Official Quant thread for CAT 2013

MBA CAT 2024
7601

@maddy2807 said:
i did it like this...for first card u can choose and card out of 52... so the no of options are 52Now for the second card you are restricted to the same suit, that of the first card.Now u hv 12 cards of that suit which can be choosen in any order so 12!thats how 52*12!...i dont think so its correct. wats the OA?
you need to divide it by 13! as we need not to consider order here
7602
@sauravd2001 said:
sum of first n natural numbers whose sum contain three digits n all 3 digits are same...find the number n
n=36?
7603
@rubikmath said:
A tennis tournament has 8 players (A to H). There are 3 levels : Quarterfinals , semifinals and finals . Players are grouped randomly for the Quarterfinals . Every player has a probability of winning as 1/2 against any other player. What is the probability that A and B meet in a match during the tournament ?
474/770? What is OA?
7604
@Torque024 said:
474/770? What is OA?
nope ..
7605
@rubikmath said:
nope ..
6/11?, Another try. :/
7606
@Torque024 said:
6/11?, Another try. :/
7607
@Torque024 said:
6/11?, Another try. :/
isn't it way much higher ??
7608
@rubikmath said:
A tennis tournament has 8 players (A to H). There are 3 levels : Quarterfinals , semifinals and finals . Players are grouped randomly for the Quarterfinals . Every player has a probability of winning as 1/2 against any other player. What is the probability that A and B meet in a match during the tournament ?
Is it a knockout tournament ? Means that if A means against H , H is straightaway eliminated ?
So, in quarterfinals, there will be four groups of 2 teams, out of which one team from each group will be eliminated and rest go to semi-finals and so on ?
7609
@soham2208 said:
Is it a knockout tournament ? Means that if A means against H , H is straightaway eliminated ?So, in quarterfinals, there will be four groups of 2 teams, out of which one team from each group will be eliminated and rest go to semi-finals and so on ?
yup . in semi finals --> 4 players ; in finals --> 2
total --> 7 matchs .
7610
@rubikmath said:
A tennis tournament has 8 players (A to H). There are 3 levels : Quarterfinals , semifinals and finals . Players are grouped randomly for the Quarterfinals . Every player has a probability of winning as 1/2 against any other player. What is the probability that A and B meet in a match during the tournament ?
1/4?

Probability that A and B meet in the quarterfinals = 1/7
Probability that A and B meet in the semifinals = 6/7 * 1/2 * 1/2 * 1/3 = 1/14
Probability that A and B meet in the finals = 6/7 * 1/2 * 1/2 * 2/3 * 1/2 * 1/2 = 1/28

Total = 1/4
7611

@Torque024
@grkkrg
OA - 1/4
7612
@kingsleyx said:
Let a , b, c, d be positive integers. Then the product (a-b) (a-c) (a-d) (b-c) (b-d) (c-d) is definitely divisible by a)2 b)3 c)8 d)12.select all options which satisfy it !!
For any number we have three possibilities:-
-> it is of form 3k
-> it is of form 3k + 1
-> It is of form 3k + 2
So, among a, b, c, d we will always have atleast two numbers belonging to the same category (one of the above three)
So, one of the six terms (a-b), (a-c), (a-d), (b-c), (b-d), (c-d) will be divisible by 3

Also, for a, b, c, d following cases are possible
-> all of them even or all odd
-> Three even or three odd
-> Two even and two odd
In all cases atleast two of (a-b), (a-c), (a-d), (b-c), (b-d), (c-d) will be divisible by 2

So, always divisible by 2, 3, 4, 6, 12
7613
@rubikmath said:
You are playing Bridge and with a well shuffled pack of 52 cards,what is the probability that you are dealt a perfect hand(13 of one suit)? (In Bridge , there are 4 players and each of them is dealt 13 cards )
4c1/52c13??
7614
@rubikmath said:
Eight tennis players (call them A,B,C,D,E,G,F,H) are randomly assigned to start positions in a ladder tournament. Initially, position 1 plays position 2, position 3 plays 4, 5 plays 6 and 7 plays 8. Second round has 2 matches: winner of (1,2) match plays winner of (3,4), and winner (5,6) plays winner(7,8). The winners of the two 2nd round matches play each other in the final match. Player A wins against any of the others. Player B always beats any opponent except player A. What is the probability that player B wins the 2nd place trophy in the final match?
its given that A wins all of his matches..hence A is the winner
so remaining players are B to H=7
now consider the following groups:

[(A,2),(3,4)],[(5,6),(7,8)]

now out of these groups B cannot be in places 2,3 and 4.
or B can take places 5,6,7 or 8

hence favorable cases = 4

hence prob of B winning second place = 4/7

7615
@rubikmath said:
A tennis tournament has 8 players (A to H). There are 3 levels : Quarterfinals , semifinals and finals . Players are grouped randomly for the Quarterfinals . Every player has a probability of winning as 1/2 against any other player. What is the probability that A and B meet in a match during the tournament ?
I wonder whether this question can be done like the following method:

Matches played in QF:SF:F = 4:2:1
Since each person has a winning probability of 1/2
Hence by symmetry we can say that the probabilities that A meets B will also be in the same ratio ,i.e, 4:2:1 OR 1:1/2:1/4

P(A meets B in QF) = 1/7

Hence, P(A meets B in SF) = 1/14
And, P(A meets B in F) = 1/28

Hence Total Probability = 1/7 + 1/14 + 1/28 = 1/4

If anyone finds any flaw in this, Please do so! :)
7616
@ScareCrow28 said:
I wonder whether this question can be done like the following method:Matches played in QF:SF:F = 4:2:1Since each person has a winning probability of 1/2Hence by symmetry we can say that the probabilities that A meets B will also be in the same ratio ,i.e, 4:2:1 OR 1:1/2:1/4P(A meets B in QF) = 1/7Hence, P(A meets B in SF) = 1/14And, P(A meets B in F) = 1/28Hence Total Probability = 1/7 + 1/14 + 1/28 = 1/4If anyone finds any flaw in this, Please do so!
ha case ki sample space kaise define kari?? i.e.7 14 and 28?
7617
@nick_baba said:
ha case ki sample space kaise define kari?? i.e.7 14 and 28?
I don't get what you asked.

There is a 1/7 probability of A meeting B in QF. (A can meet B-H= 7 ways)
And then as written in post.. Ratio would be 1 : 1/2 : 1/4
Hence 1/7 * 1/2 = 1/14
And, 1/7 * 1/4 = 1/28
7618
A table of infinite expanse has inscribed on it a set of parallel lines spaced 2a units apart. A needle of length 2l (less than 2a) is tossed on the table. What is the probability that when it comes to rest , it crosses a line ?
OA : 2l/(pi*a)
no clue how they got it
any explanations ?
7619
@rubikmath said:
A table of infinite expanse has inscribed on it a set of parallel lines spaced 2a units apart. A needle of length 2l (less than 2a) is tossed on the table. What is the probability that when it comes to rest , it crosses a line ?OA : 2l/(pi*a)no clue how they got it any explanations ?
seems like integration will come into play, p will depend on the distance of center from line and the angle , do u have explanation?
7620
@rubikmath said:
A table of infinite expanse has inscribed on it a set of parallel lines spaced 2a units apart. A needle of length 2l (less than 2a) is tossed on the table. What is the probability that when it comes to rest , it crosses a line ?OA : 2l/(pi*a)no clue how they got it any explanations ?
I have explanations jo mere bas k bahar hai :P