...


----------



## rst (Aug 4, 2014)

A vertical pole with height more than 100 m consists of two parts, the lower being on third of the whole. 
At a point on a horizontal plane through the foot and 40 m from it, the upper part subtends an angle whose tangent is 1/2. 
what is the height of the pole ?

a)110 m
b)200m
c)120 m
d)150m


----------



## rst (Aug 6, 2014)

If m, n are integers, then what is the value of 
int (0,1) sinmx sinnx dx (here int is integration, 0 is lower limit and 1 is upper limit).if m ≠n
A)2/π
B)4/π
C)0
D)1


----------



## rst (Aug 8, 2014)

What is the least value of 3sin^2 x + 4 cos^2 x 
A)1
B)2
C)3
D)4


----------



## anirbandd (Aug 9, 2014)

^3 for x=pi/2


----------



## rst (Aug 9, 2014)

anirbandd said:


> ^3 for x=pi/2


Answer is correct
3sin^2 x +4cos^2 x
=3+cos ^2 x

-1≤Cosx≤1
0≤cos^2 x ≤1 (as cos^2x can't be negative)
3≤cos^2 x+3≤4

So least value is 3


----------



## rst (Aug 13, 2014)

If equation ax^2+bx+c=0, a>0, has two distint real roots p and q such that p<-5 and q>5 then
A)c>0
B)c=0
C)c=(a+b)/2
D)c<0
E)c= a+b


----------



## anirbandd (Aug 15, 2014)

^thats easy..

its D. c<0


----------



## rst (Aug 18, 2014)

if angle between the curves y=2^x and y= 3^x is p then value of  tanp is
a)log(3/2) /[1+6(log2)(log(3)]
b) 6/7
c)1/7
d)log6 / [1+(log2)(log3)]


----------



## anirbandd (Aug 19, 2014)

whats the soln??

they meet at x= - ∞, but at -∞ it becomes 0..


----------



## Neo (Aug 25, 2014)

rst said:


> if angle between the curves y=2^x and y= 3^x is p then value of  tanp is
> a)log(3/2) /[1+6(log2)(log(3)]
> b) 6/7
> c)1/7
> d)log6 / [1+(log2)(log3)]



A. Simply find tan(A-B)


----------



## rst (Aug 25, 2014)

Neo said:


> A. Simply find tan(A-B)



answer is correct

can you explain a little bit


----------



## anirbandd (Aug 26, 2014)

rst said:


> answer is correct
> 
> can you explain a little bit


+1. 

Im a lil confused


----------



## Neo (Aug 26, 2014)

Find tanA = dy/dx for the graphs. Find tan (A-B) using the identity. Put x=0 at last because the graphs meet at it. 

What class you guys in btw?


----------



## rickenjus (Oct 5, 2014)

Guys, I have doubt over this Question. It is from 11th class RD Sharma. Permutation -> Exercise - 16.2 -> Q24. 

Below is a part of the Question. 

Q. How many 3 digit odd numbers can be formed using 0,3,5,7, repetition not allowed. 

Sol. What I did. 

_    _    _
3 X 3 X 1

For *Hundredth* place - It can be filled in 3 ways (any of 3,5,7), we cannot use 0. 
For *Tens* place - It can be filled in 3 ways (0,3,5,7) as one of 3,5,7 already filled in hundredth place).
For *Ones *place - It can be filled in 1 way as  two digits of 3,5,7 already used in above two places and it cannot use 0. 

And solution I found on book says this: - 

It fills *Hundredth* first, then *Ones* and then *Second*. 

*i.imgur.com/T5f2zsO.png
_    _    _
3 X 2 X 2

What is that I'm not understanding or doing wrong??


----------



## rst (Oct 6, 2014)

rickenjus said:


> Guys, I have doubt over this Question. It is from 11th class RD Sharma. Permutation -> Exercise - 16.2 -> Q24.
> 
> Below is a part of the Question.
> 
> ...



Book solution is correct 

 As Tens place can be filled in any way
 So it will be filled last

 Hundreds and Ones places can be filled in specific way
 So they are filled first


----------



## rickenjus (Oct 6, 2014)

[MENTION=158560]rst[/MENTION] - ya thanks... got perfect explanation on stackexchange.


----------



## rst (Nov 14, 2014)

let AB be diameter of circle and AC be the chord.let a tangent is drawn from C to meet AB produced at D. If angle BAC= 30 .Prove that 
BC= BD


----------



## doomgiver (Nov 14, 2014)

*i.imgur.com/kPWFYcl.png




angle ACB = right angle coz thales theorem. i dunno what else to do.

http:/ /www. math10.com/en/geometry/geogebra/geogebra. html
remove spaces.
tool used for pic ^^


----------



## rst (Nov 14, 2014)

By diagram it is clear

But is there any method to prove it


----------



## Chaitanya (Nov 14, 2014)

May be this will help...
BTW was too lazy type 

*tapatalk.imageshack.com/v2/14/11/14/a5bc45d18bf2793a838881c84ec8d5c9.jpg


----------



## rst (Nov 14, 2014)

Great

Thanks a lot


----------



## rst (Jan 29, 2015)

Solve for x

2^(x+2)- 6^x -2 [3^(2x +2)]=0


----------



## Chaitanya (Jan 29, 2015)

so just to be clear the question is 

{2^(x+2)} - {6^x} - 2{3^(2x+2)} = 0


----------



## rst (Jan 30, 2015)

^^ yes question is like  that


----------



## Niilesh (Jan 31, 2015)

rst said:


> Solve for x
> 
> 2^(x+2)- 6^x -2 [3^(2x +2)]=0



take log both sides then we will get a linear equation in one variable which can be easily solved.


----------



## Chaitanya (Feb 1, 2015)

^ not that simple.
I couldn't find any satisfactory solution yet..


----------



## nomad47 (Feb 1, 2015)

Here you go


*tapatalk.imageshack.com/v2/15/01/31/a36923181c8cf943b6832d00a29aea23.jpg


----------



## Chaitanya (Feb 1, 2015)

Buddy on LHS of the equation it will be "log(2^(x+2)-6^x)" & not "log(2^(x+2)) - log(6^x)"


----------



## nomad47 (Feb 2, 2015)

Oops. Damn. My mathematics is getting rusty sorry


----------



## rst (Feb 6, 2015)

rst said:


> Solve for x
> 
> 2^(2x+2)- 6^x -2 [3^(2x +2)]=0



4 2^(2x) - 2^x 3^x - 18 3^(2x) =0

Dividing both sides by3^(2x) we get

4 (2/3)^(2x) - (2/3)^x -18 =0

Let (2/3)^x =y

4 y^2 - y -18 =0

Solving 
y= 9/4 or -2

(2/3)^x = (3/2)^2

x = -2


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## rst (Feb 8, 2015)

If √(pq) =6 and p,q are positive integers. Then which of the following could not be a value of
 (p-q) ?

a) 0
b) -9
c) 8
d) 5


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## hari1 (Feb 8, 2015)

Option C:8


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## rst (Feb 8, 2015)

Correct.

In that case we are not getting positive integral value of p,q


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## rst (Feb 9, 2015)

Find the length of the diagonal of the largest possible cube inscribed in a hemisphere of radius 4√2 cm.

A) 12 cm
B) 6 cm
C) 8 cm
D) 4 cm


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## rst (Feb 11, 2015)

Ans is C) 8 cm


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## rst (Feb 12, 2015)

*imageshack.com/i/idJjorfrj


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## Chaitanya (Feb 12, 2015)

Ans is B


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## Chaitanya (Feb 12, 2015)

*tapatalk.imageshack.com/v2/15/02/12/dab353aafa1debeb7511e416c9665413.jpg


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## Chaitanya (Feb 12, 2015)

Well I have a question too..
Its not exactly mathematics but then there is no place else*tapatalk.imageshack.com/v2/15/02/12/2f52b28b05c017fb850a38576d41d2aa.jpg


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## rst (Feb 13, 2015)

Average velocity=  72 km/h= 20m/s
Total time= 25s
Total distance= 20*25=500m

Let t1,t2,t3 be time for acceleration, uniform motion and retardation

Time for acceleration and retardation will be same 

So t1 +t2+t1= 25
2 t1 + t2 = 25

Velocity after time t1
v1 = 0+5t1


Now we can draw v- t gragh of above motion which will be like trapezium

Distance= area of trapezium
500= 1/2 (t2+25)v1
1000=  (25-2t1+25) 5t1
Solving t1=5

So t2=15s


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## rst (Feb 14, 2015)

*imageshack.com/i/excBhjhYj


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## powerhoney (Feb 15, 2015)

rst said:


> *imageshack.com/i/excBhjhYj



B. 4.5 cmxcm


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## rst (Feb 15, 2015)

Yes answer is correct


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## powerhoney (Feb 16, 2015)

rst said:


> Yes answer is correct



Posting the solution below!!! 




- - - Updated - - -



rst said:


> Yes answer is correct



Posting the solution below!!! 




*i.imgur.com/GNIbbJd.jpg?1


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## rst (Feb 16, 2015)

*imageshack.com/i/ip9OIBd7j


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## powerhoney (Feb 17, 2015)

rst said:


> *imageshack.com/i/ip9OIBd7j



350 cmxcm


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## rst (Feb 17, 2015)

Ans is correct


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## rst (Feb 18, 2015)

*imageshack.com/i/p7bNY00cj


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## ankush28 (Feb 18, 2015)

rst said:


> Spoiler
> 
> 
> 
> *imageshack.com/i/p7bNY00cj


I don't know much about solving this with pen and paper but... I think it has something to with cyclic remainders, see the picture.
*i.imgur.com/TBChG1Q.png
And the answer is *18*


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## rst (Feb 18, 2015)

Ans is correct

Remainder is 18


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## rst (Feb 19, 2015)

*imageshack.com/i/ex4gzsY2j


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## rst (Feb 20, 2015)

Ans is option (c)


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## rst (Feb 22, 2015)

Find the number of all possible selections of one or more questions from 10 given questions, each question having an alternative.

A)3^(10)

B) 2^ (10) -1

C) 3^(10) -1

D) 2^(10 )


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## rst (Feb 23, 2015)

Ans is option (c) 3^10 -1

As for each question we have 3 options
1) do it 2) leave it 3) do alternative question


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## rst (Feb 24, 2015)

*imageshack.com/i/pcWymDxgj


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## rst (Mar 14, 2015)

Ans is A) y


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## rst (Mar 22, 2015)

*imageshack.com/i/p58FCIDAj

- - - Updated - - -

Option (B) is Cosα  cosβ cosγ


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## iittopper (Mar 23, 2015)

First expand all cos term . Then write the matrix in product form A*B . Then solve it normally .


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## rst (Mar 27, 2015)

Ans is option (D) 0


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## rst (Nov 14, 2015)

A cat chases a rat. For every 5 leaps of the rat , the cat takes 3 leaps, but the 2 leaps of the cat are the same as 3 leaps of the rat. Compare the speeds of the cat and the rat

a)4:5
b)12:13
c)16:15
d)none of these


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## rst (Nov 17, 2015)

Ans is 9:10


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## sohan_92 (Nov 20, 2015)

rst said:


> Ans is 9:10



care to explain?


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## rst (Nov 20, 2015)

Here distance of leaps of cat and rat are not same
clearly the  2 leaps of the cat are the same as 3 leaps of the rat
Let distance of rat leap be x

then distance of cat's leap = 3x/2 =1.5X

then speed ratio =(1.5x*3)/(x*5)= 4.5x/5x

9/10


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## rst (Nov 23, 2015)

A bike is sold by an automobile agency for rupess 19200 cash or for 4800 cash downpayment together with 5 equal monthly installment if rate of interest charged by the company is 12% per annum .find the amount of the installment


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## anirbandd (Nov 18, 2017)

the question that follows is quite common in Combinatorics and all the maths books arrive at the answer through a single method. i have never seen any variations so far. 
however, that method of arriving at the answer is not how you would deal with the problem in real life.

how would YOU solve it?

_A class contains 12 boys and 10 girls. From the class 10 students are to be chosen for a competition under the condition that atleast 4 boys and atleast 4 girls must be represented. The 2 girls who won the prizes last year should be included. In how many ways can the selection be made?_

Instead of giving a numeric answer, i would like to know how you approach the problem..


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## thetechfreak (Oct 2, 2018)

I guess the general approach could be doing a combination keeping the 2 girls fixed and selecting 2 more from the remaining 8 girls. Boys can be selected with 12C4. Multiplication of both the results should give the needed output.


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## anirbandd (Apr 10, 2021)

That problem was the end of this thread, wasnt it


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