The capacity of the vessel is 500 litre and its vertical height
is 150 cm. If 244 litre of water is put in the vessel, then what
is the height of the water level in the conical vessel above the
table?
please give complete solution...
A conical vessel is lying on a table with its base downwards.
The capacity of the vessel is 500 litre and its vertical height
is 150 cm. If 244 litre of water is put in the vessel, then what
is the height of the water level in the conical vessel above the
table?
The capacity of the vessel is 500 litre and its vertical height
is 150 cm. If 244 litre of water is put in the vessel, then what
is the height of the water level in the conical vessel above the
table?
@yudh its a simple problem, try concept of similar triangle...
lets say full volume is x, so volume of empty vessel would be x-244..and volume is a function of radius & height so assume r, R , h, H..try once if still u dont find the answr we will help :)
lets say full volume is x, so volume of empty vessel would be x-244..and volume is a function of radius & height so assume r, R , h, H..try once if still u dont find the answr we will help :)
A string of a certain length is cut at random at two points.Find the probabilty that the three pieces that result can be the three sides of a triangle.
@pakkapagal said:A string of a certain length is cut at random at two points.Find the probabilty that the three pieces that result can be the three sides of a triangle.
1/4??
we know that the sum of length of two sides of a triangle is greater than the third
if both the cuts are on the same half of the string than their sum can no way we greater than the third side, hence the triangle cannot be formed
the prob for both the cuts on same side = 1/2
also when the cuts are on two different halves of the stick but when the distance between them is more than or equal to half the length of the stick then also the triangle won't be possible, prob for this = 1/4
so, prob of triangle not being possible = 1/2+1/4 = 3/4
hence, prob of triangle being possible = 1 - 3/4 = 1/4
Vishal and Anu decide to meet at a cafe between 5 p.m and 6 p.m.They agree that the person who arrives first at the cafe would wait for exactly 15 minutes for the other.If each of them arrives at a random time between 5.00 pm and 6.00 pm ,what is the probabilty that the meeting takes place?
@Logrhythm said:1/4??also when the cuts are on two different halves of the stick but when the distance between them is more than or equal to half the length of the stick then also the triangle won't be possible, prob for this = 1/4
bhai bold part thoda aur explain kar do...1/4 kaise aaya?
@pakkapagal said:Vishal and Anu decide to meet at a cafe between 5 p.m and 6 p.m.They agree that the person who arrives first at the cafe would wait for exactly 15 minutes for the other.If each of them arrives at a random time between 5.00 pm and 6.00 pm ,what is the probabilty that the meeting takes place?
7/16?
@pakkapagal said:yes.Explain pls
See the figure.
Grey part is the area where meeting will takes place.
and white part is where no meeting takes place
As the duration is of 60 min. and one can wait for 15 min. So we are left with 45 min. in which no meeting will happen.
So the probabilty where meeting should happen will be
1- (proabilty of no meeting )= 1- (45*45/60*60) = 7/16
PS: see we assume 60 min event as a sqaure because sqaure has both the sides equal shows that every event has equal probabilty.
see the figure
I m in need of a really good explanation of the procedure to solve---In how many distinct ways can 2 identical coins be placed on a chess board such that the squares on the board on which there are positioned neither share a common side or a common vertex.
I couldn't find a very clear solution.
And yes the answer is 1806 .
I couldn't find a very clear solution.
And yes the answer is 1806 .
@ScareCrow28 said:Fingers -> a + b + c + d Rings-> r1, r2, r3, r4, r5 a+ b + c + d = 5Total 8C3*5! ways = 8!/3!.. ? ( 5! because r1r2 is diff from r2r1) ..
why a+b+c+d =5 ?? share approach
@pakkapagal said:Vishal and Anu decide to meet at a cafe between 5 p.m and 6 p.m.They agree that the person who arrives first at the cafe would wait for exactly 15 minutes for the other.If each of them arrives at a random time between 5.00 pm and 6.00 pm ,what is the probabilty that the meeting takes place?
@pakkapagal said:bhai bold part thoda aur explain kar do...1/4 kaise aaya?
take any random length of the string and calculate this....1/4 aa jayegi probability
@abhishek.2011 said:my take 100 mh/6 = y/24 y = 4hsimilarly h/6 = x/30x =5hnow y^2 + 300^2 =x^2300 = 3hh=100m
@Buck.up said:Is Height of Light house 100 m ? If yes then I will attach the solution
The answer should be 106 .. You guys are missing a small thing ... Identify it :)
And dont forget to post your figures as well :)
Team BV-- Pratik Gauri
@vishcat said:I m in need of a really good explanation of the procedure to solve---In how many distinct ways can 2 identical coins be placed on a chess board such that the squares on the board on which there are positioned neither share a common side or a common vertex.I couldn't find a very clear solution. And yes the answer is 1806 .
See any square which is not on the edge or corner of the board (36 in number) will have 8 squares which are touching it (atleast one common vertex)
Squares which are at corners (only 4 in number) will have 3 squares which are touching them
Squares which are at edges but not at the corners (24 in number) will have 5 squares which are touching them
So, 36*8 + 4*3 + 24*5 = 420 pairs of squares
But we have counted every pair twice, so actually there will be 210 pairs of squares such that they have atleast one common vertex
So, required answer will be:-
C(64, 2) - 210 = 1806
@pakkapagal said:bhai bold part thoda aur explain kar do...1/4 kaise aaya?
See let us assume to break the stick into 2 parts. Now if you put 2 cuts on same side of half stick ...you cant get a triangle because sum of 2 side > 3rd side.
so PROB(TRIANGLE NOT FORMED)=1/2
now even if 2 cuts are on differnt sides of halves of stick ,distance between these 2 should be less than 1/2(stick)..If this distance >1/2(stick)-->triangle not formed . so PROB(TRIANGLE NOT FORMED)=1/4 from this case
So total PROB(TRIANGLE NOT FORMED)=1/2+1/4=3/4
So PROB(TRIANGLE FORMED)=1/4 :)
Team BV-- Pratik Gauri
@pakkapagal said:Vishal and Anu decide to meet at a cafe between 5 p.m and 6 p.m.They agree that the person who arrives first at the cafe would wait for exactly 15 minutes for the other.If each of them arrives at a random time between 5.00 pm and 6.00 pm ,what is the probabilty that the meeting takes place?
GATE 2012 ECE 2 mark aptitude question , 7/16 option (c) 
@trip said:A Man standing ona boat south of a light house observes his shadow to be 24 m long as measured at the sea level. on sailing 300 m eastwards , he finds his shadow as 30 m long, measured in a similar manner. the ht of the man is 6 m abve sea leavel. what is the ht of light house?Puys dosto.....answer figure ka diagram batao
On drawing the figure ( Which is a 3D figure ), we have two equations
(30 + Root(x^2 + 300^2) )/30 = h/6 = (24 + x)/24 ( Triangles being similar )
OR (30 + Root(x^2 + 300^2) )/30 = (24 + x)/24
=> x^2/24^2 = (x^2 + 300^2)/30^2
From here, x = 400m
Hence, h = 6*(1 + 400/24) = 106 metres..
P.S. I have been trying to make a figure but my mouse doesn't work properly. This is what I could make with my screwed up mouse..
@CrookDinu said:why a+b+c+d =5 ?? share approach
a, b, c and d denote nothing but the 4 diff fingers. Since we have 5 rings to be placed in these 4 fingers, hence a + b + c + d = 5
No of ways of doing this is given by the formula (n+r-1)C(r-1)
Here, n =5 , r = 4
Hence total ways = 8C3
But, the rings can be picked up in 5! ways since we are assuming the rings are diff.
hence Ans would be: 8C3*5! ..
Solve