From above graph ,we can see that
Since , For A course the number of boys is minimum.
Hence Answer will be A course.
Number of men working in night shifts in different industries will be as below.
As per given graphs in question,
Number of total people working in night shifts in IT = 40250 x 12%
Number of men working in night shifts in IT = 40250 x 12% x (100% - 20% )
Number of men working in night shifts in IT = 40250 x 12% x 80%
? Number of men working in night shifts in IT = 40250 x (12/100) x 80/100
? Number of men working in night shifts in IT = 3864
Similarly,
Number of men working in night shifts in Sports = 80/100 x 18/100 x 40250 = 5796
Number of men working in night shifts in Call Center = 55/100 x 32/100 x 40250 = 7084
Number of men working in night shifts in Sales = 40/100 x 8/100 x 40250 = 1288
Number of men working in night shifts in Banking = 60/100 x 14/100 x 40250 = 3381
Number of men working in night shifts in Chemical Industries = 85/100 x 16/100 x 40250 = 5474
Hence, total number of men working in night shifts in different industries = 3864 + 5796 + 7084 + 1288 + 3381 + 5474 = 26887
Total number of men = 26887
Total number of women = 40250 - 26887 = 13363
Their difference = 26887 - 13363 = 13524
As per first graph,
The number of students for course A = Total number of students x 20%
As per second graph,
The number of girls for course A = Total number of girls x 30%
Number of boys in course 'A' = The number of students for course A - The number of girls for course A
Number of boys in course 'A' = (1200 x 20/100 ) - ( 800 x 30/100 ) = 240 - 240 = 0
Similarly,
Number of boys in course 'B' = (1200 x 15/100 ) - (800 x 10/100) = 180 - 80 = 100
Number of boys in course 'C' = (1200 x 5/100 ) - (800 x 2/100) = 60 - 16 = 44
Number of boys in course 'D' = (1200 x 35/100 ) - (800 x 30/100) = 420 - 240 = 180
Number of boys in course 'E' = (1200 x 12/100 ) - 800 x 14/100 ) = 144 - 112 = 32
Number of boys in course 'F' = (1200 x 13/100 ) - ( 800 x 14/100 ) = 156 - 112 = 44
Hence, number of boys in course 'C' and 'F' are same which is 44.
Here total experditure of both companies are given while their individual expenditures are needed to determine their incomes, Since their total income of the two companies can not be determined by the given datas
I. Using I only, we cannot determine that G stays on 4th floor.
Even by using both the statements together we cannot determine whether B has highest number of student or D.
Using I:
T 5 R and T is midway between P and Q
Using II:
R 2 Q
Using both:
P 7 T 5 R 2 Q
if the data in Statements II alone are sufficient to answer the question, while the data in Statement I alone are not sufficient to answer the question.
Let us assume E was the expenditure in 1994.
As we can see in the graph given that,
Profit = 15 % ,
Income in 1994 = 160
Then according to question,
E + E x 15 % = 160
E + E x 15 /100 = 160
(100E + E x 15 ) /100 = 160
115E / 100 = 160
115E = 160 x 100
E = 160 x 100 / 115
E = 139.13
Expenditure in 1994 E = 139.13 ? 140 Million
From I: C > A > B and - > D
From II: -> - > C. Comblining, we get E > D > C > A > B.
Using either of the statements alone we cannot find the code, and even by using both the statements together we can only find that 'never ever' is coded as 'na ja' the code for 'never' cannot be uniquely determined even by using both the statements together.
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