| Message |
|
|
|
Two persons A and B agree to meet at a place between 11 to 12 noon .The first one
to arrive waits for 20 minutes and the leaves .If the time of their arrival be
independent and at random ,then what is the probability that A and B meet? 
a)1/3 b)2/3 c)4/9 d)5/9
|
|
|
|
answer is 3/5 
|
|
|
|
nadeem can u pls explain a bit more...i think ur solution is shortest but not getting it really
|
|
|
|
The altitude through A of triangle ABC meets BC at D(2,3) and the circumscribed circle at E(5,5).The ordinate of orthocentre being a natural number,what is the probability that orthocentre lie on the lines
y=1
y=2
y=3
.........
.........
y=10 ?
|
|
|
|
|
Two small squares on a chess board are chosen at random.What is the probability that they have a common side ?
|
|
|
|
3.nC0 - 8.nC1 + 13.nC2 - 18.nC3 + .......upto (n+1)
|
|
|
|
|
a particle moving in a circle with increasing velocity
|
|
|
|
velocity with respect to still water means that we(observer) are flowing with the water flow , its not necessary that velocity of water should be zero.
like for example if u are in a car moving at 5 km/hr and a man is moving opposite to car with velo 2km/hr then velocity of man as seen from still car is 7km/hr ......when we subtract the velocity vectors to get relative velocity than the object whose velocity is the second term is considered to be at rest.
for ex. Vman/car = Vman-Vcar here car is assumed to be at rest and man to be moving with velocity Vman/car
|
|
|
|
Is the answer n=50 and numbers are 7,8
if its correct i will post the solution
|
|
|
|
|
Full form of BF and GF is wrong !
|
|
|
|
|
|
We have R=abc/4$ (where $ is the area of the triangle )
$=abc/4R
$=K(sinAsinBsinC).....(i) [where K=k3/4R]
$=K(sinAsinBsin(A+B))
keeping B constant and diffrentiating for extremum value we get 2A+B=pi (or 0 which gets rejected)
keeping A constant and diff for extremum value we get 2B+A=pi
this implies A+B= 2(pi)/3.....(ii)
similarly using equation (i) we can get A+C=2(pi)/3....(iii) and B+C=2(pi)/3....(iv)
using (ii),(iii) and (iv) we can see that triangle is equilateral
|
|
|
|
