Three circles of equal radii overlap each other such that the centre of each circle lies on the circumference of the other two circles. Find the ratio of the area of the region commomto all three circles and the area of the region common to exactly two circles approx is -
0.516 0.382 0.448 0.298

No doubt that the answer would be 0.448.
when u draw thw figure than u can see that the area common to the three circles will be just less than the half of the area common to the three circles. Is will get enclosed in the line of symmetry of the second part. udi baba my 1000th post.......

So , first sqaure the first root and form the eqn , u still have a 1/2 power .
x^2-1 =2 .2^1/2
now with one more suaring this 1/2 power will also go ...
cheers :)
DHRUV..!!!

Why do we need to square the first root waala part twice? There might be a case that it may get cancelled by some other negative term in the second equation!!!

Say in the equation for the root 2^1/2 + 3^1/3, whih will be a 6 degree polynomial, thr is a possibility that one of the term in there just cancels out the term of 6^1/2 in the eqn for the first part. OR it can also be a posibility that the equation which u get for the 2nd part(a 6 degree polynomial), might have one of the roots as 2^1/2 + 3^1/2.

You will have to prove that this is not possible!!

If the 3rd term is 10^100, and the pattern is an increase in one zero per term in the power, then the answer is 4.

10 / 7 gives you a reminder of 3.

Hence 10s can be written as 3s while dividing by 7 for finding out reminders.

10^10 / 7 gives us 3^10 / 7 which gives a reminder of 4 or -3.

Thus 10^100 / 7 gives a reminder of (-3)^10 i.e 3^10 again i.e. 4 again.
All successive powers till 10^9 give the same reminder of 4.

Hence the final reminder is..

3 - 3 + 8*4 = 32 / 7 that gives a reminder of 4.

If the zeros are increasing by 1 everytime then the answer will be remainder of 3 + 9*10^4 divided by 7, i.e. 4.

I thought that the question states that the term is only for odd powers.

From fermats little theorem,
10^6=1mod7

so the question given to us reduced to
finding the remainder of 3+ 5*10^4

So the answer is 2.

After n days, if n=even
Number of men = n/2
boys= n/2-1

every man added does extra work of 1/400 per day. and every boy added does lesser work of 1/800 per day.

so after 18 days, the extra work done is
1/400*(1+3+...+17)-1/800(2+4+6...+16) = 81/400 - 36/400 = 45/400 which is just greater than 2 days of work done by 20 women, i.e. 40/400.

Now we will find the work done in 17 days,

for n=odd,
no. of men = (n-1)/2
no. of boys = (n-1)/2

so extra work done in 17 days is
1/400*(2+4+...+16) - 1/800*(1+3+...+15) = 36/200 - 64/800 = 10/100 which is less than the work done by 20 women in 3 days = 60/400 = 15/100

So on the 18th day,
1/20 of the work will be done by 9 men, 8 boys and 3 women in days = 9/200+8/800+3/400 = 25/400
So they will finish the work on 20/25 i.e. 4/5 th of the 18th day.

17.8 is the answer.

This article has left me dumbstruck! I dont know what to say!

The only thing I know and which i had said since the start is that I dont want to live in this country! and now I too have many sane people in my group like DG etc.

Some of you have questioned the outlash which we will face in other countries but I think that will be much better to accept than this humiliation. I regard these points as baselss and people say these just for the sake of adding one more point to their worhtless rationale.

I am repeating again, the quotas will increase, come what may! Whosoever party will come to power, it is gonna go for the vote bank and we cannot do anything!

The hunger strike etc. is not going to change anything, the magnitude of opposition to the quotas in nothing as compared to that in 1989. At that time, nothing happened, what can happen now???

very well put amardeep!

Hi,

A request to all of you, whosoever is solving the puzzles, plz try to note down the time you take to solve the puzzle and put it along with the answer. It will help us determine the difficulty level and the estimated number of marks attached to the questions of the puzzle.

For the 30th question , step change for odd values will occur at ,
a=1/2 , (1/2)^3 , (1/2)^5 .......

and , b = 1/5 , (1/5)^3 , (1/5)^5 .......


So there can be infinite number of rectangles formed with (a,b)

I am unable to find the product wise sum of various a*b , as they can take any pair varying in powers ..

Has any one done this ??

If u have reached here, then it is very simple, the answer is =
summation of all possible values of a*b
= 1/2(1/5+1/5^3+...) +1/2^3*(1/5+1/5^3+...) +.....
=(1/2 + 1/2^3+...)*(1/5+1/5^3+...)
=2/3*5/24
=5/36

but the given answer is 5/9

The Last Quit India movement is just beginning

I believe its time to go for the REAL FREEDOM movement too... I dont see any solution for the multi faceted problems that lie ahead for India other than creation of a separate state (like Maharashtra, Gujarat etc) transparent and free of the barriers of RELIGION, CASTE and LANGUAGE.

Jim Rogers one of the master investors of our current era predicted a while back that India will divide into multiple countries... me too

Other than all of this obviously a MASS EXODUS of the REALLY SMART people from India looks inevitable... God help us :)

K

what?? r U not gonna leave??

(20) Let X be the set of non-negative integers. Let functions f: X -> X such that
x f(y) + y f(x) = (x + y) f(x^2 + y^2) for all x, y. Then which among the following
is always true?
(a) f(x) >= 0 (b) f(x) = f(1) (c) f(0) = 0 (d) Atleast 2 of the foregoing

soln:

Put y =x , u get , f(x) = f(2x^2)
Put y =0 , u get , f(0) = f(x^2)

Hence periodicity for the function is x^2
Now , put y=1 ,
x f(1) + f(x) = (x + 1) f(x^2 + 1)
=>x(f(1) - f(x^2 +1)) = f(x^2 +1) - f(x)

clearly , since periodicity is x^2 , LHS =0

hence ,RHS = f(x^2 +1) - f(x) = f(1) - f(x) =0
=> f(1) = f(x)
We cant confirm on the fact that f(x) cant be negative ...

Hene , option (b)

9) the answer is 7:20, u can make the calculations easier by making a big cone of radius 3r and height 3h and then proceeding with the question. (just remember the volume will be proportional to r^2 * h )

11) I also found than the only possibility is that T must coincide with Q, but i ddnt attempt it as i was not confident.

for the 20the question, i ddnt get the explanation. can u plz explain it in detail. I got the answer right in the tesat but that was by putting in random values and all were coming as equal to f(1)