@kingsleyx said:If 3+5=19, 5+9=61 and 9+12=117 then 13+14 = ?
183 :s
A line of length L is divided (broken) into 3 parts randomly. Find the probability that these parts are the possible sides of a triangle !!
@kingsleyx said:A line of length L is divided (broken) into 3 parts randomly. Find the probability that these parts are the possible sides of a triangle !!
http://www.pagalguy.com/forums/quantitative-ability-and-di/official-quant-thread-cat-2011-part-2-t-67094/p-2761065/r-2812492
Great solution by @Angadbir..have bookmarked this link for further reference..:)
@kingsleyx said:A line of length L is divided (broken) into 3 parts randomly. Find the probability that these parts are the possible sides of a triangle !!
The probability that an interval broken at n-1 points chosen uniformly at random is
broken into pieces which can be rearranged to form an n-gon is :
So P =1- (n/2^(n-1))
here n=3
so P= 1-3/2^2=1/4
broken into pieces which can be rearranged to form an n-gon is :
So P =1- (n/2^(n-1))
here n=3
so P= 1-3/2^2=1/4
@kingsleyx said:A line of length L is divided (broken) into 3 parts randomly. Find the probability that these parts are the possible sides of a triangle !!
condition: a+b>c ,
let the sample space be 9cm;
(a,b,c) cant be (_ , _ ,5 ) ( _ , _ , 6 ) n till c=7.. considering integers only n atleast 3 parts..
so p(not triangle) = 3/7 ,
p(positiv)= 4/7.. ??
let the sample space be 9cm;
(a,b,c) cant be (_ , _ ,5 ) ( _ , _ , 6 ) n till c=7.. considering integers only n atleast 3 parts..
so p(not triangle) = 3/7 ,
p(positiv)= 4/7.. ??
@dragster said:condition: a+b>c ,let the sample space be 9cm;(a,b,c) cant be (_ , _ ,5 ) ( _ , _ , 6 ) n till c=7.. considering integers only n atleast 3 parts..so p(not triangle) = 3/7 , p(positiv)= 4/7.. ??
apart frm formula, wats the aproach in these kinds of questions.. ??
@sauravd2001 said:@dragster bhai @krum ki baat manne mein hi bhalai hoti hai
@dragster said:apart frm formula, wats the aproach in these kinds of questions.. ??
there is another way-
There are 2 cuts here , giving 3 parts , for them to be triangle the cuts need to be in 2 halves, if both are in the same half, sum of 2 sides won't be greater than third
so P(first cut in one of the halves) = 1/2
P(2nd cut in one of the other half)=1/2
now u can say that we can cut the line at center, but a line has infinite many points, so that case can be ignored
so 1/2*1/2=1/4
There are 2 cuts here , giving 3 parts , for them to be triangle the cuts need to be in 2 halves, if both are in the same half, sum of 2 sides won't be greater than third
so P(first cut in one of the halves) = 1/2
P(2nd cut in one of the other half)=1/2
now u can say that we can cut the line at center, but a line has infinite many points, so that case can be ignored
so 1/2*1/2=1/4
@sauravd2001 said:@dragster bhai @krum ki baat manne mein hi bhalai hoti hai
haha m not a follower.. 😛 but yeah.. respect to winners.. 
@krum said:there is another way-There are 2 cuts here , giving 3 parts , for them to be triangle the cuts need to be in 2 halves, if both are in the same half, sum of 2 sides won't be greater than thirdso P(first cut in one of the halves) = 1/2P(2nd cut in one of the other half)=1/2now u can say that we can cut the line at center, but a line has infinite many points, so that case can be ignoredso 1/2*1/2=1/4
gud1..
Q. C.I for 3rd year = 1260 ; CI for 4th year = 1512 , compounded annually.
principal??
(don remember the answr.. but cudnt solve.. need aproach.. )
@dragster said:Q. C.I for 3 years = 1260 ; CI for 4 years = 1512 , compounded annually.principal??(don remember the answr.. but cudnt solve.. need aproach.. )
729.16
@dragster said:aproach..?
form two eqn divide (1+r/100)=1512/1260=1.2
P*(1.2)^3=1260
p=729.16
feels like m back home sabko bhai ka ____/\____