Official Quant thread for CAT 2013

MBA CAT 2024
6261
@LeoN88 said:
Yes, even I have checked the same on WA but 25! gives ______4000000, all the zeroes will get cancelled..4 will be left behind..I am a bit confused here !
Bhai according to remainder theorem, cancelled number has to be multiplied to reminder to get the final remainder...

eg...REM(18/16)=ans has to be 2, but in case we cancel 2 then REM(9/8)=1, so multiply 1 with 2 to get final rem.

6262
@pankaj1988 said:
Three squares are chosen at random on a chessboard. Find the probability that they lie in a diagonal line.
7/1488?
6263
@pankaj1988 said:
Three squares are chosen at random on a chessboard. Find the probability that they lie in a diagonal line.
Total number of cases =C(64, 3)
favorable cases= 2*[C(8, 3)+2{C(3, 3) + C(4, 3) + C(5, 3) + C(6, 3) + C(7, 3)}]

Answer=392/C(64, 3)=7/744
6264

Areeee sorry for the confusion, thanks guyz !

6265
@pankaj1988 said:
Three squares are chosen at random on a chessboard. Find the probability that they lie in a diagonal line.
4(3C3+4C3+5C3+6C3+7C3) + 2*8C3 / 64C3 ??
That is coming out to be 7/744 ??
6266
@vijay_chandola said:
Total number of cases =C(64, 3)favorable cases= 2*[C(8, 3)+2{C(3, 3) + C(4, 3) + C(5, 3) + C(6, 3) + C(7, 3)}]Answer=392/C(64, 3)=7/744
favorable cases explain kariyo
6267
@vijay_chandola said:
Total number of cases =C(64, 3)favorable cases= 2*[C(8, 3)+2{C(3, 3) + C(4, 3) + C(5, 3) + C(6, 3) + C(7, 3)}]Answer=392/C(64, 3)=7/744
@ScareCrow28 said:
4(3C3+4C3+5C3+6C3+7C3) + 2*8C3 / 64C3 ??That is coming out to be 7/744 ??
@19rsb said:
7/1488?
OA- 7/744..
@19rsb Bhai diagonals can be in any direction NE to SW,,,so u have to multiply by 4,not 2
6268
@Brooklyn said:
favorable cases explain kariyo
Bhai chk out soln in attachemnt
6269
@Brooklyn said:
favorable cases explain kariyo
3 squares are to be selected...
first take a 3x3 matrix on the chessboard.. favourable cases = 2*3C3
Take 4x4 matrix = 2*4C3
and so on...till the main diagonal of chessboard (leave it for now)
Then again 2*3C3 and 2*4C3 ...and so on till the main diagonal..
Then take diagonal .. 2 ways of selecting : 2*8C3
Hence total = 4(3C3+...7C3) + 2*8C3
6270
@Brooklyn said:
favorable cases explain kariyo
Diagonals dekho.

total number of diagonals=15

1st diagonals won't make any cases.
2nd diagonal won't also make any cases.

3rd diagonal= C(3, 3)=1 case
4th diagonal=C(C, 3)=4 cases
and so on,

for 8th diagonal=C(8, 3)

for 9th diagonal=C(7, 3)

so number of cases=C(8, 3)+2{C(3, 3) + C(4, 3) + C(5, 3) + C(6, 3) + C(7, 3)}

same is the case for the cross diagonals.

Hence, total number of cases=2*[C(8, 3)+2{C(3, 3) + C(4, 3) + C(5, 3) + C(6, 3) + C(7, 3)}]

6271
@pankaj1988 said:
OA- 7/744..@19rsb Bhai diagonals can be in any direction NE to SW,,,so u have to multiply by 4,not 2
oh yaa yaa yaa....
6272
@pankaj1988 said:
Bhai chk out soln in attachemnt
Mast question (mast because i could solve it!! :P)
6273
'A' has three share in a lottery in which there are 3 prizes and 6 blanks' 'B' has one share in a lottery in which there is 1 prize and 2 blanks. What is ratio of A's chance of winning a prizes to B's chance of winning a prizes.
6274

there are 2*9 diagonals from which we can select 3 squares, a1- h8 diagonal, a2-g8 similar to b1 h7,a3-f8 similar to c1-h6 ,a4-d8 ||| d1-h5, a5-d8 ||| to e1-h4,a6-c8||| to h3, We have to double it, since we have exact set of diagonals still remain due to symmety..
From main diagonals, we can choose in 2*8C3
similarly, 4*7C3, 4* 6c3 , 4*5C3, 4*4C3, 4*3C3 so,

2*8C3+4(3C3+4C3+5C3+6C3+7C3)
6275
@ScareCrow28 said:
A,B,C started to reach varanasi from delhi 300 km away they have only one bike which travel at a speed of 60 km/hr and every one can walk at a speed of 15 km/hr condition is every one has to reach at the same time and only two person can travel on the bike at the same time. wat is the minimum time required for them to reach there.P.S. All of them start at same time and onlyA can drive the bikeP.P.S.. : Question was posted yesterday..but i Couldn't reach to the answer matching with OAJAAGO GRAHAK JAAGO

---(5x/8)-----|--(3x/8)--|-----(3x/2)---|---x----
=>(5x/8) + (3x/8) + (3x/2) + x = 300
=> x = 600/7

hence time taken = (5x/2)/60 + x/15
=> 25/7 + 40/7 = 65/7

ATDH.
6276
@pankaj1988 said:
'A' has three share in a lottery in which there are 3 prizes and 6 blanks' 'B' has one share in a lottery in which there is 1 prize and 2 blanks. What is ratio of A's chance of winning a prizes to B's chance of winning a prizes

3:1??
6277
@pankaj1988 said:
'A' has three share in a lottery in which there are 3 prizes and 6 blanks' 'B' has one share in a lottery in which there is 1 prize and 2 blanks. What is ratio of A's chance of winning a prizes to B's chance of winning a prizes.
No of ways of A winning :
1 prize : 3C1 * 6C2
2 prizes: 3C2 * 6C1
3 prizes: 3C3 * 6C0
Total no of ways = 9C3
Prob = 3C1*6C2 + 3C2*6C1 + 3C3*6C0 / 9C3 = 16/21
No of ways of B winning:
1 Prize : 1C1 * 2C0
Total ways = 3C1
Prob = 1C1*2C0/3C1 = 1/3
Ratio = 16*3/21 = 16:7 ??
6278
@pankaj1988 said:
'A' has three share in a lottery in which there are 3 prizes and 6 blanks' 'B' has one share in a lottery in which there is 1 prize and 2 blanks. What is ratio of A's chance of winning a prizes to B's chance of winning a prizes.
16:7?

A's chance = 1 - 6C3/9C3 = 16/21
B's chance = 1 - 2/3 = 1/3

Ratio = (16/21)/(1/3) = 16/7
6279
@anytomdickandhary said:
is the OA around 9 hrs 11 mins 37 secs. Will post my approach if this looks correct. (Hope I haven't done any calc error.)ATDH.
Sir OA: 9 2/7 that should be 9 hours 17 minutes and 60/7 sec.. I think some error creeped in But ignoring it... You are absolutely correct 😃 After posting i tried again..and could solve it!! :)
Thanks a lot Sir!
6280
@ScareCrow28 said:
A,B,C started to reach varanasi from delhi 300 km away they have only one bike which travel at a speed of 60 km/hr and every one can walk at a speed of 15 km/hr condition is every one has to reach at the same time and only two person can travel on the bike at the same time. wat is the minimum time required for them to reach there.P.S. All of them start at same time and onlyA can drive the bikeP.P.S.. : Question was posted yesterday..but i Couldn't reach to the answer matching with OAJAAGO GRAHAK JAAGO
65/7 hours