[MaskedMenace]

Originally Posted by chandan.mnnit

Is the question complete?

Coz if the car is travelling in a straight line from Alpha to Beta along Alpha Beta then the max possible distance would be 7....

correct me if wrong!!!

The question is ambiguous.

what i did was, (assuming that we travel in a straight line)If it is 3 km away from A and 4 km away from B then distance between A and B is 7. so from C the distance 7+4 is 11.

[chandan.mnnit]

Originally Posted by MaskedMenace

The question is ambiguous.

what i did was, (assuming that we travel in a straight line)If it is 3 km away from A and 4 km away from B then distance between A and B is 7. so from C the distance 7+4 is 11.

Its mentioned that the 3 points are equidistant

but yes the question is ambiguous

[akhilmisra]

Originally Posted by Toughluck2009

Ya,answer is 7.Please post your approach.

my approach is a bit weird....
I considered A,B,C as the vertices of eqilateral triangle....
since 3,4 form phythagorean triplet therefore the side AB forms the diameter of the triangle A,B, car.
now drop a perpendicular from C to AB (which would also be line bisector) which would be (5^2-2.5^2) ^(1/2) =4.3 approx. so the max distance form C will be perpendicular + radius(2.5) = 4.3 + 2.5 =7 approx

i m not sure someone please cross check and tell me if i m wrong...

[chandan.mnnit]

Originally Posted by akhilmisra

my approach is a bit weird....
I considered A,B,C as the vertices of eqilateral triangle....
since 3,4 form phythagorean triplet therefore the side AB forms the diameter of the triangle A,B, car.
now drop a perpendicular from C to AB (which would also be line bisector) which would be (5^2-2.5^2) ^(1/2) =4.3 approx. so the max distance form C will be perpendicular + radius(2.5) = 4.3 + 2.5 =7 approx

i m not sure someone please cross check and tell me if i m wrong...

First of all we dont know all the 3 distances to know for sure if its a Pythagorean triplet or not!!!

and even if we did then taking the point car as D....pythagorean triplet would give angle D = 90

this would give you the side AB as 5 miles...

if you will draw the diagram you will come to know that the the distance CD wont be equal to the distance calculated by you....

[MaskedMenace]

Originally Posted by chandan.mnnit

Its mentioned that the 3 points are equidistant

but yes the question is ambiguous

yes i noticed that they are equidistant.

only then the distance from C is 7 + 4 ( assuming that car remains stationary at that point), btw y did you assume that C travelled to B ??

[chandan.mnnit]

Originally Posted by MaskedMenace

yes i noticed that they are equidistant.

only then the distance from C is 7 + 4 ( assuming that car remains stationary at that point), btw y did you assume that C travelled to B ??

For just one reason -- no condition was mentioned in the question stating the direction

[dipsleo88]

Originally Posted by dipsleo88

how many different ways can 8 persons sit in 10 chairs in a row??
how many different ways can 8 persons sit in 4 chairs in a row,where more then one person can occupy a chair??

how many different ways can 5 letters be posted in 8 letter-box??

please xplain me each solution,by comparing these questions with each other....as i feel, they are solved with the same formulae,but the conditions are different.

cheers,
dips

yaar kkoi tho bata do plss....its really essential for me to understand them.. to proceed further..

[chandan.mnnit]

Originally Posted by dipsleo88

how many different ways can 8 persons sit in 10 chairs in a row??
how many different ways can 8 persons sit in 4 chairs in a row,where more then one person can occupy a chair??

how many different ways can 5 letters be posted in 8 letter-box??

please xplain me each solution,by comparing these questions with each other....as i feel, they are solved with the same formulae,but the conditions are different.

cheers,
dips

I am not gud in P&C althogh i will try

how many different ways can 8 persons sit in 10 chairs in a row??

-> 1st person can sit in any of the 10 given chairs so 10 ways
2nd can sit on any of the remaining 9 chairs so 9 ways
similarly for 3rd,4th...till 8th person for which 3 chairs remain.

so total number of ways = 10*9*8......*4*3

if you see formula wise its just arrangement of 8 people in 10 chairs.

how many different ways can 8 persons sit in 4 chairs in a row,where more then one person can occupy a chair??

-->1st person can sit on any of the 4 chairs so 4 ways
2nd can sit on any of the 4 chairs so 4 ways...similarly for remaining 6 persons also

so total number of ways = 4^8

how many different ways can 5 letters be posted in 8 letter-box??

1st letter can be posted in any of the 8 letterbox so 8 ways

2nd letter can be posted in any of the 8 box so again 8 ways
.
.
.
.similarly for other 3 letters also

hence the total number of ways = 8^5

Hope this helps...!!

[MaskedMenace]

Originally Posted by dipsleo88

yaar kkoi tho bata do plss....its really essential for me to understand them.. to proceed further..

how many different ways can 5 letters be posted in 8 letter-box??

Each letter can be dealt in 8 different ways so it is 8^5.

[akhilmisra]

if U,V,W and M are natural numbers such that U^M + V^M = W^M , then which of the following is true....
(a)m>= min(u,v,w)
(b)m>=max(u,v,w)
(c)m<min(u,v,w)
(d)N.O.T

pls explain the approach also....