Official Quant thread for CAT 2013

@sonamaries7 said:
Co. B hosted a party for 8 members of co A. In the party no member of A had interacted with more than 3 members of B. Out of all the members of B, 3 members-each interacted with 4 members of A nd remaining members-each inetracted with 2 members of co.A. The greatest possible no of members of COB in the party is:9101112none@krum
i will also go with 9...is it correct...?
@sonamaries7 said:
@catahead , @krumPlz elaborate..mujhe ni samajh aaya
Think you have 8 dots(Co A) on one side and 3+x dots(Co B) on other side
So no of line connecting one dot from each side is maximum 8.3=24
Also we have that 3 on the other can have max 4 i.e 12 and and the remaining x will have 2 so 2x
Hence 24=12+2x=>x=6
So 9

@krum said:
x=1 29 => 0 12 24 36 48 60 => 11+8+5+2+12 = 38 17 => 1 16 31 46 61 => 1+4+7+10+0 = 22x=229 => 0 12 24 36 48 60 => 10+7+4+1+12 = 3417 => 2 17 32 47 62 => 2+5+8+11+0 = 26so 1/2(22/60+26/60) = 24/60 = 2/5 ?
krum bhai thoda explain karna
kuch samajh me nahi aya :(
@sonamaries7 said:
P.Q.R=X.Y.Z=Q.A.YThe 7 letters correspond to the 7 unique digits chosen from 0 to 9.THe value of A is:0236none
3*4*6 = 1*9*8 = 4*2*9

So, A = 2
@sonamaries7 said:
P.Q.R=X.Y.Z=Q.A.YThe 7 letters correspond to the 7 unique digits chosen from 0 to 9.THe value of A is:0236none


1 2 3 4 6 8 9 are used

2 3 2*2 2*3 2*2*2 3*3

8*9*1 = 6*4*3 = 9*2*4

so 2?

@joyjitpal said:
krum bhai thoda explain karna kuch samajh me nahi aya
answer kya hai? i have some doubts
@sonamaries7 said:
P.Q.R=X.Y.Z=Q.A.YThe 7 letters correspond to the 7 unique digits chosen from 0 to 9.THe value of A is:0236none
Clearly 0 is out
Also prime numbers 7 and 5 cannot be used for obvious reasons
Now we have 1,2,3,4,6,8,9

If 9 is present in 1 product then both 3 and 6 shud be present in the other
So let the values be
9 _ _
3 6 _
_ _ _

Now put 1 with the largest i.e 9 and with little iteration we get,
9 1 8 = 72
3 6 4 = 72
9 2 4 = 72

Hence 2
PS: Dont start iterations from the very begining or will waste much time.
@krum , @vijay_chandola , @maneeshp :
There is a fam with father , mother and 6 children A thru F. They have to attend a party for which only one among the mother/father can attend along with any no of kids subject to the foll constraints:
1) B or C won't go along with their mother
2) F goes only if A or D goes
In how many ways can the fam attend the party?
71
67
79
70
100

plz share the approach also..
@sonamaries7 said:
There is a fam with father , mother and 6 children A thru F. They have to attend a party for which only one among the mother/father can attend along with any no of kids subject to the foll constraints:1) B or C won't go along with their mother2) F goes only if A or D goesIn how many ways can the fam attend the party?71677970100plz share the approach also..
when Father goes
when A goes, not D - 2^4
when D goes, not A - 2^4
both A and D go - 2^4
when both don't go - 2^3

when Mother goes
when A goes, not D - 2^2
when D goes, not A - 2^2
both A and D go - 2^2

when both don't go - 2

so total = 48+12+8+2=70 ?
@sonamaries7 said:
There is a fam with father , mother and 6 children A thru F. They have to attend a party for which only one among the mother/father can attend along with any no of kids subject to the foll constraints:1) B or C won't go along with their mother2) F goes only if A or D goesIn how many ways can the fam attend the party?71677970100plz share the approach also..
i am getting 70
@krum : Similar to the qs on Co A and B that I posted above...using the same approach am not getting the rite ans ...see,
A certain no of girls and 10 boys attended a party. No boy shakes hands with more than 4 grls.Out of all the grls, 4 of the grls shook hands with 4 boys each and each of the rem girls shook hands with 3 boys. the greatest possible no of grls?
using the app above: 4+6C3=24 which is wrong!!

@sonamaries7 said:
There is a fam with father , mother and 6 children A thru F. They have to attend a party for which only one among the mother/father can attend along with any no of kids subject to the foll constraints:1) B or C won't go along with their mother2) F goes only if A or D goesIn how many ways can the fam attend the party?71677970100plz share the approach also..
We can select the Father/Mother in 2 ways and one or more from 6 in 2^6
So total 2.2^6 = 2^7

Now, say mother goes and B also goes then the remaining can be selected in 2^5 ways
Same when C goes 2^5
So we need to subtract these cases
So Net =2^7-2^5-2^5
But we have subtracted the case where both B and C go which will be 2^4 ways
so add this .
Net = 2^7-2^5-2^5+2^4

Now say F goes with father, then we will assume A and D are not going this makes 2^3 ways
If F goes with mother can only E acn go so 1 way
Hence Net = 2^7-2^5-2^5+2^4 - 8 -1 = 71

@sonamaries7 said:
@krum : Similar to the qs on Co A and B that I posted above...using the same approach am not getting the rite ans ...see,A certain no of girls and 10 boys attended a party. No boy shakes hands with more than 4 grls.Out of all the grls, 4 of the grls shook hands with 4 boys each and each of the rem girls shook hands with 3 boys. the greatest possible no of grls?using the app above: 4+6C3=24 which is wrong!!
is it 12?

cant generalize it, as there is other constraint that each boy shakes hand with max. 4 girls

@sonamaries7 said:
@krum : Similar to the qs on Co A and B that I posted above...using the same approach am not getting the rite ans ...see,A certain no of girls and 10 boys attended a party. No boy shakes hands with more than 4 grls.Out of all the grls, 4 of the grls shook hands with 4 boys each and each of the rem girls shook hands with 3 boys. the greatest possible no of grls?using the app above: 4+6C3=24 which is wrong!!
I am getting 12 as the answer :(
@krum said:
is it 12?
yup?!...but kaise?...and same hi to hai qs?!
@soumitrabengeri : 12 is rite 😛 approach...
@sonamaries7 said:
yup?!...but kaise?...and same hi to hai qs?! @soumitrabengeri : 12 is rite approach...
Since there are 10 guys, lets assume that the 4 girls shake hands with the first 4 boys resp.
So remaining are 6 boys. Now the remaining girls have to shake hands with 3 boys.
So start the iteration for each girl and go sequence wise.
Starting from the 5th guy to the 7th for 1 girl and then 8th to the 10th for the other.
Similarly continue for the other girls
I dunno if my approach is correct, but this is how i did it
@sonamaries7 said:
yup?!...but kaise?...and same hi to hai qs?!
say a1 a2 a3...a10 tak hain

a1 to a4 are fixed

ab 6c3 kia, matlab, each one of them is being counted 10 times, but limit is 4
@sonamaries7 said:
@krum : Similar to the qs on Co A and B that I posted above...using the same approach am not getting the rite ans ...see,A certain no of girls and 10 boys attended a party. No boy shakes hands with more than 4 grls.Out of all the grls, 4 of the grls shook hands with 4 boys each and each of the rem girls shook hands with 3 boys. the greatest possible no of grls?using the app above: 4+6C3=24 which is wrong!!
shud be 12.
@krum said:
say a1 a2 a3...a10 tak haina1 to a4 are fixedab 6c3 kia, matlab, each one of them is being counted 10 times, but limit is 4
to aagey kaise count kiya?