Official Quant thread for CAT 2013

MBA CAT 2024
29626
@vbhvgupta o @saurav205nly 63
29627
@varathawins said:
@vbhvgupta o @saurav205nly 63
check again....you missed out on something..
29628
@saurav205

i cant get 82. (8)^2 + (2)^2 = 68 which is not equal to 82-18=64!

29629
@varathawins said:
@saurav205 i cant get 82. (8)^2 + (2)^2 = 68 which is not equal to 82-18=64!
its the other way round i guess...
82+18 = (8+2)^2
also,
63+18 = (6+3)^2...

29630
@ronync said:
The OA is 30But can you please tell me how did you arrive at this : "30 when q is perpendicular to p, close to 90 when q is almost parallel to p..."Having a tough time visualising it.
I was just trying to do it in my mind....it cannot be less than 30 anyway since the two folded sheets are themselves at an angle 30. Max still not able to visualise.

regards
scrabbler
29631
@vbhvgupta said:
N is an 8 digit no and s(N) denotes the sum of its digits. IF N +S(N) = 100000000 Find ten's digit of N.
100000000 is of form 9k + 1

=> N has to be of form 9a + 5, as both N and S(N) will leave same remainder when divided by 9

Hence 99999941 is the only such number (as sum has to be less than or equal to 67)

Tens digit is 4
@vbhvgupta said:
A two digit no is 18 less than the square of the sum of its digits. How many such no exists?
(a + b)^2 - 18 = 10a + b
a^2 + a(2b - 10) + b^2 - b - 18 = 0

a = (5 - b) ± √(43 - 9b)

√(43 - 9b) is integer for b = 1, 2, 3

(a, b) = (10, 1), (8, 2), (6, 3)

So, 82 and 63 are only such numbers
29632

@chillfactor ..BRO ye 9a+5 ka funda smjh nai aaya could u plz elaborate !
29633
@priyank333333 said:
@chillfactor ..BRO ye 9a+5 ka funda smjh nai aaya could u plz elaborate !
Suppose N = 9k + r, then S(N) = 9n + r

N + S(N) = 9k + 9n + 2r = 9a + 2r

But 100000000 is of form 9m + 1 or 9m + 10

=> 2r = 1 or 10
2r = 1 is not possible

=> 2r = 10
=> r = 5

So, N is of form 9k = 5
29634
Q) 23,24 :-)
29635
@albiesriram said:
Q) 23,24
23) 10+b=n
45/x=0.15
x=4500/15=300
nc2=300
n*(n-1)/2 = 300
n(n-1)=600
n=25
b=15
29636
@albiesriram said:
Q) 23,24
for 24 is it B=4?
29637
@Subhashdec2 said:
for 24 is it B=4?
No. :-(
29638

Guys help me out with this pls---->
Three of the six vertices of a regular hexagon are chosen at random. The probability that the triangle with these vertices is equilateral is?

29639
@vishcat said:
Guys help me out with this pls---->Three of the six vertices of a regular hexagon are chosen at random. The probability that the triangle with these vertices is equilateral is?
Only 2 formations are possible, so prob = 2/6C3 ?
Edited
29640
@vishcat said:
Guys help me out with this pls---->Three of the six vertices of a regular hexagon are chosen at random. The probability that the triangle with these vertices is equilateral is?
1/10?
29641
@vishcat said:
Guys help me out with this pls---->Three of the six vertices of a regular hexagon are chosen at random. The probability that the triangle with these vertices is equilateral is?
0.1?
29642
@ScareCrow28 said:
Only 1 formation is possible, so prob = 1/6C3 ?
bhai 2 possible honge na
29643
@Subhashdec2 said:
bhai 2 possible honge na
Yeah, I thought that they both are same so we won't take them. And I realize that I am wrong here. Sorry..
29644
@ScareCrow28
@heylady
@Subhashdec2 ---yes its 1/10 explain pls
29645
How many seven-digit numbers divisible by 11 have the sum of their digits equal to 59?
1)40 2)55 3)34C3 — 23C2 4)38C3 — 26C2 5)None of these