Official Quant thread for CAT 2013

MBA CAT 2024
15043
@Cat.Aspirant123 said:
when 3179^4241 divided by 4, d Remainder ?
3
15044
The minimum value of |x-1| + |2x-1| + |3x-1| ............ + |119x -1| ?
a) 48
b)49
c) 52
d)53
15045
@Cat.Aspirant123 said:
when 3179^4241 divided by 4, d Remainder ?
3?
3179^4241 mod 4= 3^4241 mod 4 = (-1)^4241 mod 4 = -1 mod 4 =3
15046
@Cat.Aspirant123 said:
when 3179^4241 divided by 4, d Remainder ?
3......4,3179 are co-primes.....therefore 3179^(2x) will give remainder 1 with 4...as 2=4(1-1/2) where 2 in the bracket represents the only prime factor of 4.


so....3179^4240 will give rem 1 with 4.....thus we only have to find the rem of 3179 with 4 which is 3
15047
@The_Loser |x-a| + |x-b| + |x-c| has minimum value at x = b ( median value)
so our ques
|x-1| + |2x-1| + |3x-1| .........|119x-1|
rewriting as
|x-1| + |x - 1/2| + |x- 1/2| + |x - 1/3| + |x - 1/3| + |x - 1/3| + ......| x - 1/119|
total terms = 119*120/2 = 7140
middle term = 3571 / 3570
so, median term will be n(n+1)/2 = 3570
n = 84
so it will be min at x = 1/84
|1/84 -1| + | 2/84-1| + ......| 84/84-1| + |85/84 - 1| + .......| 119/84 -1|
83/84 + 82/84 + ....0 + 1/84 + 2/84 ......35/84
=49
Courtesy :- PG quant thread
15048
@The_Loser
|x-1| + |x - 1/2| + |x- 1/2| + |x - 1/3| + |x - 1/3| + |x - 1/3| + ......| x - 1/119|
total terms = 119*120/2 = 7140
middle term = 3571 / 3570

so, median term will be n(n+1)/2 = 3570
n = 84
so it will be min at x = 1/84

|1/84 -1| + | 2/84-1| + ......| 84/84-1| + |85/84 - 1| + .......| 119/84 -1|

83/84 + 82/84 + ....0 + 1/84 + 2/84 ......35/84
=49
15049
@Cat.Aspirant123 said:
when 3179^4241 divided by 4, d Remainder ?
3179^4241/4 = 3^4241/4 cycle of 3/4 => rem=3, 3^2/4 => rem =1 for odd powers rem= 3 and for even power rem= 1
so ans = 3
15050
@ravi.theja said:
@The_Loser |x-a| + |x-b| + |x-c| has minimum value at x = b ( median value)so our ques|x-1| + |2x-1| + |3x-1| .........|119x-1|rewriting as|x-1| + |x - 1/2| + |x- 1/2| + |x - 1/3| + |x - 1/3| + |x - 1/3| + ......| x - 1/119|total terms = 119*120/2 = 7140middle term = 3571 / 3570so, median term will be n(n+1)/2 = 3570n = 84so it will be min at x = 1/84|1/84 -1| + | 2/84-1| + ......| 84/84-1| + |85/84 - 1| + .......| 119/84 -1|83/84 + 82/84 + ....0 + 1/84 + 2/84 ......35/84=49 Courtesy :- PG quant thread
that was one beautiful explanation....hats off to the puy who formulated this explanation.......
15051
@adityaknsit : that solution is by @anytomdickandhary sir 😃
15052
@Brooklyn said:
@adityaknsit : that solution is by @anytomdickandhary sir
not any tom dickk or harry for sure.....
15053
@adityaknsit said:
not any tom dickk or harry for sure.....
sorry for violating the thread 😃 bt he's an IIM A and IIT Grad 😃
15054
How many ways are there to split 14 people into seven pairs
15055

@Brooklyn said:
sorry for violating the thread bt he's an IIM A and IIT Grad
IIT city bhi btaa dete
15056
@The_Loser said:
IIT city bhi btaa dete
vo yaaad nhi sir ka 😛
15057
@The_Loser said:
How many ways are there to split 14 people into seven pairs
14c7 * (7!)^2
15058
@Brooklyn said:
14c7 * (7!)^2
sar ye kya kiya hai?
7 pairs bnaane hai?
15059
@The_Loser said:
sar ye kya kiya hai?7 pairs bnaane hai?
ohh divided people into 2 groups :P

14c2 * 12c2* 10c2* 8c2 *6c2 * 4c2* 2c2* 2^7 ??
15060
14 people need to be split into 7 pairs.
This can be done in 14C2*12C2*10C2*....2C2/7! ways
Here we divide by 7! to weed out the repetitions in the pairs
15061
@Brooklyn said:
ohh divided people into 2 groups 14c2 * 12c2* 10c2* 8c2 *6c2 * 4c2* 2c2* 2^7 ??
abhi bhi kuch loch hai. ye 2^7 ku liya aapne?
it's simple divide the pairs. nt d arrangement
15062
There are 6 students, and you want to form 3 groups of 2 each. How many groups are possible?