- Created by @Simba

- Tagged in :
- CAT

I think u have to modify this. Because the years 1700,1800,1900 are not leap year but 2000 is a leap year so for 400 years u will have 25+24*3 leap years.

so, the probabilty of some year being a leap year is 97/400. If i am wrong please forgive me and please tell me where i have gone wrong.

Oh, i forgot the same, but i will still go with te nitial answer because that seems relevant in the context of the question. The question is about some people meeting in a party in 21st century(if i can say that). As far as I can say(withh 100% probability), they must have been born after 1900 and they will surely die before 2000+1700 = 3700AD.

and hey buddy no need to write forgive and stuff. Feel free to post your answers/questions etc. Other junta will certaily appreciate good efforts and will surely correct any wrong answers etc.

The extent of the period does not allow us to go that far! But In CAT the period will most probably be specified/ the answers will be too far to consider us to go in that deep a concept, until and unless the question demands us to.

rajsher Says3 years out of evry four years have 365 days and i has 366 days. thats y 3/4 for the first term, and 1/4 for the second term.

I think u have to modify this. Because the years 1700,1800,1900 are not leap year but 2000 is a leap year so for 400 years u will have 25+24*3 leap years.

so, the probabilty of some year being a leap year is 97/400. If i am wrong please forgive me and please tell me where i have gone wrong.

Commenting on this post has been disabled.

School of Inspired Leadership - Admissions Open!

SOIL has been co-created by a consortium of 32 leading companies. Admissions Open for the 7th Batch!

500 + Alumni working in top positions across industry.

Average Package : 9.0 Lacs, Highest Package : 18.0 Lacs

hi,

Probabilty of 2 students in a class of 50 having the same birthday?

+ is the probability ofonly2 students having the same bday in a class of 50.

i am unable to follow how u got 3/4 and 1/4 rajsher ,

Can u pls explain ur approach ?

bye,

A.ash

3 years out of evry four years have 365 days and i has 366 days. thats y 3/4 for the first term, and 1/4 for the second term.

Commenting on this post has been disabled.

I found this link for the birthday problem

http://mathworld.wolfram.com/BirthdayProblem.html.

Hope this link is useful for all of us for the birthday problem

http://mathworld.wolfram.com/BirthdayProblem.html.

Hope this link is useful for all of us for the birthday problem

Commenting on this post has been disabled.

hi,

Probabilty of 2 students in a class of 50 having the same birthday?

+ is the probability of**only** 2 students having the same bday in a class of 50.

i am unable to follow how u got 3/4 and 1/4 rajsher ,

Can u pls explain ur approach ?

bye,

A.ash

Probabilty of 2 students in a class of 50 having the same birthday?

+ is the probability of

i am unable to follow how u got 3/4 and 1/4 rajsher ,

Can u pls explain ur approach ?

bye,

A.ash

Commenting on this post has been disabled.

vahgar SaysProbabilty of 2 students in a class of 50 having the same birthday?

can we have the solution and answer plz...

Commenting on this post has been disabled.

vahgar SaysProbabilty of 2 students in a class of 50 having the same birthday?

+ is the probability of

Commenting on this post has been disabled.

Probabilty of 2 students in a class of 50 having the same birthday?

Commenting on this post has been disabled.

Sorry for the easy one

How to solve this:

20. 20 (raise to 0.1). 50 (raise to 0.9)

Answer is 912.44

20^1.1 * 20 ^0.9 *(2.5)^0.9 = 400 * (2.5)^0.9 =400 * (1.5)^0.9* (1+.9*2/3 -.1 *.9 *4/1 =632 *(1.5)^.9 = 632 *(1+.45 - .01125+.00206 ) = 910 (approx)

to say the exact value ...answer options if provided can be worked out ....

My Problem solving page : https://www.facebook.com/pages/Bodhi-Vriksha/620231464668997?ref=hl ............................................................. My youtube channel: http://www.youtube.com/user/vineetnitd

Commenting on this post has been disabled.

Sorry for the easy one

How to solve this:

20. 20 (raise to 0.1). 50 (raise to 0.9)

Answer is 912.44

How to solve this:

20. 20 (raise to 0.1). 50 (raise to 0.9)

Answer is 912.44

Commenting on this post has been disabled.

vijay317 SaysWell 2a+(p+q-1)d = -2 I suppose..and therefore sum is -(p+q)

So sorry vijay,

I hope u will forgive me and I will not commite such sucidal mistakes.

Commenting on this post has been disabled.

vijay317 SaysSum of first p terms in a AP is q and sum of first q terms in the same AP is p. what is the sum of first p+q terms?

given-p/2=q ; q/2=p substracting the two eqns

2a(p-q)+(p-q)(p+q-1)d=2(q-p)=>2a+(p+q-1)d+2=0

multiply the eqn by (p+q)/2 we get S+(p+q)=0

=>S=-(p+q)

DISCLAIMER-answer could be wrong as i m still a novice.

Commenting on this post has been disabled.

Hello Vijay,

The solution is quite simple as follows :

Sum of first p terms : q=(p/2) and

Sum of first q terms : p=(q/2) where a and d have usual meanings.

Then on substracting the two equations u will get:

2a+(p+q-1)d=-1.

Hence the sum of p+q terms = -(p+q)/2.

Hence the answer.

Bye

Well 2a+(p+q-1)d = -2 I suppose..and therefore sum is -(p+q)

Commenting on this post has been disabled.

vijay317 SaysSum of first p terms in a AP is q and sum of first q terms in the same AP is p. what is the sum of first p+q terms?

Hello Vijay,

The solution is quite simple as follows :

Sum of first p terms : q=(p/2) and

Sum of first q terms : p=(q/2) where a and d have usual meanings.

Then on substracting the two equations u will get:

2a+(p+q-1)d=-1.

Hence the sum of p+q terms = -(p+q)/2.

Hence the answer.

Bye

Commenting on this post has been disabled.

Sum of first p terms in a AP is q and sum of first q terms in the same AP is p. what is the sum of first p+q terms?

Commenting on this post has been disabled.

When you follow a discussion, you receive notifications about new posts and comments. You can unfollow a discussion anytime, or turn off notifications for it.

103 people follow this discussion.
0Commentsnew comments