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Re: mathematic related questions

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

)

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)

Re: mathematic related questions

Is the answer "L" ?

I'm bad at this

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"

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?

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 ??

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).

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

Re: mathematic related questions

rst said:
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

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.

V

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?

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

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

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?

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?

EDIT :

not counting original square in both

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

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?

EDIT :

not counting original square in both

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

V

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.

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?

EDIT :

not counting original square in both

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

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

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

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.

V

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!

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

V

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?

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

V

Re: mathematic related questions

rst said:
315 point of intersection

How? . . .

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

Re: mathematic related questions

rst said:
315 point of intersection

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

V

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

but it is not related with point of intersection

it requires lot of explanation

its really a tough question

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.

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)

V

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.

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)

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

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
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

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

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

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

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

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

Re: mathematic related questions

rst said:
ok
how did you find area of Δpcq and Δpaq

1/2 * pa * 3

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

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

Re: mathematic related questions

rst said:
objective question
View attachment 10057

Also plz show some work

ans is 15/4

Can anyone explain ??

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

Re: mathematic related questions

NO

It can't be solved in this way

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

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
x=7/8
now,
PQ² = 3² + (√9+x² - x)²
PQ= 3*5/4=15/4

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

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.
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.

umm.. What? Can someone explain that?

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.

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)

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)

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)

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

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

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)

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

Re: mathematic related questions

Absolutely right
great work

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

Also plz explain your answer

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we will try to solve them)

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])

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

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

Re: mathematic related questions

Alternate method for you

I know my writing is not that good :razz:

*i.imgur.com/RuGGjlT.jpg

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

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we will try to solve them)

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

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

Re: mathematic related questions

Silly me

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)

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)?

Re: mathematic related questions

answer is 91

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

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 ??

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
If A ≥ R
B ≥ R

can we say A ≥ B ??

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

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')
now 7a=21 => a=3
So ans = 7*9/7 = 9

I am also not sure

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

Also plz explain your answer

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we will try to solve them)

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)

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 ??

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?

Re: mathematic related questions

^ Haha, you beat me to it.

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

Re: mathematic related questions

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

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

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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

Re: mathematic related questions

Yeah
answer is N+1

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

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

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

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we will try to solve them)

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

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)

Re: mathematic related questions

is it 49/100?

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

Re: mathematic related questions

no
ans is 100 minutes

Re: mathematic related questions

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

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

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"
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
That's what i said/meant but giving a general solution directly

rst said:
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

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

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

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

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

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

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

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

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
It wont be a perfect square as odd no./2=a fraction(not an integer).
Like for example take K=3(or 2n+1)

Chars

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

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

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"

Re: mathematic related questions

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

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

Re: mathematic related questions

ohh...
I shouldn't have missed it

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.

Re: mathematic related questions

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

Re: mathematic related questions

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

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).

Now I wanted to find any occurances where the nth digit followed the 1.5nth digit.

The repeating digits in 1/7 and their corresponding 'n' where the digits are the first digit after the decimal point are:

142857 - repeating digits
132645 - n

n=2 follows n=3, so the lowest number is 285714. Also, n=4 follows n=6, so the next number is 571428.

Out of curiosity, I checked all the prime numbers up to 50. The ones with p-1 repeating digits are:

17, 19, 23, 29, 31, 47

None of these has two adjacent digits with the desired property, so there is no other solution with less than 50 digits.

I'd be curious if there was a way to either prove that no other solutions exist, or determine which prime numbers would have the required relationship. I think other solutions do exist, but I'll write a program to check them any furthur.

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

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we will try to solve them)

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

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we will try to solve them)

Ans is 1?

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

Also plz explain your answer

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

One, and no explanation required I guess.

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

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?

Re: mathematic related questions

^ All the conditions have to be satisfied together

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.

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.

Re: mathematic related questions

Yes that prime thing removes most of the combinations.

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
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

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

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

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 ??

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

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

Re: mathematic related questions

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

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

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

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

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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.

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.
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.

ans is 5736912480

Re: mathematic related questions

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

Re: mathematic related questions

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

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

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Re: mathematic related questions

^ Can you explain how are they arranged?

Re: mathematic related questions

I think they are arranged joining each other to form cuboid

Re: mathematic related questions

Ans is 2?

Re: mathematic related questions

Can you show your work ???

Re: mathematic related questions

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

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

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---------------------------------------------------------------------------------------------------------------------
Hint:-
answer is not 8 and 12

Re: mathematic related questions

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

Re: mathematic related questions

I also think the same
with 6 sticks tetrahedron is possible(which can enclose 3D volume)
----------------------------------------------------------------------------------------------------------------------
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*img9.imageshack.us/img9/6169/boysandcircle.png

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Re: mathematic related questions

4 min(I Used the concept of relative velocity)

Re: mathematic related questions

Can you show your work ??

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

Re: mathematic related questions

Yeah
you are right

Also this method is shorter than mine

Re: mathematic related questions

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

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.

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

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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)

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

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

Re: mathematic related questions

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

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

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

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

Re: mathematic related questions

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

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

A

Re: mathematic related questions

rst said:
----------------------------------------------------------------------------------------------------------------------
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*img191.imageshack.us/img191/6536/formulaj.png

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

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....)

Re: mathematic related questions

answer is absolutely right
Also thanks for the explanation

Re: mathematic related questions

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

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

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

Re: mathematic related questions

ans is b. 10 %

Can someone explain it ??

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

Re: mathematic related questions

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

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

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

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

Re: mathematic related questions

rst said:
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

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

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

Can you explain that?

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

Also plz explain your answer

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

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

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

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)

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

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 (?)

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

M

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..

Re: mathematic related questions

You are welcome

Re: mathematic related questions

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

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 :mrgreen:

M

Re: mathematic related questions

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

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

M

Re: mathematic related questions

Vignesh B said:
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.

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

Re: mathematic related questions

^Is it Sunday ?

M

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 :wink:

Re: mathematic related questions

mastercool8695 said:
this one :
July 3, 1997 is a sunday, then July 3,1970 was ??

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

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

M

Re: mathematic related questions

^^ sorry, posted 1977 as 1997
sorry..

post edited now..

anyways, your explaination was great.. thanks..
:clap2:

Re: mathematic related questions

mastercool8695 said:
this one :
July 3, 1977 is a sunday, then July 3,1970 was ??

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

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

M

Re: mathematic related questions

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

Re: mathematic related questions

But it doesnot look like 10+2 level question

M

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..

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)
----------------------------------------------------------------------------------------------------------------------

M

Re: mathematic related questions

*i.imgur.com/FMKAwY1.png

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)

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.

Re: mathematic related questions

rst said:
Ans is (D)
----------------------------------------------------------------------------------------

A) It is not true
Counter example :-
Let a=3 ,b=5 and n=2
aⁿ +bⁿ =34 (not prime)

B) It is not true
Counter example :-
Let a=2 ,b=4 and n=8(power of 2)
aⁿ +bⁿ =even number(not prime)

C)It is not true
Counter example :-
Let a=2 ,b=4 and n=6(even number)
aⁿ +bⁿ =even number(not prime)

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

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

M

Re: mathematic related questions

so you guys suggest hit and trial ??

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

M

Re: mathematic related questions

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

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

Re: mathematic related questions

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

don't do it reverse manner

M

Re: mathematic related questions

ok...
forgot that..
thanks..

any example ?

Re: mathematic related questions

take example which I gave in previous post

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

M

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..

Re: mathematic related questions

rst said:
----------------------------------------------------------------------------------------------------------------------
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)
----------------------------------------------------------------------------------------------------------------------

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

M

Re: mathematic related questions

55 is the 9th number,

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

Re: mathematic related questions

ok, edited.

M

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

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
B)EXACTLY 2
C)>2 BUT FINITELY MANY
D)INFINITELY MANY

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

Re: mathematic related questions

rst said:
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 b² +................+(-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

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

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 )

Re: mathematic related questions

rst said:
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

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
a³+b³ =(a+b) (a²-ab+b²)

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

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

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

M

Re: mathematic related questions

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

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
x² + y² = 4K
or x² + y² = 4K +1
or x² + y² = 4K +2
Hence, no integral values of (x,y) will be possible.

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

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

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) :duh2:

I will explain the solution tomorrow when i will be online from PC

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

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

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"

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"

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"

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)

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)
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

Now your proof looks nice

M

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

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)

M

Re: mathematic related questions

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

Re: mathematic related questions

Yeah
I use copy and paste

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
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

Is it 9C2 = 36?

Re: mathematic related questions

yeah
it is the correct answer

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

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 ??

M

Re: mathematic related questions

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

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
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

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 ]

Re: mathematic related questions

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

Can you explain your work ??

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.

M

Re: mathematic related questions

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

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

A

Re: mathematic related questions

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

M

3.3^2001 + 25^1001
=>3.27^667 + 25^1001
=>3.1^667 + (-1)^1001
=> 3-1
=> 2 which should be the answer .

the answer for the second one is 28 .
it would convert to 28 ! / 29 .....
also (p-1)!+1 is divisible by p and so on .

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

Re: mathematic related questions

I think remainder is 1.

M

Re: mathematic related questions

^^ Explain..

Re: mathematic related questions

remainder is 1

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.

M

Re: mathematic related questions

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

Re: mathematic related questions

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

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

A

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

M

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 .. :mrgreen:

rst said:
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
=[1 + 2001C1 x 6 + 2001C2 x 6^2 ..... +2^2001 x 3^2001 ] x7

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

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

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

M

Re: mathematic related questions

^^ thanks. got it now. :goodjob:

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.)

M

yup..
buddy..
i know that :lol:

ok leave it, i found the link

*www.python.org/download/

found this too.. *www.learnpython.org

will learn prog lingis after june 2

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can someone give me the solution to the following question?

Find range of (2-x)^½ + (1+x)^½.

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

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
domain =[-1,2]

Put values in function and find range

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)

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 said:
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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

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

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}

rst said:
√ [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}

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

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

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

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

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

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

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

How did you get side 7 and 3 ??

^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.

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

^What's up with "2007" ?

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

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.

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

Niilesh said:
Ans is (4-3)*9 = 9

Both ans and explanation are correct
well done !

@rst: Are you doing B.Tech?

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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

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

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}

Niilesh said:
The Question was right

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}

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

^ Ohh you are right actually it is the reverse

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

Answer is 1640

rst said:
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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

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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

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

Can you care to show your proof?

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 said:
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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

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
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)

Thanks

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Ans is Median (I think no explanation is required)

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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.

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

=>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.

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])

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

This thread is only for maths questions

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

rst said:
This thread is only for maths questions

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

P&C comes under maths.

I thought he mean "physics and chemistry"

I mean Permutation and combinations

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

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

rst said:
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

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]

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^ 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 said:
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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

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 form garland in 7! ways
Also white flowers can be arranged in 3! ways
total ways =7! x 3!

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.

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

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

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

rst said:
your method is correct
It will also give same result

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

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)

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?

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

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

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 said:
What's that extra 5!3! for?

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

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^ 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

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rst said:
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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)

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 -

3

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)

Answer -

3

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

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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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 said:
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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 ₊ ₋

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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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 ₊ ₋

Evaluate sin(cot^(-1)x)

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 said:
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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 ₊ ₋

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)

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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 ₊ ₋

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 //

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.

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

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.

You're welcome - btw I love maths.

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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 ₊ ₋

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

Well, I don't know ans

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

Give ans in short

Write only main steps or just give hint

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.

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.

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.

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

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 said:
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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 ₊ ₋

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

rst said:
We can also solve this question by putting P = a + t

By solving with LH rule ,we get

1/ √a =1

which is possible for a=1

You get 1 even by doing what I said

A bit tedious though :wink:

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----------------------------------------------------------------------------------------------------------------------
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 said:
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NEW OBJECTIVE QUESTION

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----------------------------------------------------------------------------------------------------------------------
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 ₊ ₋

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)

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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 said:
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----------------------------------------------------------------------------------------------------------------------
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 ₊ ₋

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

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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 ₊ ₋

M

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.

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

M

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

thanks..

rst said:
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----------------------------------------------------------------------------------------------------------------------
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 ₊ ₋

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)

V

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

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!

V

rst said:
n! =n (n-1)!

n! /n =(n-1)!

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

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.

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NEW OBJECTIVE QUESTION

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Plz explain your answer

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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 said:
----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION

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Plz explain your answer

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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 ₊ ₋

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

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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 said:
----------------------------------------------------------------------------------------------------------------------
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 ₊ ₋

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)

----------------------------------------------------------------------------------------------------------------------
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 ₊ ₋

^ 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

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

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 ₊ ₋

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

----------------------------------------------------------------------------------------------------------------------
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 ₊ ₋

^
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

Niilesh said:
^
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

Nice method

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

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 .

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

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

If g is inverse function of f

Then f(x)=y and g(y)=x Not valid always

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

or f(g(x))=x 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

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

rst said:
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

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

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

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

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

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)

rst said:
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)

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

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]

----------------------------------------------------------------------------------------------------------------------
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 ₊ ₋

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)

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]

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

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

^ 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

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?

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

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

@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..?

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.

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?

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?

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

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)

Niilesh said:
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)

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

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.......

----------------------------------------------------------------------------------------------------------------------
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 ₊ ₋

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?

rst said:
----------------------------------------------------------------------------------------------------------------------
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 ₊ ₋

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
=> 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 said:
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 2(only one solution)

How is it -1 ??

rst said:
How is it -1 ??

It's 1, typo

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
=>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 2(only one solution)

I mean only solution is not x=2
as at x=2, 2x²-5x+3=1>0 which contradicts case (2)

rst said:
So only one solution is 1 or 2

I mean only solution is not x=2
as at x=2, 2x²-5x+3=1>0 which contradicts case (2)

wth happened to me
another typo, its 1 only

----------------------------------------------------------------------------------------------------------------------
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 ₊ ₋

rst said:
----------------------------------------------------------------------------------------------------------------------
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 ₊ ₋

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.....

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)
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.....

f '(x) represent ordinary diff.

But you used it for partial diff.

T

rst said:
f '(x) represent ordinary diff.

But you used it for partial diff.

No, he did it correctly AFAIK.

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 said:
----------------------------------------------------------------------------------------------------------------------
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 ₊ ₋

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)

----------------------------------------------------------------------------------------------------------------------
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 ₊ ₋

V

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

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"

V

^^ 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. :/

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.,

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)

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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 ₊ ₋

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)

Niilesh said:
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)

Absolutely right

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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 said:
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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 ₊ ₋

120 ants?

rohanz said:
120 ants?

May be possible.
I don't know the answer.
Can you explain your answer ??

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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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 ₊ ₋

^ 17.5
I think this expression would be enough for explanation -> (360/12) * 35/60 = 17.5

rst said:
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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 ₊ ₋

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

rst said:
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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 ₊ ₋

Ans is 50 mins. (<1 hr)
explanation

no by pythagoras theorem 3 rd side is 50 meters so it means 50 min..

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.)

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

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

Ans is correct
What are the assumptions ?

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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 said:
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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
₇ ₈ ₉ ₊ ₋

Ans : 20

1 term for each of x1.. x20

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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 said:
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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 ₊ ₋

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

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

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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 ₊ ₋

^ and is the third option(find ratio of areas)

Niilesh said:
^ and is the third option(find ratio of areas)

correct

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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 ₊ ₋

^ 9 m i will let someone else explain it as your questions are getting easy

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 ?

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'

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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 ₊ ₋

^^154

Chaitanya said:
^^154

Yeah,its ans is 154

But I also need explanation

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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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 ₊ ₋

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..

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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 ₊ ₋

*i.imgur.com/aZV9YSD.png

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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 ₊ ₋

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

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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 said:
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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 ₊ ₋

Domain of sin⁻¹x is [-1,1]

So -1≤log₂(x/2) ≤ 1

=> 2⁻¹≤(x/2) ≤ 2¹

=> 1≤ x ≤ 4

Couldn't solve this one -
∫ (tan(2x)/√(cos[SUP]6[/SUP]x+sin[SUP]6[/SUP]x))dx

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

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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 ₊ ₋

^^by mod you mean remainder after division ri8??

Chaitanya said:
^^by mod you mean remainder after division ri8??

Yeah, you got it right

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...

rst said:
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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 ₊ ₋

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

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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 said:
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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 ₊ ₋

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)

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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 said:
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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 ₊ ₋

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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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 ₊ ₋

^ 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...

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

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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 ₊ ₋

It's easy π² ≈ g ≈ 9.8
[π²] = 9
[-π²] = -10
so ans will be (2)

Niilesh said:
It's easy π² ≈ g ≈ 9.8
[π²] = 9
[-π²] = -10
so ans will be (2)

Correct

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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 said:
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----------------------------------------------------------------------------------------------------------------------
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 ₊ ₋

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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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 ₊ ₋

i am facing problems in learning vedic maths
is there any forum where i can discuss it ??

Post here
We will try to solve them

rst said:
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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 ₊ ₋

Put x=0,then we get
e=|0 -1 3|
|1 0 -4|
|-3 4 0|

e=0

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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 ₊ ₋

K

I luv dis section !!

rst said:
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----------------------------------------------------------------------------------------------------------------------
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 ₊ ₋

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

T

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

We will solve this question by BODMAS rule
So 48÷2*12 = 24*12=288

T

rst said:
We will solve this question by BODMAS rule
So 48÷2*12 = 24*12=288

& this 48*2÷12

theterminator said:
& this 48*2÷12

48*2÷12 = 48 *(1/6)=8

T

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 )

Did you first looked for ÷ sign in your keyboard? :-D

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

T

∫sec²x dx/(sec x + tan x )^9/2

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

theterminator said:
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

Since there is no bracket 48÷2*12 = 24*12 = 288 = 48*12÷2 = 12÷2*48

View attachment 12070
With AO || CD, find area of shaded region.

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

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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 ₊ ₋

A gift for all members

Math Help Forum

rst said:
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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 ₊ ₋

I am getting ans = -2
By using L'Hospital Rule then opening the Determinant and using limits respectively on each summation term

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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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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 ₊ ₋

if f(x)=|x-2| + |x+1|-x then find value of f ' (-10)

Is it given x belong to which domain?

Superayush said:
Is it given x belong to which domain?

domain is not given

options are
(a) -3 (b) -2 (c) -1 (d) 0

A

(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:
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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 ₊ ₋

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.

^^ Thanks a lot avg

What will be remainder if (16 ! +1) is divided by 17 ?

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

what is inverse of y = 5 ^ logx

rst said:
what is inverse of y = 5 ^ logx

y=Antilog((log x)/(log 5))???

Or is it Y=5^x???

^^Asking or answering?

seamon said:
^^Asking or answering?

Answering but sorry, y=5^x is wrong... :-(

You could do???

I could only do till log x Base 10= log y Base 5
:-(

Is the answer y=10^(log x base 5)???

powerhoney said:
Answering but sorry, y=5^x is wrong...
Could you solve???

.....

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

rst said:
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

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 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...

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

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
(C)16 ( a^2 +b^2) = - c^2
(D)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.

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)

[MENTION=158560]rst[/MENTION]

You got PM...

powerhoney said:
[MENTION=158560]rst[/MENTION]

You got PM...

Yeah, I got it

rst said:
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)

a or b is not equal to zero. Clear?

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
(C)16 ( a^2 +b^2) = - c^2
(D)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

rst said:
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

That's one way to do that. And am really bad with all these formulas.

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

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)

powerhoney said:
D. 18/(Square Root of 5)

Answer is correct

Can you explain it ?

rst said:
Answer is correct

Can you explain it ?

Solved using formula...

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

Hey, how to find limit of ((2^x-1)/x)^2 as x tends to zero???

Btw, your total posts have reached 666!!! ;-)

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

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

Find the total number of squares and rectangles in an NxN chessboard

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

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

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

Good... There were no new questions being posted so I posted that one...

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

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

What is the least value of 3sin^2 x + 4 cos^2 x
A)1
B)2
C)3
D)4

^3 for x=pi/2

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

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

^thats easy..

its D. c<0

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)]

whats the soln??

they meet at x= - ∞, but at -∞ it becomes 0..

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)

Neo said:
A. Simply find tan(A-B)

answer is correct

can you explain a little bit

rst said:
answer is correct

can you explain a little bit

+1.

Im a lil confused

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?

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??

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.

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??

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

[MENTION=158560]rst[/MENTION] - ya thanks... got perfect explanation on stackexchange.

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

*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 ^^

By diagram it is clear

But is there any method to prove it

May be this will help...
BTW was too lazy type :-D

*tapatalk.imageshack.com/v2/14/11/14/a5bc45d18bf2793a838881c84ec8d5c9.jpg

Great

Thanks a lot

Solve for x

2^(x+2)- 6^x -2 [3^(2x +2)]=0

so just to be clear the question is

{2^(x+2)} - {6^x} - 2{3^(2x+2)} = 0

^^ yes question is like that

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.

^ not that simple.
I couldn't find any satisfactory solution yet..

Here you go

*tapatalk.imageshack.com/v2/15/01/31/a36923181c8cf943b6832d00a29aea23.jpg

Buddy on LHS of the equation it will be "log(2^(x+2)-6^x)" & not "log(2^(x+2)) - log(6^x)"

Oops. Damn. My mathematics is getting rusty sorry

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

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

Option C:8

Correct.

In that case we are not getting positive integral value of p,q

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

Ans is C) 8 cm

*imageshack.com/i/idJjorfrj

Ans is B

*tapatalk.imageshack.com/v2/15/02/12/dab353aafa1debeb7511e416c9665413.jpg

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

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

*imageshack.com/i/excBhjhYj

rst said:
*imageshack.com/i/excBhjhYj

B. 4.5 cmxcm

Yes answer is correct

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

*imageshack.com/i/ip9OIBd7j

rst said:
*imageshack.com/i/ip9OIBd7j

350 cmxcm

Ans is correct

*imageshack.com/i/p7bNY00cj

rst said:
*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

Ans is correct

Remainder is 18

*imageshack.com/i/ex4gzsY2j

Ans is option (c)

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 )

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

*imageshack.com/i/pcWymDxgj

Ans is A) y

*imageshack.com/i/p58FCIDAj

- - - Updated - - -

Option (B) is Cosα cosβ cosγ

First expand all cos term . Then write the matrix in product form A*B . Then solve it normally .

Ans is option (D) 0

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

Ans is 9:10

rst said:
Ans is 9:10

care to explain?

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

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

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..

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.

That problem was the end of this thread, wasnt it