Official Quant thread for CAT 2013

MBA CAT 2024
21646
@raopradeep said:
five balls of different colours are to be placed in three different boxes such that any box contains atleast one ball . find maximum number of different ways to do this90120150180
cases
113=5C3*2C1/2! * 3!=60
122 =5c1*4c2/2! *3!=5*3*6=90
150
21647
@raopradeep said:
five balls of different colours are to be placed in three different boxes such that any box contains atleast one ball . find maximum number of different ways to do this90120150180
150
21648
@iLoveTorres said:
bro consider the case of (1,2,2)so it is (5c1*3[3 ways of selecting a box])*(4c2*2[2 ways of selecting a box])*2c2*1=180.@scrabbler bhai wats the mistake i am doing?
(1,2,2)--->5!/(2!*2!*1!*2!) equal to 10
(1,1,3)--->5!/(1!*1!*3!*2!) equal to 15

sum 25

now 3 different boxes so 25*3! equal to 150
21649
@joyjitpal said:
(1,2,2)--->5!/(2!*2!*1!*2!) equal to 10 (1,1,3)--->5!/(1!*1!*3!*2!)equal to 15sum 25now 3 different boxes so 25*3! equal to 150
bhai bold kiya hua part samjha do.. pls
21650
@raopradeep said:
how many different sums can be formed with the following coins5rs ,1 rs ,50ps, 25ps, 10ps, 3ps ,2ps ,1ps
2^8-1-9c2 (atleast two is taken)
256-1-36=219?
21651
@iLoveTorres said:
bhai bold kiya hua part samjha do.. pls
refer to this link

samajh me a jayega

http://totalgadha.com/mod/forum/discuss.php?d=1051
21652
@Subhashdec2 said:
2^8-1-9c2 (atleast two is taken)256-1-36=219?
why atleast two taken

21653
If p > 4 is a prime number, then 24 divides p2 - 1 without remainder
(A) never
(B) sometimes only
(C) always
(D) only if p = 5
(E) none of these
21654

how is BYJU's classes 4 cat preparation ?

21655
@iLoveTorres said:
If p > 4 is a prime number, then 24 divides p2 - 1 without remainder (A) never (B) sometimes only (C) always (D) only if p = 5 (E) none of these
always
21656
@iLoveTorres always
21657
@prayas20 said:
how is BYJU's classes 4 cat preparation ?
dude this is not the place ............ i think
21658
@iLoveTorres said:
If p > 4 is a prime number, then 24 divides p2 - 1 without remainder (A) never (B) sometimes only (C) always (D) only if p = 5 (E) none of these
(p+1)(p-1)
cleary it will be divisble by 8
since p+1 or p-1 will be a multiple of 4 and other one will be a multiple of 2

since generally p is of the form 6k+1 or 6k-1

(6k+2)6k or (6k)(6k-2)

hence it will also be a factor of 3

hence always
21659
@iLoveTorres said:
If p > 4 is a prime number, then 24 divides p2 - 1 without remainder (A) never (B) sometimes only (C) always (D) only if p = 5 (E) none of these
c ) always checked by putting values

21660
@iLoveTorres said:
If p > 4 is a prime number, then 24 divides p2 - 1 without remainder (A) never (B) sometimes only (C) always (D) only if p = 5 (E) none of these
(C) always
21661
@prayas20 said:
how is BYJU's classes 4 cat preparation ?
good

but wrong thread to discuss

refer here http://www.pagalguy.com/posts/4730813
21662
@raopradeep said:
why atleast two taken
i dont know i just thought the question implied that

since we are talking abt the sums

anyways what is the OA?
21663

a box contains five sets of balls while there are three balls in each set . each set of balls has one colour which is different from every other set . what is the least number of balls need to be removed from the box so as to assure that a pair of same coloured balls has been removed

6
7
9
11

21664
@Subhashdec2 2^8 -1 -1 ( beacuse this includes every case ) and other -1 is because (1 ,2 ) can generate 3 i.e.
2^8 -1 -1 =254
21665
@raopradeep said:
a box contains five sets of balls while there are three balls in each set . each set of balls has one colour which is different from every other set . what is the least number of balls need to be removed from the box so as to assure that a pair of same coloured balls has been removed67911
7??