Official Quant thread for CAT 2013

MBA CAT 2024
10261
@tmohan02 said:
Remainder of (173*261)/13 + (248*249*250)/15 =???
4?

173 mod 13 = 4
261 mod 13 = 1
248 mod 15 = 8
249 mod 15 = 9
250 mod 15 = 10

Required answer = (4*1) mod 13 + (8*9*10) mod 15
= 4 mod 13 + 0 mod 15
= 4
10262

@anantn said:
Difficulty: very difficult q3: There are 7 different classes that are held in a day, to be held on 7 days of the week, and there 10 students in the class who attend(or not attend) any class in any way. How many different such student combinations are possible for the entire week.
guys, i dont have the OA for this one... any idea on how to got about it? im getting 2^490 as the answer....but...i may be misinterpreting the question i feel o.o
10263
@anantn said:
Difficulty: very difficult q3: There are 7 different classes that are held in a day, to be held on 7 days of the week, and there 10 students in the class who attend(or not attend) any class in any way. How many different such student combinations are possible for the entire week.
My take:
7 classes are held in a day :
(7!)^7
10 students in the class :
((2^10)^7)^7

So total = 7!^7 * 2^490
10264

But it says here the student can attend or not attend the classes, and not asking about arrangement of classes...thats the confusing part.

Here's what i interpreted:
taking students one at a time
1st dy: 7 classes student can attend 0-7 classes so 2^7. similar for all days so it ends up as
(2^7)^7=2^49.
This is for one student, same logic and be extrapolated for all 10 students so:
(2^49)^10= 2^490.

since its talking about the number of ways the students attend classes...is the arrangement of classes using 7! really necessary?

10265
How (51^203)/7 is converted to (2^203)7 ???
10266
@tmohan02 said:
How (51^203)/7 is converted to (2^203)7 ???
Euler of 7=6
Now (51^6)^33*51^5/7
Remainder (51^6)^33/7=1
Now remainder 51/7=2
So Remainder 2^5/7=4..
10267

@tmohan02 : 51= 7*2 +2
so (51)^203 = (7*2+2)^203
now do bionomial expansion, and all terms will have a multiple of 7 except the last term which is 7*2^0*(2^203) so it reduces to finding the remainder of 2^203/7
10268

er i meant 7^2 +2 not 7*2 +2

10269
@anantn said:
Difficulty: very difficult q3: There are 7 different classes that are held in a day, to be held on 7 days of the week, and there 10 students in the class who attend(or not attend) any class in any way. How many different such student combinations are possible for the entire week.
say,1 st guy can choose 1,2,3,4,5,6,7 days in week in
7C0 + 7C1+7C2 ++... = 2^7 ways
now he can attend 0 or more classes..(2^7) ways so,2^7.(2^7)
so for total 10 guys,,
(2^7.(2^7))^10 ways
10270
@sujamait : yeah thought so too 2^490 is what i got
10271
@anantn said:
But it says here the student can attend or not attend the classes, and not asking about arrangement of classes...thats the confusing part.Here's what i interpreted:taking students one at a time1st dy: 7 classes student can attend 0-7 classes so 2^7. similar for all days so it ends up as(2^7)^7=2^49.This is for one student, same logic and be extrapolated for all 10 students so:(2^49)^10= 2^490.since its talking about the number of ways the students attend classes...is the arrangement of classes using 7! really necessary?
Students A,B,C attending class C1 and D,E attending class C2 is different from A,B,C attending C2 and D,E attending C1.
That's how I interpreted the student combinations to be.
10272

difficulty: moderate
q4: The professor plays a game of "pass the baton" with his class of N students. He randomly selects one student, say A, and that student again randomly selects another student (say B)and passes him the baton, then B selects another student and passes him the baton and the game ends. What is the probability that the first student who got the baton from the professor ends up with the baton in the end.

1. 1/(n-1)
2. 1/2(n-1)
3. 1/n
4. none of these

10273
@grkkrg : I see. yeah i had assumed the class schedule to be fixed( since it hadnt mentioned that it was variable). Your approach makes logical sense.
10274
@anantn said:
difficulty: moderateq4: The professor plays a game of "pass the baton" with his class of N students. He randomly selects one student, say A, and that student again randomly selects another student (say B)and passes him the baton, then B selects another student and passes him the baton and the game ends. What is the probability that the first student who got the baton from the professor ends up with the baton in the end.1. 1/(n-1)2. 1/2(n-1)3. 1/n4. none of these
3. 1/n ?

(n-1)/n * 1/(n-1) + 1/n * 0 = 1/n

Not sure though. :|
10275

OA says 2.

total number of ways= n(n-1)(n-1)
favourable ways: nc2
hence nc2/(n(n-1)(n-1) = option 2

10276

The remainder of (54^124)/17 = ??

Plz explain.
10277
@tmohan02 answer: 4
Euler of 17= 16
so 124 = 16*7 +12
(54^112)*54^12= (54^112) gives remainder one according to euler
(54^12)mod 7= 3^12 mod 7= 3^(4*3) mod 7 = (-4)^3 mod 7 = -64 is remainder which is 17k-13 or 17k+4
hence remainder is 4
10278
@grkkrg said:
3. 1/n ?(n-1)/n * 1/(n-1) + 1/n * 0 = 1/nNot sure though.
i personally feel answer is 1 though. *dunno*
10279
@anantn said:
difficulty: moderateq4: The professor plays a game of "pass the baton" with his class of N students. He randomly selects one student, say A, and that student again randomly selects another student (say B)and passes him the baton, then B selects another student and passes him the baton and the game ends. What is the probability that the first student who got the baton from the professor ends up with the baton in the end.1. 1/(n-1)2. 1/2(n-1)3. 1/n4. none of these
Sample space:
Choosing 1 student from N students, C(N, 1) ways.
He can pass the baton to anyone of the remaining (N-1) students.
again, the second student can pass the baton to anyone of the (N-1) students.

So, total number of ways=C(N, 1)*(N-1)*(N-1)

Favorable cases: Select any 2 random students who can pass the baton to each other,
So, C(N, 2) cases.

=> probability=C(N, 2)/{C(N, 1)*(N-1)*(N-1)}=1/2*(N-1)
10280
@tmohan02 said:
The remainder of (54^124)/17 = ??Plz explain.


54^124 = 3^124 Mod 17 , Now Since HCF(3,17) = 1 We can apply Euler

3^8 = -1 Mod 17 , which gives (3^8)^15 * 3^4 = -1 * 13 = -13 or 4 Mod 17 ;)