From above given table , we can see that
from year 1989 to 1994 ,( P + Q ) :
In 1989 = 8 + 16 = 24 ,
In 1990 = 20 + 10 = 30 ,
In 1991 = 16 + 14 = 30 ,
In 1992 = 17 + 12 = 29 ,
In 1993 = 21 + 12 = 33 ,
In 1994 = 6 + 14 = 20
simillary , ( R + S ) :
In 1989 = 21 + 4 = 25 ,
In 1990 = 17 + 6 = 23 ,
In 1991 = 16 + 10 = 26 ,
In 1992 = 15 + 16 = 31 ,
In 1993 = 13 + 20 = 33 ,
In 1994 = 8 + 31 = 39
Hence, In 1993 year , the total production of cars of types P and Q together will be equal to the total production of cars of types R and S together
As per given above table , we have
Required difference = Total Sales ? Gross Profit
In 1990 ,
= 351.6 ? 155.5 = 196.1;
In 1991 ,
= 407.9 ? 134.3 = 273.6
In 1992
= 380.1 ? 149.9 = 230.2;
In 1993 ,
= 439.7 ? 160.5 = 279.2
In 1994 ,
= 485.9 ? 203.3 = 282.6
Hence required answer will be 1990 .
From above given table ,we can see that
in 1992 , total sales = 380.1
gross profit = 149.9
and net profit = 38.9
in 1993 , total sales = 439.7
gross profit = 160.5
and net profit = 50.3
in 1994 , total sales = 485.9
gross profit = 203.3
and net profit = 65.8
Hence 1993 and 1994 both years show increase in all the categories simultaneously, that is total sales, gross profit and net profit as compared to the previous year .
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 .
? 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%
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