Official Quant thread for CAT 2013

MBA CAT 2024
29045
@Tusharrr said:
ABC is an isosceles triangle where AB=AC and BC=60. D is a point on BC such that the perpendicular distance from D to AB and AC is 16 and 32, respectively. What is the length of AB?
area of triangle ABC = ABD + ACD = 1/2*16*AB + 1/2*32*AC = 8AB + 16AC = 24x
also area = 1/2*60*root(x^2 - 30^2) = 30*root(x^2 - 30^2) = 24x
30^2(x^2 - 30^2) = 24^2*x^2
54*6x^2 = 30^4
x = 30^2 / 18 = 50.... AB = 50

29046
@sbharadwaj said:
4^10 - 1.??
@albiesriram said:
@Buck.up 806236 ways?
ans is => 4^10 - 4 = 1048572 cases

Plz explain this
29047
@Buck.up said:
Find the no of ways in which 10 different balls can be placed in 4 different boxes such that at max 2 boxes are empty.
total number of ways = 4^10
this will include the ways in which 3 boxes are empty. we need to remove them.
number of such ways = 4

so required ways = 4^10 - 4


29048
@Buck.up said:
ans is => 4^10 - 4 = 1048572 casesPlz explain this
well, 4^10 is the total distribution and is the number of ways in which no boxes are empty + one box is empty +two boxes are empty + three boxes are empty. and ofcourse 4 boxes cannot be empty.

So what we do is find the number of distributions in which 3 boxes are empty. which implies the cases A,B,C,D boxes receives all the ten balls at a time. hence we subtract the 4 cases. to arrive at the solution.

29049

@Buck.up said:
Find the no of ways in which 10 different balls can be placed in 4 different boxes such that at max 2 boxes are empty.
4^10-4

u just have remove the case when 3 boxes are empty
that means all balls in one box
that is 1 case

and that 1 box can be selected in 4c1 ways=4

4^10-4

29050
@Buck.up said:
Find the no of ways in which 10 different balls can be placed in 4 different boxes such that at max 2 boxes are empty.
No of ways in which 4 boxes can be filled = 4^10

But here we counted those cases also when 3 boxes are empty (which is possible only in 4 cases, when all 10 goes to same box)

that why answer will be 4^10 - 4
29051
@techgeek2050 said:

number of such ways = 1

Number of such ways is 4, isnt it?

29052
@albiesriram said:
Number of such ways is 4, isnt it?
oh ya right! my bad...
29053

29054
@albiesriram said:
for q2
is it 1002+1/2=2005/2 ????
29055
@albiesriram said:
1. D.(5,7) Not sure

2. 1002 + f(1003/2006)
1002 + .5=1002.5
29056
@albiesriram said:
for the second one - 1002.5?

f(x)+f(1-x) = 1
29057
@albiesriram said:
first one A (-5,-7) ?

common roots are roots of 2x^2 + (q-r) = 0
hence their sum is 0
sum of all 3 roots of 1st eq is -5, two of these sum to 0, hence rem one is -5, similarly for the other eqn
29058
@RDN said:
for the second one - 1002.5?
f(1003) mid me bachega.
Iski value .5 kaise hogi
29059
@Buck.up said:
f(1003) mid me bachega. Iski value .5 kaise hogi
f (1003/2006) = f (.5) = .5
29060
@RDN said:
f (1003/2006) = f (.5) = .5
oh... my bad
29061
What is the probability of getting a word with all 'T's together for the word ATTEMPT ?
29062
@albiesriram said:
for 1st -5,-7
29063
@Buck.up 3! X 5!/7! = 1/7?
29064

for 13 A hoga.


Aur for 14 shayad 1002.5, No OA availale but the the approach is f(x)+f(1-x) =1

hence ,
f(1/2006) + f (2005/2006) =1 and so on, and f (1003/2006) =f(0.5) = 3/3+3 =1/2
so 1002.5