Official Quant thread for CAT 2013

MBA CAT 2024
12081
http://en.wikipedia.org/wiki/Barometer_question
12082
@joethaliath said:
@shadowwarrior why do u have to throw the stone when you already have the barometer ? drop the barometer and follow the same procedure , my friend was asked the very same question for Mu Sigma interview
ohh, and i got placed in mu sigma by answering this and 4-5 others like this :p

There are 3 bulbs in a room, u can enter the room only once, and room is completely shut off (not visible from outside) . There are 3 switches outside the room, each 1 for each bulb. How would you find the correct combination if you are allowed to enter the room only once ? (sound logic , no begari :p)
12083
@krum said:
ohh, and i got placed in mu sigma by answering this and 4-5 others like this There are 3 bulbs in a room, u can enter the room only once, and room is completely shut off (not visible from outside) . There are 3 switches outside the room, each 1 for each bulb. How would you find the correct combination if you are allowed to enter the room only once ? (sound logic , no begari )
Switch 1 => on for a few minutes, then off
Switch 2 => on
Switch 3 => off

Bulb 1 => Off but hot
Bulb 2 => On
Bulb 3 => Off and cold :D
12084

Another favourite among recruiters:

Five pirates have 100 gold coins. they have to divide up the loot in order of seniority (suppose pirate 5 is most senior, pirate 1 is least senior), the most senior pirate proposes a distribution of the loot. They vote and if at least 50% accept the proposal, the loot is divided as proposed. Otherwise the most senior pirate is executed, and they start over again with the next senior pirate. What solution does the most senior pirate propose? Assume they are very intelligent and extremely greedy (and that they would prefer not to die).

(to be clear on what 50% means, 3 pirates must vote for the proposal when there are 5 for it to pass, 2 if there are 4, 2 if there are 3, etc. )

12085

You have 2 ropes, A and B. Each takes 60s to burn completely, how would u measure 90s?

45s?

u have match box and not allowed to bend the rope or do any other freaky stuff :p

12086
@krum said:
You have 2 ropes, A and B. Each takes 60s to burn completely, how would u measure 90s?45s?u have match box and not allowed to bend the rope or do any other freaky stuff
for 90 sec: Burn the first rope with one end. As soon as it finishes burn the other rope from both the ends. so total 60+30=90

for 45 sec: Burn the first one from both the end and the other one from one end. Rope with two ends burning finishes off in 30sec and as soon as it finishes burn the second rope from second end. so it will take 30 + 15= 45
12087
@grkkrg said:
Another favourite among recruiters:Five pirates have 100 gold coins. they have to divide up the loot in order of seniority (suppose pirate 5 is most senior, pirate 1 is least senior), the most senior pirate proposes a distribution of the loot. They vote and if at least 50% accept the proposal, the loot is divided as proposed. Otherwise the most senior pirate is executed, and they start over again with the next senior pirate. What solution does the most senior pirate propose? Assume they are very intelligent and extremely greedy (and that they would prefer not to die).(to be clear on what 50% means, 3 pirates must vote for the proposal when there are 5 for it to pass, 2 if there are 4, 2 if there are 3, etc. )
98, 0, 1, 0, 1?
12088
@grkkrg said:
Another favourite among recruiters:Five pirates have 100 gold coins. they have to divide up the loot in order of seniority (suppose pirate 5 is most senior, pirate 1 is least senior), the most senior pirate proposes a distribution of the loot. They vote and if at least 50% accept the proposal, the loot is divided as proposed. Otherwise the most senior pirate is executed, and they start over again with the next senior pirate. What solution does the most senior pirate propose? Assume they are very intelligent and extremely greedy (and that they would prefer not to die).(to be clear on what 50% means, 3 pirates must vote for the proposal when there are 5 for it to pass, 2 if there are 4, 2 if there are 3, etc. )
E and C are dead meat :p

so 98 0 1 0 1 ,

a bhaiya asked me this 1 4 years ago :D
12089
@grkkrg said:
Another favourite among recruiters:Five pirates have 100 gold coins. they have to divide up the loot in order of seniority (suppose pirate 5 is most senior, pirate 1 is least senior), the most senior pirate proposes a distribution of the loot. They vote and if at least 50% accept the proposal, the loot is divided as proposed. Otherwise the most senior pirate is executed, and they start over again with the next senior pirate. What solution does the most senior pirate propose? Assume they are very intelligent and extremely greedy (and that they would prefer not to die).(to be clear on what 50% means, 3 pirates must vote for the proposal when there are 5 for it to pass, 2 if there are 4, 2 if there are 3, etc. )
98 0 1 0 1?
12090
I can go to work on either the number 17 bus or the number 29 bus. Number 17 buses runs every 15 mins. Number 29 buses runs every 12 mins. The first number 17 bus leaves x mins after the first number 29 bus, where 0
12091
We play the coin tossing game in which if tosses match, I get both coins; if they differ, you get both. If you have m coins and I have n, what is the expected length of the game (i.e., the number of tosses until one of us is wiped out)?
12092
@joyjitpal said:
We play the coin tossing game in which if tosses match, I get both coins; if they differ, you get both. If you have m coins and I have n, what is the expected length of the game (i.e., the number of tosses until one of us is wiped out)?
If both continue to win/loss alternately, then no one will wipe out ?
12093
@joyjitpal said:
We play the coin tossing game in which if tosses match, I get both coins; if they differ, you get both. If you have m coins and I have n, what is the expected length of the game (i.e., the number of tosses until one of us is wiped out)?
infinity?
12094
Co. B hosted a party for 8 members of co A. In the party no member of A had interacted with more than 3 members of B. Out of all the members of B, 3 members-each interacted with 4 members of A nd remaining members-each inetracted with 2 members of co.A. The greatest possible no of members of COB in the party is:
9
10
11
12
none
@krum
12095
@sonamaries7 said:
Co. B hosted a party for 8 members of co A. In the party no member of A had interacted with more than 3 members of B. Out of all the members of B, 3 members-each interacted with 4 members of A nd remaining members-each inetracted with 2 members of co.A. The greatest possible no of members of COB in the party is:9101112none@krum
I am getting 9.
3*4 + x.2 = 8.3 = >x=6
So 6+3=9

Dont know if i am on the rite track
12096
@sonamaries7 said:
Co. B hosted a party for 8 members of co A. In the party no member of A had interacted with more than 3 members of B. Out of all the members of B, 3 members-each interacted with 4 members of A nd remaining members-each inetracted with 2 members of co.A. The greatest possible no of members of COB in the party is:9101112none@krum
9?

3+4c2
12097
@catahead , @krum
Plz elaborate..mujhe ni samajh aaya
12098
@joyjitpal said:
I can go to work on either the number 17 bus or the number 29 bus. Number 17 buses runs every 15 mins. Number 29 buses runs every 12 mins. The first number 17 bus leaves x mins after the first number 29 bus, where 0
x=1
29 => 0 12 24 36 48 60 => 11+8+5+2+12 = 38
17 => 1 16 31 46 61 => 1+4+7+10+0 = 22

x=2
29 => 0 12 24 36 48 60 => 10+7+4+1+12 = 34
17 => 2 17 32 47 62 => 2+5+8+11+0 = 26

so 1/2(22/60+26/60) = 24/60 = 2/5 ?
12099
@sonamaries7 said:
@catahead , @krumPlz elaborate..mujhe ni samajh aaya
say

x1 x2 x3 x4 x5 x6 x7 x8 are members of A

now 3 members of B meet 4 of A

so take 4 common members so that remaining members can be maximized

so 4 members of A are left, and each remaining member of B meets 2 of them, so we need all the possible combinations of 4 members of A taking 2 at a time

==> 3+4c2=9
12100
P.Q.R=X.Y.Z=Q.A.Y
The 7 letters correspond to the 7 unique digits chosen from 0 to 9.
THe value of A is:
0
2
3
6
none