Job Roles

Let total population = p
number of females = 5p/9
number of males =(p-5p/9) = 4p/9
married females = 30% of 5p/9 = 30x5p/100x9 = p/6
married males = p/6
unmarried males =(4p/9-p/6) = 5p/18
Percentage of unmarried males in the total males = {(5p/18)/4p/9}x100 = (5p/18)x(9/4p) = 125/2 % = 62.5%

Ratio C1:C2 = 3:5 and C2:C3 = 7:11
So C1:C2:C3 = 21 : 35 : 55
Let the strength of three classes are 21x, 35x and 55x respectively, then
Given that 21x + 35x + 55x = 333
=> 111x = 333 or x=3
So strength of the class with highest number of students =55x = 55x3 = 165.

Let wholesaler dealers marked price = 100%, Retailer's C.P = 80%
And the retailer sells at 5% less than the marked price => S.P = 95%
If S.P of 95% of the retailer costs Rs.19 to customer,so its C.P of 80% will cost 80 x 19/95 = 16
Profit made by the retailer = 19-16 =  Rs.3

The statements  a,e,c  are logically sequenced.

Let us assume S as number of shirts and T as number of trousers
Given that each trouser cost = Rs.70 and that of shirt = Rs.30
Therefore, 70 T + 30 S = 810
=> 7T + 3S = 81......(1)
T = ( 81 - 3S )/7

We need to find the least value of S which will make (81 - 3S) divisible by 7 to get maximum value of T
Simplifying by taking 3 as common factor i.e, 3(27-S) / 7
In the above equation least value of S as 6 so that 27- 3S becomes divisible by 7

Hence T = (81-3xS)/7 = (81-3x6)/7 = 63/7 = 9

Hence for S, put T in eq(1), we get
S = 81-7(9)/3 = 81-63 / 3 = 18/3 = 6.
The ratio of T:S = 9:6 = 3:2.

Paging is a memory management technique in which process address space is broken into blocks of the same size called pages.

Let last digit is 2
when second last digit is 4 remaining 4 digits can be filled in 120 ways, similarly second last digit is 6 remained 4 digits can be filled in 120 ways.
so for last digit = 2, total numbers=240

Similarly for 4 and 6
When last digit = 4, total no. of ways =240
and last digit = 6, total no. of ways =240
so total of 720 even numbers are possible.