Official Quant thread for CAT 2013

MBA CAT 2024
15403
@Deb.csen said:
Ram, Shyam and Hari went out for a 100 km journey. Ram and Hari started the journey in Ram's car at the rate of 25 kmph, while shyam walked at 5 kmph. After sometime, Hari got off and started walking at the rate of 5kmph and Ram went back to pick up shyam. All three reached the destination simultaneously. The number of hours required for the trip was:A. 8B. 7C. 6D. 5E. 4Solve this with explanation...
8
15404
@gs4890 said:
8
How plz xplain..
15405
@Deb.csen said:
Ram, Shyam and Hari went out for a 100 km journey. Ram and Hari started the journey in Ram's car at the rate of 25 kmph, while shyam walked at 5 kmph. After sometime, Hari got off and started walking at the rate of 5kmph and Ram went back to pick up shyam. All three reached the destination simultaneously. The number of hours required for the trip was:A. 8B. 7C. 6D. 5E. 4Solve this with explanation...
the point at which ram turns back and goes to collect shyam is 75kms.. so shyam wil cover 15kms in the process the distance b/w ram n shyam is 60.. after two hours ram meets shyam at 25km point while hari is at 85km point.. so from der ram, hari will both require 3 hours at their respective speed. so the total time required will be 8 hours
15406
@Deb.csen said:
Ram, Shyam and Hari went out for a 100 km journey. Ram and Hari started the journey in Ram's car at the rate of 25 kmph, while shyam walked at 5 kmph. After sometime, Hari got off and started walking at the rate of 5kmph and Ram went back to pick up shyam. All three reached the destination simultaneously. The number of hours required for the trip was:A. 8B. 7C. 6D. 5E. 4Solve this with explanation...
starting point be A, end point be B. let the point where hari gets off be C, at a distance d from the A.
write all d equations and solve to get d = 75 km
on solving u'l get
total time taken = d/25 + 2d/75 + (100 - d/3)/25
putting d = 75 , time = 8 hrs
15407
@Estallar12 said:
=> 2*[mC2 - mC1*nC1] = 66=> mC2 - m*n = 33=> m*[(m-1)/2 - n] = 11*3m can only be 11 and so n = 2.1.) No. od Participants = 13.2.) Total Games Played = 2*13C2 = 13*12 = 156.
bhai waise to answer cannot be determined hona chahiye tha .. matlab m=33 v to ho sakta tha .. and n= 15 .. but chalo yahan aisa option nhi tha .. :)
15408
A fair coin is tossed 10 times. find the prob that two heads do not occur consecutively??
15409
@bs0409 said:
A fair coin is tossed 10 times. find the prob that two heads do not occur consecutively??
http://www.pagalguy.com/posts/4466158
15410
@bs0409

Total Sample Space=2^10= 1024..

Now, When the coin is tossed once, there are 2 outcomes, H n T..with 2 events suggesting no consecutive head..

When the coin is tossed twice, there are 4 outcomes, HH HT TH TT..with 3 events suggesting no consecutive head..

When the coin is tossed thrice, there are 8 outcomes, HHH HTH THH HHT TTH THT HTT TTT...with 5 events suggesting no consecutive head..

When, the coin is tossed four times, there are a total of 16 outcomes..with 8 events suggesting no consecutive head..

Here, a close inspection reveal that the caselets form a Fibonacci Series..

Thus,Summing it up = 2(2^1), 3(2^2), 5(2^3), 8(2^4), 13(2^5), 21(2^6), 34(2^7), 55(2^8), 89(2^9) n 144(2^10)

Thus, P(E)= 144/1024= 9/64..
15411
There are 10 pair of socks in a cupboard from which 4 individual socks are picked at random. the prob that there is at least one pair is?
15412
@bs0409 said:
There are 10 pair of socks in a cupboard from which 4 individual socks are picked at random. the prob that there is at least one pair is?
1- 16C4 / 20C4
15413
@bs0409 said:
There are 10 pair of socks in a cupboard from which 4 individual socks are picked at random. the prob that there is at least one pair is?
10c2+10c1*9c2/20c4
=45+360 /5*19*3*17
=405/5*19*51
=81/19*15
=27/19*17
=27/323
15414
@swapnil4ever2u said:
bhai waise to answer cannot be determined hona chahiye tha .. matlab m=33 v to ho sakta tha .. and n= 15 .. but chalo yahan aisa option nhi tha ..
Haan but then m + n = 33 + 15 = 48 ho jaata and that is not in options. :)
15415
@bs0409

Total Sample Space= 20C4 = 4845..

Now, Ways to select 4 pairs of socks = 10C4..

In each of these ways, there can be 2C1*2C1*2C1*2C1 ways to choose non-pair Individual socks..

Thus, Total ways = 10C4*2^4= 3360 Ways..

Thus, P(E) = 1 - P(where all non-pair socks r choosen)

=> 1 - 3360/4845= 99/323...
15416
@ravi.theja said:
1- 16C4 / 20C4
@techsurge said:
10c2+10c1*9c2/20c4=45+360 /5*19*3*17=405/5*19*51=81/19*15=27/19*17=27/323
@pyashraj said:
@bs0409Total Sample Space= 20C4 = 4845..Now, Ways to select 4 pairs of socks = 10C4..In each of these ways, there can be 2C1*2C1*2C1*2C1 ways to choose non-pair Individual socks..Thus, Total ways = 10C4*2^4= 3360 Ways..Thus, P(E) = 1 - P(where all non-pair socks r choosen)=> 1 - 3360/4845= 99/323...
OA=99/323
15417
@pyashraj said:
@bs0409Total Sample Space= 20C4 = 4845..Now, Ways to select 4 pairs of socks = 10C4..In each of these ways, there can be 2C1*2C1*2C1*2C1 ways to choose non-pair Individual socks..Thus, Total ways = 10C4*2^4= 3360 Ways..Thus, P(E) = 1 - P(where all non-pair socks r choosen)=> 1 - 3360/4845= 99/323...
Another way:

Choose any sock first up./ The chance that second will not match it is 18/19.

Given that, the chance that the third will not match either is 16/18 and similarly for the 4th one it is 14/17.

So Prob of no pair is 18/19 x 16/18 x 14/17 = 224/323 which means the prob of a pair is 1 - 224/323 = 99/323.

regards
scrabbler
15418
V anand and G kasparov play a series of 5 chess games. the prob that anand wins a game is 2/5 and kasparov is 3/5. there is no prob of draw . the series will be won by a person who wins 3 matches . find prob that Anand wins the series( The series ends the moment w/n any of the two wins 3 matches)
15419

Here is one question from my side..
How Many Two Digit Numbers Have Exactly Four Factors
a. 29
b. 30
c. 31
d. 32

15420
@bs0409 said:
V anand and G kasparov play a series of 5 chess games. the prob that anand wins a game is 2/5 and kasparov is 3/5. there is no prob of draw . the series will be won by a person who wins 3 matches . find prob that Anand wins the series( The series ends the moment w/n any of the two wins 3 matches)
Is it 1064/3125??
case1 - winning streak :D
www = 2^3/5^3 = 8/125
case2 - 1 loss
wwwl -> arrangement possible = 4!/3! = 4 but we also need to remove the case "www" hence 3 cases
=> 2/5*2/5*2/5*3/5*3 = 72/625
case3 - 2 loss and 3 win
arrangements possible = 5!/3!*2! = 10 but we need to remove the 3 cases counted above, hence total 7 cases
=> 2^3/5^3*3^2/5^2*7 = 504/3125

total prob = case1+case2+case3 = 1064/3125...
15421
@DespicableA said:
Here is one question from my side..How Many Two Digit Numbers Have Exactly Four Factorsa. 29b. 30c. 31d. 32
should be 29...
2*5 till 2*47 -> 13 numbers
3*5 till 3*31 -> 9 numbers
5*7 till 5*19 -> 5 numbers
7*11 till 7*13 -> 2 numbers
hence, total 29 numbers...

need to count the cubes as well
3^3 and 4^3 (i wish the world just ends) :p

so it makes 30..
15422
@Logrhythm said:
should be 29...2*5 till 2*47 -> 13 numbers3*5 till 3*31 -> 9 numbers5*7 till 5*19 -> 5 numbers7*11 till 7*13 -> 2 numbershence, total 29 numbers...
have u considered 8,27 ?