Let Anushka age = X
Charu latha 12 yrs older than Anushka so Charu latha age = x+12
Kajal is thrice as old as Anushka so Kajal age = 3x
The ages of Anushka, Kajal and Charu latha together are 57 years
i.e x + (x+12) + 3x = 57
=> 5x = 45
:- x = 9
Anushka age = 9
Kajal age = 3x = 27
Charu latha age = x+12 = 21

Clealry, C and D sit opposite to each other. So, if A sits opposite to cook, B shall be the cook.
Now, B is to the right of barber. So, one of the rest, say C will be barber.
Clealry, then D on the opposite side shall be washerman or tailor. But, washerman is left of tailor and D on the left of A.
So, D is washerman and A is tailor.

Thus, A and B are Tailor and Cook.

Ratio of the share of a lady, a gents and a girl is 7 : 4 : 3
No of ladies, gents and girls are 5, 3, 3
Thus effective ratio of ladies, gents and girls is 7 x 5 : 4 x 3 : 3 x 3 = 35 : 12 : 9
so part of gents = (12/56) x 7444 = Rs. 1595.142
so part of 1 gents = 1595.142/3 = Rs. 531.71.

Here, S = {1, 2, 3, 4, ...., 19, 20}=> n(s) = 20
Let E = event of getting a multiple of 4 or 15
=multiples od 4 are {4, 8, 12, 16, 20}
And multiples of 15 means multiples of 3 and 5
= {3, 6 , 9, 12, 15, 18, 5, 10, 15, 20}.
= the common multiple is only (15).
=> E = n(E)= 6
Required Probability = P(E) = n(E)/n(S) = 6/20 = 3/10.

There are 600 Kabaddi players
4% wear knee band on one leg
Remaining players = 600 x 0.96 = 576
Of the remaining, 25% wear knee bands on both leg
and Of the remaining, 75% do not wear knee bands .
So players not wearing knee bands = 0.75 x 576 = 576 x 3/4 = 144 x 3 = 432.

let 't' be the time after which they met since L starts.

Given K is 50% faster than L

50 t + 1.5*50(t-1) = 300

50 t +75 t = 300 + 75

t = 375 / 125 = 3 hrs past the time that L starts

So they meet at (9 + 3)hrs = 12:00 noon.

Given the company has 8 SD + 6 SRP + 7 IN = 21 cars + 'x' small cars.
Given SD makes 1/4 th of the total fleet.
=> 8 = 1/4 (21 + x)
=> 32 = 21 + x
=> x = 11
Therefore, the company has 11 small cars.

Let the number of males be 'M'
Let the number of females be 'F'
From the given conditions,
(F-15)=2M.....(i)
F=5(M-45).....(ii)
By solving the equations (1) & (2) we get M = 80
Then substitute M = 80 in (1)
F = 2M + 15
F = 2 x 80 + 15
=> F = 175
Therefore, number of males = 80 and females = 175 in the party.