The total number of students who have opted the course of Business Management and the course of Stenography exceeds the number of students studying in typewriting course by:

Required percentage = (25 + 120) x 100/536 = 100? 27

Total number of boys = 45 + 65 + 186 + 32 + 120 + 58 + 100 + 5 = 611
Total number of girls = 25 + 23 + 25 + 12 + 25 + 20 + 12 + 3 = 145
? Required percentage = (611 - 145/145) x 100 ? 321%

? Total boys = 45 + 186 +120 +100 + 65 + 32 + 58 + 5 = 611
? Number of boys who pass in their respective courses = 60% of 611 = 611 x 60/100 = 366.60
Total number of girls = 25 + 23 + 25 + 12 + 25 + 20 + 12 + 3 = 145
Number of girls who pass in their respective courses =70% of 145= 145 x 70/100 = 101.50
? Combined pass percent = (366.6 + 101.5) x 100/(611 + 145) = 46810/756 = 61.92 = 62%

From the values given in above table , it is clear that during the period 1989?94 , the production of S's car is continuous increase . Hence answer will be S type of cars .

Percentage of girls in Business Management course = (25 + 25) x 100/160 = 31.25
Percentage of girls in typewriting course = (23 + 20) x 100/261 = 16.47
Percentage of girls in stenography course = (25 + 12) x 100/215 =12.56
And percentage of girls in typewriting and stenography course = (12 + 3) x 100/120 = 12.50
Hence, it is clear from the above that percentage of girls in Business Management course is the highest,

In society A:
Number of children = 25/100 x (250 + 350) = 1/4 x 600 = 150
Number of male children = 40/100 x 150 = 60
Number of female children = 150 - 60 = 90
In society B:
Number of children = 40/100 x (400 + 150) = 2/5 x 550 = 220
Number of male children = 75/100 x 220 = 165
Number of female children = 220 - 165 = 55
In society C:
Number of children = 16/100 x (300 + 275) = 4/25 x 575 = 92
Number of male children = 25/100 x 92 = 23
Number of female children = 92 - 23 = 69
In society D:
Number of children = 25/100 x (280 + 300) = 1/4 x 580 = 145
Number of male children = 80/100 x 145 = 116
Number of female children = 145 - 116 = 29
In society E:
Number of children = 40/100 x (180 + 250) = 2/5 x 430 = 172
Number of male children = 50/100 x 172 = 86
Number of female children = 172 - 86 = 86
In society F :
Number of children = 24/100 x (325 + 300) = 6/25 x 625 = 150
Number of male children = 46/100 x 150 = 69
Number of female children = 150 - 69 = 81
Hence, total number of female children = 90 + 55 + 69 + 29 + 86 + 81 = 410
Total Number of adult female = (350 + 150 + 275 + 300 + 250 + 300) - 410
= 1625 - 410 = 1215
Hence, required ratio = 1215 : 410 = 283 : 82

Total number of adult males in the society A and B= (250 + 400) - (60 + 165) = 650 - 225 = 425
Total number of adult males in the society E and F = (280 + 325) - (86 + 69) = (180 + 325) - (86 + 69) = 505 - 155 = 350
Hence, required ratio = 425 : 350 = 17 : 14

Total number of female children = 410

Number of all members
= (250 + 350) + (400 + 150) + (300 + 275) + (280 + 300) + (180 + 250) + (325 + 300) + 150 + 220 + 92 + 145 + 172 + 150
= 600 + 550 + 575 + 580 + 430 + 625 + 929
= 4289