Official Quant thread for CAT 2013

MBA CAT 2024
26028
@bodhi_vriksha said:
Solve this one..If 13 people lose their wallets in a Cinema hall, and the guard searches for and returns the wallets at random, what is the chance that no one receives his / her own wallet?Team BV - Vineet
.367??? we have to use dearrangement na?? any other short method
26029
@bodhi_vriksha said:
Solve this one..If 13 people lose their wallets in a Cinema hall, and the guard searches for and returns the wallets at random, what is the chance that no one receives his / her own wallet?Team BV - Vineet
(1-1+1/2!-1/3!+1/4!-1/5!+...-1/13!)
26030
@Logrhythm said:
13!(1-1+1/2!-1/3!+1/4!-1/5!+...-1/13!)
Read the problem statement again :)

Team BV - Vineet
26031
@bodhi_vriksha said:
Solve this one..If 13 people lose their wallets in a Cinema hall, and the guard searches for and returns the wallets at random, what is the chance that no one receives his / her own wallet?Team BV - Vineet
13!(1-1+1/2!-1/3!+1/4!-1/5!+...-1/13!) = total no of deaarangement and divide by 13! to get the chance

Sir ji, Find the remainder when 35! is divided by (15!)(16!).
 logic ?
26032
@bodhi_vriksha said:
Solve this one..

If 13 people lose their wallets in a Cinema hall, and the guard searches for and returns the wallets at random, what is the chance that no one receives his / her own wallet?

Team BV - Vineet
Saar is it 1/2! - 1/3! +1/4! - .... -1/13! ?
26033
@amresh_maverick said:
13!(1-1+1/2!-1/3!+1/4!-1/5!+...-1/13!) = total no of deaarangement and divide by 13! to get the chanceSir ji, Find the remainder when 35! is divided by (15!)(16!). logic ?
Remainder is 0 because product of k consecutive natural numbers is always divisible by k! and here 35! = 15! * 16*17...*35= 15!*16!*k

Team BV - Vineet
26034
@bodhi_vriksha said:
Read the problem statement again Team BV - Vineet
ahh "chance"

need to divide it by 13! as well

editing
26035
@amresh_maverick said:

Sir ji, Find the remainder when 35! is divided by (15!)(16!). logic ?
15!*16! = 2^26*3^12*5^6*7^4*11^2*13^2
35! has every power greater than these for the primes, so it is divisible. hence 0
26036
@Budokai001 said:
Saar is it 1/2! - 1/3! +1/4! - .... -1/13! ?
Indeed :)

Interesting to note here is the fact that, as number of people becomes even larger the probability tends towards e^(-1), and so our value would be somewhat in the proximity of e^(-1).

This happens because e^-1=1-1/1!+1/2! - 1/3! +1/4! .... + (-1)^n/n! +...infinity

Team BV - Vineet
26037

Anoop managed to draw 7 circles of equal radii with their centres on the diagonal of a square such that the two extreme circles touch two sides of the square and each middle circle touches two circles on either side. Find the ratio of the radius of the circles to the side of the square.

26038
@bodhi_vriksha said:
Indeed Interesting to note here is the fact that, as number of people becomes even larger the probability tends towards e^(-1), and so our value would be somewhat in the proximity of e^(-1).This happens because e^-1=1-1/1!+1/2! - 1/3! +1/4! .... + (-1)^n/n! +...infinityTeam BV - Vineet
i would like to post a question on similar lines

There are 7 envelopes and 7 letters, in how many ways can these be arranged such that:

a) exactly 4 letters are are posted in the wrong envelopes (this one is easy)
b) not more than 3 letters are posted in the correct envelopes
c) exactly 6 letters are posted in the wrong envelope

PS - Self made question, hence no answer
26039
@Budokai001 said:
Anoop managed to draw 7 circles of equal radii with their centres on the diagonal of a square such that the two extreme circles touch two sides of the square and each middle circle touches two circles on either side. Find the ratio of the radius of the circles to the side of the square.
1/4*(1+rt18)
26040
@Dexian said:
1/4*(1+rt18)
No bro.. If you assume radius of circle=1,
Youl get 12+sqrt(2) + sqrt(2) . so its 1 : 2(6+sqrt(2))
26041
@Logrhythm said:
i would like to post a question on similar linesThere are 7 envelopes and 7 letters, in how many ways can these be arranged such that: a) exactly 4 letters are are posted in the wrong envelopes (this one is easy)b) not more than 3 letters are posted in the correct envelopes c) exactly 6 letters are posted in the wrong envelopePS - Self made question, hence no answer


a) exactly 4 letters are are posted in the wrong envelopes (this one is easy)

7C4 4! [1/2!-1/3!+1/4!]

b) not more than 3 letters are posted in the correct envelopes
0 correct + 1 correct + 2 correct + 3 correct
7! [1/2!-.......-1/7!] + 7C1* 6! [1/2!-.......+1/6!] +7C2 *5! [1/2!-.......-1/5!] +7C3*4! [1/2!-1/3!+1/4!]

c) exactly 6 letters are posted in the wrong envelope
7C1 * 6! [1/2!-.......+1/6!]
26042
@hatemonger said:
.367??? we have to use dearrangement na?? any other short method
This method is the shortest for this problem :)

Now try this one...

Logrhythm is dating 9 girls simultaneously in his hometown, Ranchi. On the week leading to the Valentines Day he was unfortunately stuck in Gurgaon, so he went to the Sahara mall and collected 9 gift items from the Archie's shop. He attached name tags of each girl friend to each gift and then wrapped each gift with a wrapper. However, in haste Logrhythm pasted address labels on the wrappers and forgot to double check them. In how many ways could he have pasted the address labels so that at least three of his girl friends would get to know that they were being two-timed.

Disclaimer: The problem is fictitious and has no resemblance to anyone living! :)

Team BV - Vineet
26043
@Budokai001 said:
Anoop managed to draw 7 circles of equal radii with their centres on the diagonal of a square such that the two extreme circles touch two sides of the square and each middle circle touches two circles on either side. Find the ratio of the radius of the circles to the side of the square.
let the radius of the circle be r
and side of the square be x
=> diagonal = x*rt(2)
also 12r+2r*rt(2)

=> x*rt(2) = 2r(6+rt(2))
multiply both sides by rt(2)
=> 2x = 2(6r*rt(2) + 2r)
=>x/r = 6*rt(2) + 2

hence, r/x = 1/(6*rt(2)+2)
26044
@bodhi_vriksha said:
Solve this one..If 13 people lose their wallets in a Cinema hall, and the guard searches for and returns the wallets at random, what is the chance that no one receives his / her own wallet?Team BV - Vineet
Approximately 1/e ~ 0.37.

regards
scrabbler
26045
@Logrhythm said:
i would like to post a question on similar lines

There are 7 envelopes and 7 letters, in how many ways can these be arranged such that:

a) exactly 4 letters are are posted in the wrong envelopes (this one is easy)
b) not more than 3 letters are posted in the correct envelopes
c) exactly 6 letters are posted in the wrong envelope

PS - Self made question, hence no answer
a) 3 Correct ; 4 Wrong
Number of ways of selecting these 3 is : 7C3
No of arrangements in which 4 can be wrong is : 4! *(1/2!-1/3! +1/4! )
So, 7C3 * 4! *(1/2!-1/3! +1/4! )
b) Correct
Correct=0 ; Wrong=7
Correct=1; Wrong=6
Correct=2; Wrong=5
Correct=3; Wrong=4
7! (1/2!-.......-1/7!) + 7C1* 6! (1/2!-.......+1/6!) +7C2 *5!(1/2!-.......-1/5!) +7C3*4! (1/2!-1/3!+1/4!)
c )1 Correct ; 6 Wrong
7* 6! *(1/2!-1/3! +1/4!-1/5!+1/6! )
26046
@amresh_maverick said:
Sir ji, Find the remainder when 35! is divided by (15!)(16!). logic ?
0? Note that 35! / (15!*20!) i.e. 35C15 is a natural number...so 35! is divisible by 156!*20! and hence also by 15!16!.

regards
scrabbler
26047
@Budokai001 said:
Anoop managed to draw 7 circles of equal radii with their centres on the diagonal of a square such that the two extreme circles touch two sides of the square and each middle circle touches two circles on either side. Find the ratio of the radius of the circles to the side of the square.
2r*root2 + 12r =s*root2 => r/s = root2 / (12+2*root2)

Team BV -Vineet