Official Quant thread for CAT 2013

MBA CAT 2024
15363
@pirateiim478 said:
If i understoof correctly only 3 distinct digits in 5 digit no means other 2 numbers must be repeated from these 3 distinct numbers.If 5 digit no doesnt start with 0 => 9*8*7*3c2If If 5 digit no starts with 0 =>8*8*7*3c2Total = 2856 numbers//
It cant start with 0
15364

@coonal2006 said:
can anyone please post detailed sol of this question -Q. Number of 5 digit numbers with only 3 distinct digits.Thanks
32,616 ?? what's the OA?
15365
@sundip said:
32,616 ?? what's the OA?
no options..
15366
@ravi.theja said:
with out having a zero in the selected 3 numbers... its 10C3 *[3c1* 5!/3! + 3C2*5!/2! * 2! ]am trying the cases including zero in the selected 3 digits..i ll edit it aftr i get...There are total of 200 cases if 0 is included ... so its 10C3 *[3c1* 5!/3! + 3C2*5!/2! * 2! + 200 ]@coonal2006 edited check it
in the first case if you are not including zero then how can it be 10C3??
15367
@coonal2006 said:
in the first case if you are not including zero then how can it be 10C3??
out of 10 digits..u select 3....which does not contain zero..the case is zero is not present in the 3 selected...
15368
@coonal2006 said:
no options..

I did like this:

Case1 - _ _ _ (_ _) lets consider the last two digits to be same, so last two digits - 10 ways & nxt digits in 9 ways, 8 ways, 7 ways respectively (if 0 is in middle two digits); - 10 ways & nxt digits in 9 ways, 8 ways, 6 ways respectively (if 0 is not in the middle two digits)

so case 1 - 10*9*8*7 + 10*9*8*6;

Case2 - _ _ (_ _) _ same as case1

Case3 - _ (_ _) _ _ same as case1

Case4 - (_ _) _ _ _ for the first two digits 9ways remaining, 9 ways, 8 ways , 7 ways resp.

total - 30*9*8*13 + 9*9*8*7 = 32,616 numbers.
15369
number shud be of form aaabc or aabbc
10c1*9*8*5!/3!*9/10 + 10c2*8*5!/2!2!*9/10
15370
@coonal2006 said:
can anyone please post detailed sol of this question -Q. Number of 5 digit numbers with only 3 distinct digits.Thanks
Is the answer 8280? Plz tag me in the answer...
15371

The mean of six positive integers is 15. The median is 18, and the only mode of the integers is less than 18. The maximum possible value of the six integers is:
A. 26
B. 28
C. 30
D. 32
E. 34
Please any body help me to solve with explanation..

15372
@Deb.csen said:
The mean of six positive integers is 15. The median is 18, and the only mode of the integers is less than 18. The maximum possible value of the six integers is:A. 26B. 28C. 30D. 32E. 34Please any body help me to solve with explanation..
that will be the maximum possible value of the largest of the six integers is: plz do..
15373
A car after traveling 18 km from a point A developed some problem in the engine and the speed became 4/ 5"' of its original speed. As a result, the car reached point b 45 minutes late if the engine had developed the same problem after travelling 30 km from a , then it would have reached b only 36 min late.the original speed of the car and the distance ab are ":"
a 25,130 b 30,150 c . 20,190 d none

15374
@Deb.csen said:
The mean of six positive integers is 15. The median is 18, and the only mode of the integers is less than 18. The maximum possible value of the six integers is:A. 26B. 28C. 30D. 32E. 34Please any body help me to solve with explanation..
32?edited..1,1,17,19,20,32
15375
@Deb.csen
32
15376

@stellarPG said:
32. This years XAT Q Integers will be 1,1,17,19,20,32
ya but how? plz xplain a bit..
15377
@Deb.csen said:
The mean of six positive integers is 15. The median is 18, and the only mode of the integers is less than 18. The maximum possible value of the six integers is:
A. 26
B. 28
C. 30
D. 32
E. 34
Please any body help me to solve with explanation..
abcdef
=> (c,d) = (17,19)
let mode = 1
=> (a,b) = (1,1)
(1+1+17+19+e+f)/6 = 15
e+f=52
but e>19
=> e=20 for max f.
hence f = 32....
15378
@IIM-A2013 ans c) 20,190
15379
@IIM-A2013 easier approach wil be when the problem develops at 18 and when it develops at 30kms respectively, effectively he is travelling at 4/5th original speed for 12kms(i.e 30-18=12) in the first case and in the second case he travells at the original speed for the 12kms and the subsequent time difference is 9mins
so the only equation is 12/(4s/5)-12/s=9/60.
solving this you get answer as C. 20,190
15380
@Deb.csen said:
The mean of six positive integers is 15. The median is 18, and the only mode of the integers is less than 18. The maximum possible value of the six integers is:A. 26B. 28C. 30D. 32E. 34Please any body help me to solve with explanation..
a+b+c+d+e+f=90
median =c+d/2=18..c+d=36....
a=b....as mode s
c+d=36...bt c nd d not equal to 18..as the mode ll b 18 in tht case....
a=b fr max valu of f..a=b mst be least positive intgr, a=b=1...

c+d=36..fr max valu f "f"....e mst b least.and near to d.....
c=17 nd d=19...
e=20
a=b=1

max valu f "f"=90-2-36-20=32..!!!

15381

Sara has just joined Facebook. She has 5 friends. Each of her five friends has twenty five friends. It is found that two of sara's friend are connected with each other. On her birthday, Sara decides to invite her friends and the friens of her friends. How many people did she invite for her birthday party?
A. >= 105
B. C. D. >= 100 and >=105 and plz solve this with explanation..

15382

Sara has just joined Facebook. She has 5 friends. Each of her five friends has twenty five friends. It is found that two of sara's friend are connected with each other. On her birthday, Sara decides to invite her friends and the friens of her friends. How many people did she invite for her birthday party?
A. >= 105
B. C. D. >= 100 and >=105 and plz solve this with explanation..