Let the number of students in the class be 'x'.
Given if 1 is absent rest of people can be divided into 6 equal parts i.e (x-1)/6.........(1)
And given if 2 are absent rest of people can be divided into 7 equal parts i.e (x-2)/7.....(2)
From (1)&(2), we get
x-1/6 = ((x-2)/7) + 1
=> x = 37.

From above diagram
AC represents the hill and DE represents the tower

Given that AC = 100 m

angleXAD = angleADB = 30° (∵ AX || BD )
angleXAE = angleAEC = 60° (∵ AX || CE)

Let DE = h

Then, BC = DE = h, AB = (100-h) (∵ AC=100 and BC = h), BD = CE

tan 60°=AC/CE => √3 = 100/CE =>CE = 100/√3 ----- (1)

tan 30° = AB/BD => 1/√3 = 100−h/BD => BD = 100−h(√3)
∵ BD = CE and Substitute the value of CE from equation 1
100/√3 = 100−h(√3) => h = 66.66 mts

The height of the tower = 66.66 mts.

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.

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