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  • Discussion in CAT
  • Created by @Simba
> 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. ...
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.
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> rajsher Says > > 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. 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...
rajsher Says
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.

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.
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> 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 3 years out of evr...
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


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.
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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
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
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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
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
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> vahgar Says > > Probabilty of 2 students in a class of 50 having the same birthday? can we have the solution and answer plz...
vahgar Says
Probabilty of 2 students in a class of 50 having the same birthday?


can we have the solution and answer plz...
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> vahgar Says > > 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. ::
vahgar Says
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.
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Probabilty of 2 students in a class of 50 having the same birthday?
Probabilty of 2 students in a class of 50 having the same birthday?
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> 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 t...
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 ....
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Sorry for the easy one :: How to solve this: 20\. 20 (raise to 0.1). 50 (raise to 0.9) Answer is 912.44
Sorry for the easy one

How to solve this:

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

Answer is 912.44
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> vijay317 Says > > Well 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.
vijay317 Says
Well 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.
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> vijay317 Says > > 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? :: 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 ...
vijay317 Says
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?

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.
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> 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...
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)
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> vijay317 Says > > 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? :: 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 usu...
vijay317 Says
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?


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
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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? ::
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.

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