Official Quant thread for CAT 2013

MBA CAT 2024
11561
@freebull said:
i don't know the ans yet but can u plz tell ur approach as how u came to this ?
5 powers of 2 has to be distributed among a.b and c
a+b+c=5
==> 7c2

similarly others ==> 7c2*6c2*9c2
11562
@vbhvgupta

Let, a= x+d, b= x and c= x-d..Again, a-k=y/r, b=y and c +k=yr

Thus, a-k=y/r, or, x + d -k =y/r, or, x+d = y/r + k...(i)

Again, c + k=yr, or, x - d+ k= yr, or, x - d=yr - k..(ii)

Thus, Solving (i) and (ii), we have, 2x= y/r + yr..(iii)

Now, b=y and b=x..Thus, x=y

Therefore, 2 = 1/r + r, or, 2r = 1 +r^2, or, r^2 + 1 - 2r =0

=>(r-1)^2=0, or, r=1

Looking at the Options, Only Option(c) suffice..Thus, 0 always hold true for the above said conditions...
11563
@pyashraj said:
@vbhvguptaLet, a= x+d, b= x and c= x-d..Again, a-k=y/r, b=y and c +k=yr
Or better let a,b,c be 3,2,1 and k=1
then a-k=2,b=2,c+k=2
Common ratio=1
Only 0 satisfies :)
11564
@sparklingaubade said:
In the sequence 1,9,7,7,4,7,5,3,9,4,1, €Ś every digit from the fifth on is the sum of the preceding 4 digits mod 10.Does one of the following set ever occur in the sequence?a) 1,2,3,4 b) 3,2,6,9 c) 0,1,9,8 d) 7,9,5,3
checked upto 2000 terms, none of the options are there in the sequence
11565
@krum said:
checked upto 2000 terms, none of the options are there in the sequence
2000 terms???
I checked only about 50 :P
11566
@soumitrabengeri said:
2000 terms??? I checked only about 50
c++ \m/
11567
@soumitrabengeri said:
2000 terms??? I checked only about 50
I checked 20 terms then put away my pen sighed.... and moved on to the next quest with hope somebody might post the soln :P:P
11568
@krum said:
checked upto 2000 terms, none of the options are there in the sequence
good one..
11569
@soumitrabengeri said:
2000 terms??? I checked only about 50
@deedeedudu said:
I checked 20 terms then put away my pen sighed.... and moved on to the next quest with hope somebody might post the soln
@nick_baba said:
good one..
series starts repeating after 1560 terms :splat:
11570
@sparklingaubade said:
Two identical marked dices are brought together and kept with one of their faces in full contact. How many different arrangements are possible? (1) 36 (2) 60 (3) 72 (4) 84 (5) none of these
unable to understand it ..do we need to take care of which faces are aligned here ?
11571
@krum said:
series starts repeating after 1560 terms

\\___0//
11572
@sparklingaubade b -210..when the year starts with tuesday and the one following it is a leap year
11573
@krum said:
series starts repeating after 1560 terms
True.
but how/why.
and how does this help us in the paper?
:D
11574
@sparklingaubade 1)36
11575
@sujamait said:
unable to understand it ..do we need to take care of which faces are aligned here ?
maine aise kia sirji __/\__

take each face of dice 1 one by one

1st face can be aligned with 6 faces of dice 2, for each of these arrangements, there are four sides
2nd face can be aligned with 5 remaining faces...

similarly, 4*(6+5+4+3+2+1)=4*21=84

@Maxray2 , i went upto 30 terms and didn't get any, so i was like what the heck, just wanted to see for myself, can't let the question win over me :p, not for exams :p
11576
@krum said:
maine aise kia sirji __/\__take each face of dice 1 one by one1st face can be aligned with 6 faces of dice 2, for each of these arrangements, there are four sides 2nd face can be aligned with 5 remaining faces...similarly, 4*(6+5+4+3+2+1)=4*21=84@Maxray2, i went upto 30 terms and didn't get any, so i was like what the heck, just wanted to see for myself, can't let the question win over me , not for exams
yar samjha nhn yar logic 😐
11577
@sparklingaubade said:
What is the maximum possible sum of the number of Tuesdays and Wednesdays in two consecutiveyears?(a) 209 (b) 210 (c) 208 (d) 212
365*2=730
730 mod 7
104*2=208+2=210
11578
@antodaya said:
Q Š There are 20 guavas,10apples,5chikos.How many canuselect 25 fruits.?*show approach , no answer*
sorry for replying so late.......just wanted to know whether the following approach to solve this problem works or not......if it does then are there any preconditions to it......
a+b+c=25
atherefore
35-(a1+b1+c1)=25
=>a1+b1+c1=10
therefore required ans = 12C2=66??
11579
@sparklingaubade said:
Two identical marked dices are brought together and kept with one of their faces in full contact. How many different arrangements are possible? (1) 36 (2) 60 (3) 72 (4) 84 (5) none of these
72

6*6

Each can be arranged in 2 ways

11580
@manasvr said:
sorry for replying so late.......just wanted to know whether the following approach to solve this problem works or not......if it does then are there any preconditions to it......a+b+c=25atherefore35-(a1+b1+c1)=25=>a1+b1+c1=10therefore required ans = 12C2=66??
no

count or remove individual cases, like when a>20, but this method would be very tedious here