Official Quant thread for CAT 2013

MBA CAT 2024
14583
@sbharadwaj said:
_/\_ Bhai, can you explain chinese remiander theorem.? N mod k.,we segregate k into 2 coprimes a and b., then.?
bhai see 3rd post http://www.pagalguy.com/forums/quantitative-ability-and-di/official-quant-thread-c-t-67094/p-2761065?page=212
14584
A 51-digit number N consists of fifty x's and one y, where x and y are both digits from among 0 to 9.
1. If x = 2, N is divisible by 17 and y is the 17th digit from the right, y =
(1) 1 (2) 2 (3) 3 (4) 0
14585
@TootaHuaDil said:
A 51-digit number N consists of fifty x's and one y, where x and y are both digits from among 0 to 9. 1. If x = 2, N is divisible by 17 and y is the 17th digit from the right, y = (1) 1 (2) 2 (3) 3 (4) 0
10^8 mod 17=-1

let xxxxxxxx=A
xxx-A+A-A+xxxxxxxy-A+A mod 17
xxx+xxxxxxxy-A mod 17 should be zero

A mod 17=9

1+2222222*10+y -9 mod 17=0
1-10+y-9 mod 17=0
y-18 mod 17=0
y-1 mod 17 =0

y=1

14586
@TootaHuaDil
@TootaHuaDil said:
Final concentration = initial concentration ( 1- voulme replaced / initial volume )^n25 / ( 231 + 25 ) = ( 2 / ( 2 + 3 ) ) ( 1 - 15 / x )^3x = 40
sorry bhai, books says answer 25 ltrs
14587
@TootaHuaDil said:
A 51-digit number N consists of fifty x's and one y, where x and y are both digits from among 0 to 9. 1. If x = 2, N is divisible by 17 and y is the 17th digit from the right, y = (1) 1 (2) 2 (3) 3 (4) 0
10^8 + 1 =17k
so make groups of 8 and apply alternate sum procedure.
we'll be left with y+220 at the end.
this shoudl be divisible by 17.
so y=1
14588
@x2maverickc
plss explain it i didnt get it
14589
@TootaHuaDil said:
A 51-digit number N consists of fifty x's and one y, where x and y are both digits from among 0 to 9. 1. If x = 2, N is divisible by 17 and y is the 17th digit from the right, y = (1) 1 (2) 2 (3) 3 (4) 0
question is nothing but if 22x is divisible by 17, then find x .. x = 1
14590
@naga25french
how?
14591
@ishu1991 said:
@x2maverickcplss explain it i didnt get it
http://www.pagalguy.com/news/divide-and-conquer-3-simple-rules-find-out-divisibility-numbers-a-16684
14592

What is the min no. of people in a group of 5 people who have an identical no. of friends within the group, provided if A is friend of B, then B is also friend of A.

14593
@naga25french

_/\_ saar!
please show some light over post #14846
14594
@ishu1991 said:
@naga25frenchhow?
Euler number of 17 is 16 .. so number formed by writing a unit digit 16 times consecutively will be divisible by 16 ..

so 22222222 ( up to 32 times will be divisible by 17 and also 22222222 ( up to 16 times will be divisible by 17

1 - 32 --> digit divisible and 36 - 51 divisible by 17

so we are left with 22x or x22 or 2x2

Its now obvious .. Just check other posts for knowing what is euler number if you are aware of it .. hope this helps :mg:
14595
@x2maverickc said:
@naga25french_/\_ saar!please show some light over post #14846
Thats xat 2013 question 😛
14596
@sunnychopra89 said:
What is the min no. of people in a group of 5 people who have an identical no. of friends within the group, provided if A is friend of B, then B is also friend of A.
my take.
ans should be 2.
1 is not possible.

A-------B-------C
______/\
______/ \
_____D--E

A and C have one friend each.

This is when no one has zero friends.
14597
@x2maverickc said:
@naga25french_/\_ saar!please show some light over post #14846
how do we know the number of posts?? :)
14598
@sunnychopra89 said:
What is the min no. of people in a group of 5 people who have an identical no. of friends within the group, provided if A is friend of B, then B is also friend of A.
Is it 2??
14599
@sunnychopra89 said:
What is the min no. of people in a group of 5 people who have an identical no. of friends within the group, provided if A is friend of B, then B is also friend of A.
Is it 2??
A -> (B,C,D,E)
B -> (A,C,D)
C -> (D,A,B)
D -> (A,B,C,E)
E -> (A,D)
hence, min 2 ppl hv the same number of frnds...
14600
@TootaHuaDil said:
how do we know the number of posts??
Doesnt your post says reply #xyz under your profile name ;)
14601
@bs0409 said:
Is it 2??
Yes
14602

A no. has exactly 32 factors out of which 4 are not composite. Product of these 4 factors(not composite) is 30. How many such nos. are possible?